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+\section{Cayley-Menger Determinant}
+Tags: Linear Algebra, Distance Geometry
+
+\begin{theorem}
+Let $S$ be a [[Definition:Simplex|$j$-simplex]] in $\R^n$.
+Let the [[Definition:Vertex of Polyhedron|vertices]] of $S$ be $v_1, v_2, \ldots, v_{j+1}$.
+Let $B = \left[{\beta_{ij} }\right]$ denote the $\left({j + 1}\right) \times \left({j + 1}\right)$ [[Definition:Square Matrix|matrix]] given by:
+:$\beta_{ij} = \left\vert{v_i - v_j}\right\vert_2^2$
+where $\left\vert{v_i - v_j}\right\vert_2$ is the [[Definition:P-Norm|vector 2-norm]] of the vector $v_i - v_j$.
+Then the [[Definition:Content|content]] $V_j$ is given by:
+{{begin-eqn}}
+{{eqn | l = V_j^2 \left({S}\right)  
+      | r = \frac {\left({-1}\right)^{j+1} } {2^j \left({j!}\right)^2} \det C 
+}}
+{{end-eqn}}
+where $C$ is the $\left({j + 2}\right) \times \left({j + 2}\right)$ [[Definition:Square Matrix|matrix]] obtained from $B$ by bordering $B$ with a top [[Definition:Row of Matrix|row]] $\left({0, 1, \ldots, 1}\right)$ and a left [[Definition:Column of Matrix|column]] $\left({0, 1, \ldots, 1}\right)^\intercal$.
+\end{theorem}<|endoftext|>
+\section{Real Natural Logarithm is Restriction of Complex Natural Logarithm}
+Tags: Natural Logarithms
+
+\begin{theorem}
+Let $\ln: \C_{\ne 0} \to \C$ be the [[Definition:Complex Natural Logarithm|complex natural logarithm]].
+Let $\ln': \R_{>0} \to \R$ be the [[Definition:Real Natural Logarithm|real natural logarithm]].
+Then:
+:$\ln' = \ln \restriction_{\R_{>0} \times \R}$
+That is, the [[Definition:Real Natural Logarithm|real natural logarithm]] is the [[Definition:Restriction of Relation|restriction]] of the [[Definition:Complex Natural Logarithm|complex natural logarithm]].
+\end{theorem}
+
+\begin{proof}
+From [[Domain of Real Natural Logarithm]]:
+: $\operatorname{Dom} \left({\ln'}\right) = \R_{>0}$
+From [[Image of Real Natural Logarithm]]:
+: $\operatorname{Im} \left({\ln'}\right) = \R$
+Let $z \in \C$ such that $z = x + i y$.
+Let $z$ be expressed in [[Definition:Exponential Form of Complex Number|exponential form]] as $z = r e^{i \theta}$.
+Let $x > 0$ and $y = 0$.
+Thus $z \in \R_{>0}$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \ln z
+      | r = \left\{ {\ln' r + i \theta + 2 \pi k i}\right\}
+      | c = 
+}}
+{{eqn | r = \left\{ {\ln' x + 2 \pi k i}\right\}
+      | c = as $r = x$ and $\theta = 0$
+}}
+{{end-eqn}}
+In order for $\ln z \in \R$ it is necessary that $k = 0$.
+Thus $\ln z = \ln' x$.
+Hence the result.
+{{qed}}
+[[Category:Natural Logarithms]]
+j4h0xmkgtlc49xsftaqfnqphpm2d3yj
+\end{proof}<|endoftext|>
+\section{Rule of Commutation/Disjunction/Formulation 2/Forward Implication}
+Tags: Rule of Commutation
+
+\begin{theorem}
+:$\vdash \left({p \lor q}\right) \implies \left({q \lor p}\right)$
+\end{theorem}
+
+\begin{proof}
+Using a [[Definition:Tableau Proof (Formal Systems)|tableau proof]] for [[Definition:Gentzen Proof System/Instance 1|instance 1 of a Gentzen proof system]]:
+{| border="1"
+|+$\vdash \left({p \lor q}\right) \implies \left({q \lor p}\right)$
+|-
+! Line !!
+! Pool
+! Formula
+! Rule
+! Depends upon
+! Notes
+|-
+| 1 || ||
+| $\neg p, q, p$
+| Axiom || ||
+|-
+| 2 || ||
+| $\neg q, q, p$
+| Axiom || ||
+|-
+| 3 || ||
+| $\neg \left({p \lor q}\right), q, p$
+| [[Definition:Gentzen Proof System/Instance 1/Alpha-Rule|$\alpha$-Rule: $\alpha\lor$]]
+| 1, 2 ||
+|-
+| 4 || ||
+| $\neg \left({p \lor q}\right), q \lor p$
+| [[Definition:Gentzen Proof System/Instance 1/Beta-Rule|$\beta$-Rule: $\beta\lor$]]
+| 3 ||
+|-
+| 5 || ||
+| $\left({p \lor q}\right) \implies \left({q \lor p}\right)$
+| [[Definition:Gentzen Proof System/Instance 1/Beta-Rule|$\beta$-Rule: $\beta\implies$]]
+| 4 ||
+|}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Provable by Gentzen Proof System iff Negation has Closed Tableau}
+Tags: Propositional Logic
+
+\begin{theorem}
+Let $\mathscr G$ be [[Definition:Gentzen Proof System/Instance 1|instance 1 of a Gentzen proof system]].
+\end{theorem}<|endoftext|>
+\section{Provable by Gentzen Proof System iff Negation has Closed Tableau/Formula}
+Tags: Propositional Logic
+
+\begin{theorem}
+Let $\mathbf A$ be a [[Definition:WFF of Propositional Logic|WFF of propositional logic]].
+Then $\mathbf A$ is a [[Definition:Theorem (Formal Systems)|$\mathscr G$-theorem]] [[Definition:Iff|iff]]:
+:$\neg \mathbf A$ has a [[Definition:Closed Tableau|closed semantic tableau]]
+where $\neg \mathbf A$ is the [[Definition:Negation|negation]] of $\mathbf A$.
+\end{theorem}
+
+\begin{proof}
+This is a specific instance of [[Provable by Gentzen Proof System iff Negation has Closed Tableau/Set of Formulas|Provable by Gentzen Proof System iff Negation has Closed Tableau: Set of Formulas]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Provable by Gentzen Proof System iff Negation has Closed Tableau/Set of Formulas}
+Tags: Propositional Logic
+
+\begin{theorem}
+Let $U$ be a [[Definition:Set|set]] of [[Definition:WFF of Propositional Logic|WFFs of propositional logic]].
+Then $U$ is a [[Definition:Theorem (Formal Systems)|$\mathscr G$-theorem]] [[Definition:Iff|iff]]:
+:$\bar U$ has a [[Definition:Closed Tableau|closed semantic tableau]]
+where $\bar U$ is the [[Definition:Set|set]] comprising the [[Definition:Logical Complement|logical complements]] of all [[Definition:WFF of Propositional Logic|WFFs]] in $U$.
+\end{theorem}
+
+\begin{proof}
+Denote with $\bar{\mathbf A}$ and $\bar U$ the [[Definition:Logical Complement|logical complement]] of a [[Definition:WFF of Propositional Logic|WFF]] $\mathbf A$ and [[Definition:Set|set]] $U$, respectively.
+=== Necessary Condition ===
+We aim to prove this result using the [[Second Principle of Mathematical Induction]], applied to the minimal length of a [[Definition:Formal Proof|formal proof]] of $U$.
+Suppose first that $U$ is an [[Definition:Axiom (Formal Systems)|axiom]] of $\mathscr G$.
+Then $U$ contains a [[Definition:Complementary Pair|complementary pair]] of [[Definition:Literal|literals]].
+Hence, so does $\bar U$, by definition of [[Definition:Logical Complement|logical complement]].
+Therefore, the [[Definition:Semantic Tableau|semantic tableau]] comprising only a [[Definition:Root Node|root node]] labeled $\bar U$ is [[Definition:Closed Tableau|closed]].
+Next, suppose that the last step in proving $U$ was an instance of the [[Definition:Gentzen Proof System/Instance 1/Alpha-Rule|$\alpha$-rule]].
+That is, for some [[Definition:Alpha-Formula|$\alpha$-formula]] $\mathbf A$ and corresponding $\mathbf A_1, \mathbf A_2$:
+:$U = U_1 \cup U_2 \cup \left\{{\mathbf A}\right\}$
+where $U_1 \cup \left\{{\mathbf A_1}\right\}$ and $U_2 \cup \left\{{\mathbf A_2}\right\}$ are [[Definition:Theorem (Formal Systems)|$\mathscr G$-theorems]].
+By induction hypothesis, the [[Definition:Logical Complement|logical complements]]:
+:$\bar U_1 \cup \left\{{\bar{\mathbf A}_1}\right\}$
+:$\bar U_2 \cup \left\{{\bar{\mathbf A}_@}\right\}$
+of these [[Definition:Theorem (Formal Systems)|theorems]] have [[Definition:Closed Tableau|closed tableaus]].
+From the [[Soundness Theorem for Semantic Tableaus]], it follows that these sets are [[Definition:Unsatisfiable (Boolean Interpretations)|unsatisfiable]].
+Now by [[Superset of Unsatisfiable Set is Unsatisfiable]], so are:
+:$\bar U' := \bar U_1 \cup \bar U_2 \cup \left\{{\bar{\mathbf A}_1}\right\}$ 
+:$\bar U'' := \bar U_1 \cup \bar U_2 \cup \left\{{\bar{\mathbf A}_1}\right\}$
+By the [[Completeness Theorem for Semantic Tableaus]], these have [[Definition:Closed Tableau|closed tableaus]].
+Since $\mathbf A$ is an [[Definition:Alpha-Formula|$\alpha$-formula]], it is [[Definition:Semantic Equivalence (Boolean Interpretations)|equivalent]] to $\mathbf A_1 \land \mathbf A_2$.
+By [[De Morgan's Laws (Logic)/Disjunction of Negations|De Morgan's Laws]], it follows that:
+:$\bar{\mathbf A}$ is [[Definition:Semantic Equivalence (Boolean Interpretations)|equivalent]] to $\bar{\mathbf A}_1 \lor \bar{\mathbf A}_2$.
+Therefore, $\bar{\mathbf A}$ is a [[Definition:Beta-Formula|$\beta$-formula]], and $\bar{\mathbf A}_1, \bar{\mathbf A}_2$ correspond to it as in the [[Definition:Table of Beta-Formulas|table of $\beta$-formulas]].
+This allows the [[Semantic Tableau Algorithm]] to expand a [[Definition:Leaf Node|leaf]] labeled $\bar U$ into two leaves labeled $\bar U'$ and $\bar U''$.
+Because $\bar U'$ and $\bar U''$ have [[Definition:Closed Tableau|closed tableaus]], so does $\bar U$.
+Finally, suppose that the last step in proving $U$ was an instance of the [[Definition:Gentzen Proof System/Instance 1/Beta-Rule|$\beta$-rule]].
+That is, for some [[Definition:Beta-Formula|$\beta$-formula]] $\mathbf B$ and corresponding $\mathbf B_1, \mathbf B_2$:
+:$U = U_1 \cup \left\{{\mathbf B}\right\}$
+where $U' := U_1 \cup \left\{{\mathbf B_1, \mathbf B_2}\right\}$ is a [[Definition:Theorem (Formal Systems)|$\mathscr G$-theorem]].
+By induction hypothesis, the [[Definition:Logical Complement|logical complement]]:
+:$\bar U' = \bar U_1 \cup \left\{{\bar{\mathbf B}_1, \bar{\mathbf B}_2}\right\}$
+of this [[Definition:Theorem (Formal Systems)|$\mathscr G$-theorem]] has a [[Definition:Closed Tableau|closed tableau]].
+Since $\mathbf B$ is a [[Definition:Beta-Formula|$\beta$-formula]], it is [[Definition:Semantic Equivalence (Boolean Interpretations)|equivalent]] to $\mathbf B_1 \lor \mathbf B_2$.
+By [[De Morgan's Laws (Logic)/Conjunction of Negations|De Morgan's Laws]], it follows that:
+:$\bar{\mathbf B}$ is [[Definition:Semantic Equivalence (Boolean Interpretations)|equivalent]] to $\bar{\mathbf B}_1 \land \bar{\mathbf B}_2$.
+Therefore, $\bar{\mathbf B}$ is an [[Definition:Alpha-Formula|$\alpha$-formula]], and $\bar{\mathbf B}_1, \bar{\mathbf B}_2$ correspond to it as in the [[Definition:Table of Alpha-Formulas|table of $\alpha$-formulas]].
+This allows the [[Semantic Tableau Algorithm]] to expand a [[Definition:Leaf Node|leaf]] labeled $\bar U$ into a leaf labeled $\bar U'$.
+Because $\bar U'$ has a [[Definition:Closed Tableau|closed tableau]], so does $\bar U$.
+The result follows by the [[Second Principle of Mathematical Induction]].
+{{qed|lemma}}
+=== Sufficient Condition ===
+Let $T$ be a [[Definition:Closed Tableau|closed tableau]] for $\bar U$.
+We prove that $U$ is a [[Definition:Theorem (Formal Systems)|$\mathscr G$-theorem]] by the [[Second Principle of Mathematical Induction]], applied to the number of [[Definition:Node (Graph Theory)|nodes]] of $T$.
+Suppose $T$ has only one [[Definition:Node (Graph Theory)|node]].
+Then $\bar U$ has to contain a [[Definition:Complementary Pair|complementary pair]] of [[Definition:Literal|literals]] for $T$ to be [[Definition:Closed Tableau|closed]].
+But then $U$ also contains a [[Definition:Complementary Pair|complementary pair]].
+Hence, it is an [[Definition:Axiom (Formal Systems)|axiom]] of $\mathscr G$.
+If $T$ has more than one [[Definition:Node (Graph Theory)|node]], the first iteration of the [[Semantic Tableau Algorithm]] must have selected either an [[Definition:Alpha-Formula|$\alpha$-formula]] $\mathbf A$ or a [[Definition:Beta-Formula|$\beta$-formula]] $\mathbf B$ from $\bar U$.
+First the case that an [[Definition:Alpha-Formula|$\alpha$-formula]] $\mathbf A$ was selected.
+Define, for convenience, $\bar U' = \bar U \setminus \left\{{\mathbf A}\right\}$.
+Then the rest of $T$ is a [[Definition:Semantic Tableau|semantic tableau]] for $\bar U' \cup \left\{{\mathbf A_1, \mathbf A_2}\right\}$.
+This [[Definition:Semantic Tableau|tableau]] has fewer [[Definition:Node (Graph Theory)|nodes]] than $T$.
+For $T$ to be [[Definition:Closed Tableau|closed]], this [[Definition:Subtree|subtree]] must also be a [[Definition:Closed Tableau|closed tableau]].
+By induction hypothesis, this means that $U' \cup \left\{{\bar{\mathbf A}_1, \bar{\mathbf A}_2}\right\}$ is a [[Definition:Theorem (Formal Systems)|$\mathscr G$-theorem]].
+Now, since $\mathbf A$ is an [[Definition:Alpha-Formula|$\alpha$-formula]], it is [[Definition:Semantic Equivalence (Boolean Interpretations)|equivalent]] to $\mathbf A_1 \land \mathbf A_2$.
+Hence by [[De Morgan's Laws (Logic)/Disjunction of Negations|De Morgan's Laws]], $\bar{\mathbf A}$ is [[Definition:Semantic Equivalence (Boolean Interpretations)|equivalent]] to $\bar{\mathbf A}_1 \lor \bar{\mathbf A}_2$.
+That is, $\bar{\mathbf A}$ is a [[Definition:Beta-Formula|$\beta$-formula]], so that the [[Definition:Gentzen Proof System/Instance 1/Beta-Rule|$\beta$-rule]] may be applied to it.
+Hence, $U = U' \cup \left\{{\bar{\mathbf A}}\right\}$ is a [[Definition:Theorem (Formal Systems)|$\mathscr G$-theorem]], as desired.
+Now the case that a [[Definition:Beta-Formula|$\beta$-formula]] $\mathbf B$ was selected.
+Define, for convenience, $\bar U' = \bar U \setminus \left\{{\mathbf B}\right\}$.
+Then the rest of our [[Definition:Closed Tableau|closed tableau]] $T$ comprises two [[Definition:Semantic Tableau|semantic tableaus]] for $\bar U' \cup \left\{{\mathbf B_1}\right\}$ and $\bar U' \cup \left\{{\mathbf B_2}\right\}$, respectively.
+Each of these has fewer [[Definition:Node (Graph Theory)|nodes]] than $T$.
+For $T$ to be [[Definition:Closed Tableau|closed]], these [[Definition:Subtree|subtrees]] must also be [[Definition:Closed Tableau|closed tableaus]].
+By induction hypothesis, this means that $U' \cup \left\{{\bar{\mathbf B}_1}\right\}$ and $U' \cup \left\{{\bar{\mathbf B}_2}\right\}$ are [[Definition:Theorem (Formal Systems)|$\mathscr G$-theorems]].
+Now, since $\mathbf B$ is a [[Definition:Beta-Formula|$\beta$-formula]], it is [[Definition:Semantic Equivalence (Boolean Interpretations)|equivalent]] to $\mathbf B_1 \lor \mathbf B_2$.
+Hence by [[De Morgan's Laws (Logic)/Conjunction of Negations|De Morgan's Laws]], $\bar{\mathbf B}$ is [[Definition:Semantic Equivalence (Boolean Interpretations)|equivalent]] to $\bar{\mathbf B}_1 \land \bar{\mathbf B}_2$.
+That is, $\bar{\mathbf B}$ is an [[Definition:Alpha-Formula|$\alpha$-formula]], so that the [[Definition:Gentzen Proof System/Instance 1/Alpha-Rule|$\alpha$-rule]] may be applied to it.
+Hence, $U = U' \cup \left\{{\bar{\mathbf B}}\right\}$ is a [[Definition:Theorem (Formal Systems)|$\mathscr G$-theorem]], as desired.
+The result now follows from the [[Second Principle of Mathematical Induction]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Domain of Real Natural Logarithm}
+Tags: Natural Logarithms
+
+\begin{theorem}
+Let $\ln$ be the [[Definition:Real Natural Logarithm|natural logarithm function on the real numbers]].
+Then the [[Definition:Domain of Mapping|domain]] of $\ln$ is the set of [[Definition:Strictly Positive Real Number|strictly positive real numbers]]:
+:$\Dom \ln = \R_{>0}$
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Natural Logarithm/Positive Real/Definition 2|natural logarithm]]:
+:$\ln^{-1} = \exp$
+From [[Exponential Tends to Zero and Infinity]]:
+:$\Dom \exp = \openint {-\infty} {+\infty}$
+:$\Img \exp = \openint 0 {+\infty}$
+From [[Exponential is Strictly Increasing]]:
+:$\exp$ is [[Definition:Strictly Increasing Real Function|strictly increasing]].
+From [[Strictly Monotone Real Function is Bijective]], $\exp: \R \to \R_{>0}$ is a [[Definition:Bijection|bijection]].
+Thus:
+:$\Dom \ln = \Img \exp$
+and so $\Dom \ln = \R_{>0}$.
+{{qed}}
+[[Category:Natural Logarithms]]
+14154zzvbik35lknflmgrxyz0xxq2th
+\end{proof}<|endoftext|>
+\section{Image of Real Natural Logarithm}
+Tags: Natural Logarithms
+
+\begin{theorem}
+Let $\ln$ be the [[Definition:Real Natural Logarithm|natural logarithm function on the real numbers]].
+Then the [[Definition:Image of Mapping|image]] of $\ln$ is the set of [[Definition:Real Number|real numbers]]:
+:$\Img \ln = \R$
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Natural Logarithm/Positive Real/Definition 2|natural logarithm]]:
+:$\ln^{-1} = \exp$
+From [[Exponential Tends to Zero and Infinity]]:
+:$\Dom \exp = \openint {-\infty} {+\infty}$
+:$\Img \exp = \openint 0 {+\infty}$
+From [[Exponential is Strictly Increasing]]:
+:$\exp$ is [[Definition:Strictly Increasing Real Function|strictly increasing]].
+From [[Strictly Monotone Real Function is Bijective]], $\exp: \R \to \R_{>0}$ is a [[Definition:Bijection|bijection]].
+Thus:
+:$\Img \ln = \Dom \exp$
+and so $\Img \ln = \R$.
+{{qed}}
+[[Category:Natural Logarithms]]
+4d0rbbqnorxwqsc2hkxxvh1ektnvkzw
+\end{proof}<|endoftext|>
+\section{Soundness and Completeness of Gentzen Proof System}
+Tags: Propositional Logic
+
+\begin{theorem}
+Let $\mathscr G$ be [[Definition:Gentzen Proof System/Instance 1|instance 1 of a Gentzen proof system]].
+Let $\mathrm{BI}$ be the [[Definition:Formal Semantics of Boolean Interpretations|formal semantics of boolean interpretations]].
+Then $\mathscr G$ is a [[Definition:Sound Proof System|sound]] and [[Definition:Complete Proof System|complete proof system]] for $\mathrm{BI}$.
+\end{theorem}
+
+\begin{proof}
+This is an immediate consequence of:
+* [[Provable by Gentzen Proof System iff Negation has Closed Tableau]]
+* [[Soundness and Completeness of Semantic Tableaus]]
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Definition:Deduction Rule}
+Tags: Definitions: Propositional Logic
+
+\begin{theorem}
+Let $\mathcal L$ be the [[Definition:Language of Propositional Logic|language of propositional logic]].
+The '''Deduction Rule''' is the [[Definition:Rule of Inference|rule of inference]]:
+:From $U, \mathbf A \vdash \mathbf B$, one may infer $U \vdash \mathbf A \implies \mathbf B$
+where:
+:$\mathbf A, \mathbf B$ are [[Definition:WFF of Propositional Logic|WFFs of propositional logic]]
+:$U$ is a [[Definition:Set|set]] of [[Definition:WFF of Propositional Logic|WFFs]]
+For a given [[Definition:Proof System|proof system]], the '''Deduction Rule''' can be either a basic [[Definition:Rule of Inference|rule of inference]], or a [[Definition:Derived Rule|derived rule]].
+\end{theorem}<|endoftext|>
+\section{Deduction Theorem}
+Tags: Named Theorems, Propositional Logic
+
+\begin{theorem}
+Let $\mathscr H$ be [[Definition:Hilbert Proof System/Instance 1|instance 1 of a Hilbert proof system]].
+Then the [[Definition:Deduction Rule|deduction rule]]:
+::$\dfrac{U,\mathbf A \vdash \mathbf B}{U \vdash \mathbf A \implies \mathbf B}$
+is a [[Definition:Derived Rule|derived rule]] for $\mathscr H$.
+\end{theorem}
+
+\begin{proof}
+For any [[Definition:Formal Proof|proof]] of $U, \mathbf A \vdash \mathbf B$, we indicate how to transform it into a proof of $U \vdash \mathbf A \implies \mathbf B$ without using the [[Definition:Deduction Rule|deduction rule]].
+This is done by applying the [[Second Principle of Mathematical Induction]] to the length $n$ of the [[Definition:Formal Proof|proof]] of $U,\mathbf A \vdash \mathbf B$.
+If $n = 1$, then one of the following must occur:
+* $\mathbf B \in U$
+* $\mathbf B = \mathbf A$
+* $\mathbf B$ is an [[Definition:Axiom (Formal Systems)|axiom]] or a [[Definition:Theorem (Formal Systems)|theorem]] of $\mathscr H$
+In the first case, obviously $U \vdash \mathbf B$.
+By '''Axiom 1''', $U \vdash \mathbf B \implies \left({\mathbf A \implies \mathbf B}\right)$.
+By [[Modus Ponens]], $U \vdash \mathbf A \implies \mathbf B$.
+In the second case, $U \vdash \mathbf A \implies \mathbf A$ by the [[Law of Identity/Formulation 2/Proof 3|Law of Identity]].
+Finally, in the third case, we have $U \vdash \mathbf B$.
+As in the first case, we conclude $U \vdash \mathbf A \implies \mathbf B$.
+If $n > 1$, the only other option for arriving at $U, \mathbf A \vdash \mathbf B$ is through [[Modus Ponens]].
+That is to say, two earlier lines of the [[Definition:Formal Proof|proof]] contain:
+:$U, \mathbf A \vdash \mathbf C$
+:$U, \mathbf A \vdash \mathbf C \implies \mathbf B$
+for some [[Definition:WFF of Propositional Logic|WFF]] $\mathbf C$.
+But then these [[Definition:Sequent|sequents]] have shorter [[Definition:Formal Proof|proofs]].
+Hence, they satisfy the induction hypothesis.
+Thus, we may infer:
+:$U \vdash \mathbf A \implies \mathbf C$
+:$U \vdash \mathbf A \implies \left({\mathbf C \implies \mathbf B}\right)$
+This allows us to give the following [[Definition:Formal Proof|proof]] of $U \vdash \mathbf A \implies \mathbf B$:
+{{BeginTableau|U \vdash \mathbf A \implies \mathbf B|nohead=1}}
+{{TableauLine|n = 1
+  |f = U \vdash \mathbf A \implies \mathbf C
+  |rtxt = Hypothesis
+}}
+{{TableauLine|n = 2
+  |f = U \vdash \mathbf A \implies \left({\mathbf C \implies \mathbf B}\right)
+  |rtxt = Hypothesis
+}}
+{{TableauLine|n = 3
+  |f = U \vdash \left({\mathbf A \implies \left({\mathbf C \implies \mathbf B}\right)}\right) \implies \left({\left({\mathbf A \implies \mathbf C}\right)\implies \left({\mathbf A \implies \mathbf B}\right)}\right)
+  |rtxt = Axiom 2
+}}
+{{ModusPonens|4||U \vdash \left({\mathbf A \implies \mathbf C}\right)\implies \left({\mathbf A \implies \mathbf B}\right)|2|3}}
+{{ModusPonens|5||U \vdash \mathbf A \implies \mathbf B|1|4}}
+{{EndTableau}}
+The result follows by the [[Second Principle of Mathematical Induction]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Soundness Theorem for Hilbert Proof System}
+Tags: Hilbert Proof System Instance 1
+
+\begin{theorem}
+Let $\mathscr H$ be [[Definition:Hilbert Proof System/Instance 1|instance 1 of a Hilbert proof system]].
+Let $\mathrm{BI}$ be the [[Definition:Formal Semantics of Boolean Interpretations|formal semantics of boolean interpretations]].
+Then $\mathscr H$ is a [[Definition:Sound Proof System|sound proof system]] for $\mathrm{BI}$:
+:Every [[Definition:Theorem (Formal Systems)|$\mathscr H$-theorem]] is a [[Definition:Tautology (Boolean Interpretations)|tautology]].
+\end{theorem}
+
+\begin{proof}
+Recall the [[Definition:Axiom (Formal Systems)|axioms]] of $\mathscr H$:
+{{begin-axiom}}
+{{axiom | lc = '''Axiom 1:'''
+        | m  = \mathbf A \implies \paren {\mathbf B \implies \mathbf A}
+}}
+{{axiom | lc = '''Axiom 2:'''
+        | m  = \paren {\mathbf A \implies \paren {\mathbf B \implies \mathbf C} } \implies \paren {\paren {\mathbf A \implies \mathbf B} \implies \paren {\mathbf A \implies \mathbf C} }
+}}
+{{axiom | lc = '''Axiom 3:'''
+        | m  = \paren {\neg \mathbf B \implies \neg \mathbf A} \implies \paren {\mathbf A \implies \mathbf B}
+}}
+{{end-axiom}}
+That these are [[Definition:Tautology (Boolean Interpretations)|tautologies]] is shown on, respectively:
+:[[True Statement is implied by Every Statement/Formulation 2/Proof 2|True Statement is implied by Every Statement]]
+:[[Self-Distributive Law for Conditional/Formulation 1/Proof 2|Self-Distributive Law for Conditional]]
+:[[Rule of Transposition/Formulation 2/Proof 2|Rule of Transposition]]
+That [[Modus Ponens]] infers [[Definition:Tautology (Boolean Interpretations)|tautologies]] from [[Definition:Tautology (Boolean Interpretations)|tautologies]] is shown on:
+:[[Modus Ponendo Ponens/Sequent Form/Proof 2|Modus Ponendo Ponens]]
+Since:
+:All [[Definition:Axiom (Formal Systems)|axioms]] of $\mathscr H$ are [[Definition:Tautology (Boolean Interpretations)|tautologies]];
+:All [[Definition:Rule of Inference|rules of inference]] of $\mathscr H$ preserve [[Definition:Tautology (Boolean Interpretations)|tautologies]]
+it is guaranteed that every [[Definition:Formal Proof|formal proof]] in $\mathscr H$ results in a [[Definition:Tautology (Boolean Interpretations)|tautology]].
+That is, all [[Definition:Theorem (Formal Systems)|$\mathscr H$-theorems]] are [[Definition:Tautology (Boolean Interpretations)|tautologies]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Dominate (Set Theory)}
+Tags: Set Theory
+
+\begin{theorem}
+Let $S, T$ be [[Definition:Set|sets]].
+{{TFAE|def = Dominate (Set Theory)|view = Dominate|context = Set Theory}}
+\end{theorem}
+
+\begin{proof}
+=== Definition 1 implies Definition 2 ===
+Let $f: S \to T$ be an [[Definition:Injection|injection]].
+By [[Injection to Image is Bijection]], $f$ is a [[Definition:Bijection|bijection]] from $S$ to the [[Definition:Image of Mapping|image of $f$]].
+{{qed|lemma}}
+=== Definition 2 implies Definition 1 ===
+Let $T' \subseteq T$ such that there exists a [[Definition:Bijection|bijection]] $f: S \to T'$.
+Let $i: T' \to T$ be the [[Definition:Inclusion Mapping|inclusion]] of $T'$ in $T$.
+Then by [[Composite of Injections is Injection]], $i \circ f: S \to T$ is an [[Definition:Injection|injection]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Element of Toset has at most One Immediate Successor}
+Tags: Total Orderings
+
+\begin{theorem}
+Let $\struct {S, \preceq}$ be a [[Definition:Toset|toset]].
+Let $a \in S$.
+Then $a$ has at most one [[Definition:Immediate Successor Element|immediate successor]].
+\end{theorem}
+
+\begin{proof}
+Let $b, b' \in S$ be [[Definition:Immediate Successor Element|immediate successors]] of $a$.
+We have that $\preceq$ is a [[Definition:Total Ordering|total ordering]].
+{{WLOG}}:
+:$b \preceq b'$
+By virtue of $b'$ being a [[Definition:Immediate Successor Element|immediate successor]] of $a$:
+:$\neg \exists c \in S: a \prec c \prec b'$
+However, since $b$ is also an [[Definition:Immediate Successor Element|immediate successor]]:
+:$a \prec b$
+Hence, it cannot be the case that $b \prec b'$.
+Since $b \preceq b'$, it follows that $b = b'$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Element of Toset has at most One Immediate Predecessor}
+Tags: Total Orderings
+
+\begin{theorem}
+Let $\struct {S, \preceq}$ be a [[Definition:Toset|toset]].
+Let $a \in S$.
+Then $a$ has at most one [[Definition:Immediate Predecessor Element|immediate predecessor]].
+\end{theorem}
+
+\begin{proof}
+Let $b, b' \in S$ be [[Definition:Immediate Predecessor Element|immediate predecessors]] of $a$.
+We have that $\preceq$ is a [[Definition:Total Ordering|total ordering]].
+{{WLOG}}:
+:$b \preceq b'$
+By virtue of $b$ being a [[Definition:Immediate Predecessor Element|immediate predecessor]] of $a$:
+:$\neg \exists c \in S: b \prec c \prec a$
+However, since $b'$ is also an [[Definition:Immediate Predecessor Element|immediate predecessor]]:
+:$b' \prec a$
+Hence, it cannot be the case that $b \prec b'$.
+Since $b \preceq b'$, it follows that $b = b'$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Complex Inverse Tangent Function}
+Tags: Inverse Tangent
+
+\begin{theorem}
+{{TFAE|def = Complex Inverse Tangent}}
+Let $S$ be the [[Definition:Subset|subset]] of the [[Definition:Complex Plane|complex plane]]:
+:$S = \C \setminus \set {0 + i, 0 - i}$
+\end{theorem}
+
+\begin{proof}
+The proof strategy is to how that for all $z \in \C$:
+:$\set {w \in \C: \map \tan w = z} = \set {\dfrac 1 {2 i} \map \ln {\dfrac {i - z} {i + z} } + k \pi: k \in \Z}$
+Note that when $z = 0 + i$:
+{{begin-eqn}}
+{{eqn | l = i - z
+      | r = 0 + 0 i
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {i - z} {i + z}
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \map \ln {\dfrac {i - z} {i + z} }
+      | o = 
+      | r = \text {is undefined}
+      | c = 
+}}
+{{end-eqn}}
+Similarly, when $z = 0 - i$:
+{{begin-eqn}}
+{{eqn | l = i + z
+      | r = 0 + 0 i
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {i - z} {i + z}
+      | o = 
+      | r = \text {is undefined}
+      | c = 
+}}
+{{end-eqn}}
+Thus let $z \in \C \setminus \set {0 + i, 0 - i}$.
+=== Definition 1 implies Definition 2 ===
+It is demonstrated that:
+:$\set {w \in \C: \map \tan w = z} \subseteq \set {\dfrac 1 {2 i} \map \ln {\dfrac {i - z} {i + z} } + k \pi: k \in \Z}$
+Let $w \in \set {w \in \C: z = \map \tan w}$.
+Then:
+{{begin-eqn}}
+{{eqn | l = z
+      | r = i \frac {1 - e^{2 i w} } {1 + e^{2 i w} }
+      | c = [[Tangent Exponential Formulation]]
+}}
+{{eqn | ll= \leadsto
+      | l = e^{2 i w}
+      | r = \frac {1 + i z} {1 - i z}
+      | c = solving for $e^{2 i w}$
+}}
+{{eqn | r = \frac {i - z} {i + z}
+      | c = multiplying top and bottom by $i$, noting $i^2 = -1$
+}}
+{{eqn | ll= \leadsto
+      | l = \map \ln {e^{2 i w} }
+      | r = \ln \frac {i - z} {i + z}
+}}
+{{eqn | ll= \leadsto
+      | l = 2 i w + 2 k' \pi i: k' \in \Z
+      | r = \ln \frac {i - z} {i + z}
+      | c = {{Defof|Complex Natural Logarithm}}
+}}
+{{eqn | ll= \leadsto
+      | l = w
+      | r = \frac 1 {2 i} \ln \frac {i - z} {i + z} + k \pi: k \in \Z
+      | c = putting $k = -k'$
+}}
+{{end-eqn}}
+Thus by definition of [[Definition:Subset|subset]]:
+:$\set {w \in \C: \map \tan w = z} \subseteq \set {\dfrac 1 {2 i} \map \ln {\dfrac {i - z} {i + z} } + k \pi: k \in \Z}$
+{{qed|lemma}}
+=== Definition 2 implies Definition 1 ===
+It is demonstrated that:
+:$\set {w \in \C: \map \tan w = z} \supseteq \set {\dfrac 1 {2 i} \map \ln {\dfrac {i - z} {i + z} } + k \pi: k \in \Z}$
+Let $w \in \set {\dfrac 1 {2 i} \map \ln {\dfrac {i - z} {i + z} } + k \pi: k \in \Z}$.
+Then:
+{{begin-eqn}}
+{{eqn | ll= \exists k \in \Z:
+      | l = w 
+      | r = \dfrac 1 {2 i} \map \ln {\dfrac {i - z} {i + z} } + k \pi
+      | c = 
+}}
+{{eqn | lll=\leadsto
+      | ll= \exists k \in \Z:
+      | l = 2 i w + 2 \paren {-k} \pi i
+      | r = \map \ln {\dfrac {i - z} {i + z} }
+      | c = 
+}}
+{{eqn | lll=\leadsto
+      | l = e^{2 i w + 2 \paren {-k} \pi i}
+      | r = \dfrac {i - z} {i + z}
+      | c = {{Defof|Complex Natural Logarithm}}
+}}
+{{eqn | lll=\leadsto
+      | l = e^{2 i w}
+      | r = \dfrac {i - z} {i + z}
+      | c = [[Complex Exponential Function has Imaginary Period]]
+}}
+{{eqn | lll=\leadsto
+      | l = z
+      | r = i \dfrac {1 - e^{2 i w} } {1 + e^{2 i w} }
+      | c = 
+}}
+{{eqn | lll=\leadsto
+      | l = z
+      | r = \tan w
+      | c = [[Tangent Exponential Formulation]]
+}}
+{{eqn | lll=\leadsto
+      | l = w
+      | o = \in
+      | r = \set {w \in \C: \map \tan w = z}
+      | c = 
+}}
+{{end-eqn}}
+Thus by definition of [[Definition:Superset|superset]]:
+:$\set {w \in \C: \map \tan w = z} \supseteq \set {\dfrac 1 {2 i} \map \ln {\dfrac {i - z} {i + z} } + k \pi: k \in \Z}$
+{{qed|lemma}}
+Thus by definition of [[Definition:Set Equality|set equality]]:
+:$\set {w \in \C: \map \tan w = z} = \set {\dfrac 1 {2 i} \map \ln {\dfrac {i - z} {i + z} } + k \pi: k \in \Z}$
+{{qed}}
+[[Category:Inverse Tangent]]
+s1fxr3hj3h5330i801ydq32gce2lf49
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Complex Inverse Cotangent Function}
+Tags: Inverse Cotangent
+
+\begin{theorem}
+{{TFAE|def = Complex Inverse Cotangent}}
+Let $S$ be the [[Definition:Subset|subset]] of the [[Definition:Complex Plane|complex plane]]:
+:$S = \C \setminus \left\{{0 + i, 0 - i}\right\}$
+\end{theorem}
+
+\begin{proof}
+The proof strategy is to how that for all $z \in S$:
+:$\left\{{w \in \C: \cot \left({w}\right) = z}\right\} = \left\{{\dfrac 1 {2 i} \ln \left({\dfrac {z + i} {z - i}}\right) + k \pi: k \in \Z}\right\}$
+Note that when $z = 0 - i$:
+{{begin-eqn}}
+{{eqn | l = z + i
+      | r = 0 + 0 i
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \frac {z + i} {z - i}
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \ln \left({\dfrac {z + i} {z - i} }\right)
+      | o = 
+      | r = \text {is undefined}
+      | c = 
+}}
+{{end-eqn}}
+Similarly, when $z = 0 + i$:
+{{begin-eqn}}
+{{eqn | l = z - i
+      | r = 0 + 0 i
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \frac {z + i} {z - i}
+      | o = 
+      | r = \text {is undefined}
+      | c = 
+}}
+{{end-eqn}}
+Thus let $z \in \C \setminus \left\{{0 + i, 0 - i}\right\}$.
+=== Definition 1 implies Definition 2 ===
+It is demonstrated that:
+:$\left\{{w \in \C: \cot \left({w}\right) = z}\right\} \subseteq \left\{{\dfrac 1 {2 i} \ln \left({\dfrac {z + i} {z - i}}\right) + k \pi: k \in \Z}\right\}$
+Let $w \in \left\{{w \in \C: z = \cot \left({w}\right)}\right\}$.
+Then:
+{{begin-eqn}}
+{{eqn | l = z
+      | r = i \frac {e^{i w} + e^{-i w} } {e^{i w} - e^{-i w} }
+      | c = [[Cotangent Exponential Formulation]]
+}}
+{{eqn | ll= \implies
+      | l = e^{2 i w}
+      | r = \frac {i z - 1} {i z + 1}
+      | c = solving for $e^{2 i w}$
+}}
+{{eqn | r = \frac {z + i} {z - i}
+      | c = multiplying top and bottom by $-i$, noting $i^2 = -1$
+}}
+{{eqn | ll= \implies
+      | l = \ln \left({e^{2 i w} }\right)
+      | r = \ln \left({\frac {z + i} {z - i} }\right)
+}}
+{{eqn | ll= \implies
+      | l = 2 i w + 2 k' \pi i: k' \in \Z
+      | r = \ln \left({\frac {z + i} {z - i} }\right)
+      | c = Definition of [[Definition:Complex Natural Logarithm|Complex Natural Logarithm]]
+}}
+{{eqn | ll= \implies
+      | l = w
+      | r = \frac 1 {2 i} \ln \left({\frac {z + i} {z - i} }\right) + k \pi: k \in \Z
+      | c = putting $k = -k'$
+}}
+{{end-eqn}}
+Thus by definition of [[Definition:Subset|subset]]:
+:$\left\{{w \in \C: \cot \left({w}\right) = z}\right\} \subseteq \left\{{\dfrac 1 {2 i} \ln \left({\dfrac {z + i} {z - i}}\right) + k \pi: k \in \Z}\right\}$
+{{qed|lemma}}
+=== Definition 2 implies Definition 1 ===
+It is demonstrated that:
+:$\left\{{w \in \C: \cot \left({w}\right) = z}\right\} \supseteq \left\{{\dfrac 1 {2 i} \ln \left({\dfrac {z + i} {z - i}}\right) + k \pi: k \in \Z}\right\}$
+Let $w \in \left\{{\dfrac 1 {2 i} \ln \left({\dfrac {z + i} {z - i}}\right) + k \pi: k \in \Z}\right\}$.
+Then:
+{{begin-eqn}}
+{{eqn | ll= \exists k \in \Z:
+      | l = w 
+      | r = \dfrac 1 {2 i} \ln \left({\dfrac {z + i} {z - i} }\right) + k \pi
+      | c = 
+}}
+{{eqn | lll= \implies
+      | ll= \exists k \in \Z:
+      | l = 2 i w + 2 \left({-k}\right) \pi i
+      | r = \ln \left({\frac {z + i} {z - i} }\right)
+      | c = 
+}}
+{{eqn | lll= \implies
+      | l = e^{2 i w + 2 \left({-k}\right) \pi i}
+      | r = \frac {z + i} {z - i}
+      | c = Definition of [[Definition:Complex Natural Logarithm|Complex Natural Logarithm]]
+}}
+{{eqn | lll= \implies
+      | l = e^{2 i w}
+      | r = \frac {z + i} {z - i}
+      | c = [[Complex Exponential Function has Imaginary Period]]
+}}
+{{eqn | lll= \implies
+      | l = z
+      | r = i \frac {e^{2 i w} + 1} {e^{2 i w} - 1}
+      | c = 
+}}
+{{eqn | lll= \implies
+      | l = z
+      | r = \cot w
+      | c = [[Cotangent Exponential Formulation]]
+}}
+{{eqn | lll= \implies
+      | l = w
+      | o = \in
+      | r = \left\{ {w \in \C: \cot \left({w}\right) = z}\right\}
+      | c = 
+}}
+{{end-eqn}}
+Thus by definition of [[Definition:Superset|superset]]:
+:$\left\{{w \in \C: \cot \left({w}\right) = z}\right\} \supseteq \left\{{\dfrac 1 {2 i} \ln \left({\dfrac {z + i} {z - i}}\right) + k \pi: k \in \Z}\right\}$
+{{qed|lemma}}
+Thus by definition of [[Definition:Set Equality|set equality]]:
+:$\left\{{w \in \C: \cot \left({w}\right) = z}\right\} = \left\{{\dfrac 1 {2 i} \ln \left({\dfrac {z + i} {z - i}}\right) + k \pi: k \in \Z}\right\}$
+{{qed}}
+[[Category:Inverse Cotangent]]
+olaj1niepowkba7m70eh4ocwvpgtk23
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Complex Inverse Hyperbolic Tangent}
+Tags: Inverse Hyperbolic Tangent
+
+\begin{theorem}
+{{TFAE|def = Complex Inverse Hyperbolic Tangent}}
+Let $S$ be the [[Definition:Subset|subset]] of the [[Definition:Complex Plane|complex plane]]:
+:$S = \C \setminus \left\{{-1 + 0 i, 1 + 0 i}\right\}$
+\end{theorem}
+
+\begin{proof}
+The proof strategy is to how that for all $z \in S$:
+:$\left\{{w \in \C: z = \tanh \left({w}\right)}\right\} = \left\{{\dfrac 1 2 \ln \left({\dfrac {1 + z} {1 - z} }\right) + k \pi i: k \in \Z}\right\}$
+Note that when $z = -1 + 0 i$:
+{{begin-eqn}}
+{{eqn | l = 1 + z
+      | r = 0 + 0 i
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \frac {1 + z} {1 - z}
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \ln \left({\dfrac {1 + z} {1 - z} }\right)
+      | o = 
+      | r = \text {is undefined}
+      | c = 
+}}
+{{end-eqn}}
+Similarly, when $z = 1 + 0 i$:
+{{begin-eqn}}
+{{eqn | l = 1 - z
+      | r = 0 + 0 i
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \frac {1 + z} {1 - z}
+      | o = 
+      | r = \text {is undefined}
+      | c = 
+}}
+{{end-eqn}}
+Thus let $z \in \C \setminus \left\{{-1 + 0 i, 1 + 0 i}\right\}$.
+=== Definition 1 implies Definition 2 ===
+It is demonstrated that:
+:$\left\{{w \in \C: z = \tanh \left({w}\right)}\right\} \subseteq \left\{{\dfrac 1 2 \ln \left({\dfrac {1 + z} {1 - z} }\right) + k \pi i: k \in \Z}\right\}$
+Let $w \in \left\{{w \in \C: z = \tanh \left({w}\right)}\right\}$.
+Then:
+{{begin-eqn}}
+{{eqn | l = z
+      | r = \frac {e^w - e^{- w} } {e^w + e^{- w} }
+      | c = Definition of [[Definition:Hyperbolic Tangent|Hyperbolic Tangent]]
+}}
+{{eqn | l = z
+      | r = \frac {e^{2 w} - 1} {e^{2 w} + 1}
+      | c = multiplying top and bottom by $e^w$
+}}
+{{eqn | ll= \implies
+      | l = e^{2 w}
+      | r = \frac {1 + z} {1 - z}
+      | c = solving for $e^{2 w}$
+}}
+{{eqn | ll= \implies
+      | l = \ln \left({e^{2 w} }\right)
+      | r = \ln \frac {1 + z} {1 - z}
+}}
+{{eqn | ll= \implies
+      | l = 2 w + 2 k' \pi i: k' \in \Z
+      | r = \ln \frac {1 + z} {1 - z}
+      | c = Definition of [[Definition:Complex Natural Logarithm|Complex Natural Logarithm]]
+}}
+{{eqn | ll= \implies
+      | l = w
+      | r = \frac 1 2 \ln \frac {1 + z} {1 - z} + k \pi i: k \in \Z
+      | c = putting $k = -k'$
+}}
+{{end-eqn}}
+Thus by definition of [[Definition:Subset|subset]]:
+:$\left\{{w \in \C: z = \tanh \left({w}\right)}\right\} \subseteq \left\{{\dfrac 1 2 \ln \left({\dfrac {1 + z} {1 - z} }\right) + k \pi i: k \in \Z}\right\}$
+{{qed|lemma}}
+=== Definition 2 implies Definition 1 ===
+It is demonstrated that:
+:$\left\{{w \in \C: z = \tanh \left({w}\right)}\right\} \supseteq \left\{{\dfrac 1 2 \ln \left({\dfrac {1 + z} {1 - z} }\right) + k \pi i: k \in \Z}\right\}$
+Let $w \in \left\{{\dfrac 1 2 \ln \left({\dfrac {1 + z} {1 - z} }\right) + k \pi i: k \in \Z}\right\}$.
+Then:
+{{begin-eqn}}
+{{eqn | ll= \exists k \in \Z:
+      | l = w 
+      | r = \dfrac 1 2 \ln \left({\dfrac {1 + z} {1 - z} }\right) + k \pi i
+      | c = 
+}}
+{{eqn | lll= \implies
+      | ll= \exists k \in \Z:
+      | l = 2 w + 2 \left({-k}\right) \pi i
+      | r = \ln \left({\dfrac {1 + z} {1 - z} }\right)
+      | c = 
+}}
+{{eqn | lll= \implies
+      | l = e^{2 w + 2 \left({-k}\right) \pi i}
+      | r = \dfrac {1 + z} {1 - z}
+      | c = Definition of [[Definition:Complex Natural Logarithm|Complex Natural Logarithm]]
+}}
+{{eqn | lll= \implies
+      | l = e^{2 w}
+      | r = \dfrac {1 + z} {1 - z}
+      | c = [[Complex Exponential Function has Imaginary Period]]
+}}
+{{eqn | lll= \implies
+      | l = z
+      | r = \dfrac {e^w - e^{- w} } {e^w + e^{- w} }
+      | c = 
+}}
+{{eqn | lll= \implies
+      | l = z
+      | r = \tanh w
+      | c = Definition of [[Definition:Hyperbolic Tangent|Hyperbolic Tangent]]
+}}
+{{eqn | lll= \implies
+      | l = w
+      | o = \in
+      | r = \left\{ {w \in \C: \tanh \left({w}\right) = z}\right\}
+      | c = 
+}}
+{{end-eqn}}
+Thus by definition of [[Definition:Superset|superset]]:
+:$\left\{{w \in \C: z = \tanh \left({w}\right)}\right\} \supseteq \left\{{\dfrac 1 2 \ln \left({\dfrac {1 + z} {1 - z} }\right) + k \pi i: k \in \Z}\right\}$
+{{qed|lemma}}
+Thus by definition of [[Definition:Set Equality|set equality]]:
+:$\left\{{w \in \C: z = \tanh \left({w}\right)}\right\} = \left\{{\dfrac 1 2 \ln \left({\dfrac {1 + z} {1 - z} }\right) + k \pi i: k \in \Z}\right\}$
+{{qed}}
+[[Category:Inverse Hyperbolic Tangent]]
+mqr7r0y3zyb1dfvmowcj3eu6d0s6e0n
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Complex Inverse Hyperbolic Cotangent}
+Tags: Inverse Hyperbolic Cotangent
+
+\begin{theorem}
+{{TFAE|def = Complex Inverse Hyperbolic Cotangent}}
+Let $S$ be the [[Definition:Subset|subset]] of the [[Definition:Complex Plane|complex plane]]:
+:$S = \C \setminus \left\{{-1 + 0 i, 1 + 0 i}\right\}$
+\end{theorem}
+
+\begin{proof}
+The proof strategy is to how that for all $z \in S$:
+:$\left\{{w \in \C: \coth \left({w}\right) = z}\right\} = \left\{{\dfrac 1 2 \ln \left({\dfrac {z + 1} {z - 1} }\right) + k \pi i: k \in \Z}\right\}$
+Note that when $z = -1 + 0 i$:
+{{begin-eqn}}
+{{eqn | l = z + 1
+      | r = 0 + 0 i
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \frac {z + 1} {z - 1}
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \ln \left({\dfrac {z + 1} {z - 1} }\right)
+      | o = 
+      | r = \text {is undefined}
+      | c = 
+}}
+{{end-eqn}}
+Similarly, when $z = 1 + 0 i$:
+{{begin-eqn}}
+{{eqn | l = z - 1
+      | r = 0 + 0 i
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \frac {z + 1} {z - 1}
+      | o = 
+      | r = \text {is undefined}
+      | c = 
+}}
+{{end-eqn}}
+Thus let $z \in \C \setminus \left\{{-1 + 0 i, 1 + 0 i}\right\}$.
+=== Definition 1 implies Definition 2 ===
+It is demonstrated that:
+:$\left\{{w \in \C: \coth \left({w}\right) = z}\right\} \subseteq \left\{{\dfrac 1 2 \ln \left({\dfrac {z + 1} {z - 1} }\right) + k \pi i: k \in \Z}\right\}$
+Let $w \in \left\{{w \in \C: z = \coth \left({w}\right)}\right\}$.
+Then:
+{{begin-eqn}}
+{{eqn | l = z
+      | r = \frac {e^w + e^{-w} } {e^w - e^{-w} }
+      | c = Definition of [[Definition:Hyperbolic Cotangent|Hyperbolic Cotangent]]
+}}
+{{eqn | ll= \implies
+      | l = e^{2 w}
+      | r = \frac {z + 1} {z - 1}
+      | c = solving for $e^{2 w}$
+}}
+{{eqn | ll= \implies
+      | l = \ln \left({e^{2 w} }\right)
+      | r = \ln \left({\frac {z + 1} {z - 1} }\right)
+}}
+{{eqn | ll= \implies
+      | l = 2 w + 2 k' \pi i: k' \in \Z
+      | r = \ln \left({\frac {z + 1} {z - 1} }\right)
+      | c = Definition of [[Definition:Complex Natural Logarithm|Complex Natural Logarithm]]
+}}
+{{eqn | ll= \implies
+      | l = w
+      | r = \frac 1 2 \ln \left({\frac {z + 1} {z - 1} }\right) + k \pi i: k \in \Z
+      | c = putting $k = -k'$
+}}
+{{end-eqn}}
+Thus by definition of [[Definition:Subset|subset]]:
+:$\left\{{w \in \C: \coth \left({w}\right) = z}\right\} \subseteq \left\{{\dfrac 1 2 \ln \left({\dfrac {z + 1} {z - 1} }\right) + k \pi i: k \in \Z}\right\}$
+{{qed|lemma}}
+=== Definition 2 implies Definition 1 ===
+It is demonstrated that:
+:$\left\{{w \in \C: \coth \left({w}\right) = z}\right\} \supseteq \left\{{\dfrac 1 2 \ln \left({\dfrac {z + 1} {z - 1} }\right) + k \pi i: k \in \Z}\right\}$
+Let $w \in \left\{{\dfrac 1 2 \ln \left({\dfrac {z + 1} {z - 1} }\right) + k \pi i: k \in \Z}\right\}$.
+Then:
+{{begin-eqn}}
+{{eqn | ll= \exists k \in \Z:
+      | l = w 
+      | r = \dfrac 1 2 \ln \left({\dfrac {z + 1} {z - 1} }\right) + k \pi i
+      | c = 
+}}
+{{eqn | lll= \implies
+      | ll= \exists k \in \Z:
+      | l = 2 w + 2 \left({-k}\right) \pi i
+      | r = \ln \left({\frac {z + 1} {z - 1} }\right)
+      | c = 
+}}
+{{eqn | lll= \implies
+      | l = e^{2 w + 2 \left({-k}\right) \pi i}
+      | r = \frac {z + 1} {z - 1}
+      | c = Definition of [[Definition:Complex Natural Logarithm|Complex Natural Logarithm]]
+}}
+{{eqn | lll= \implies
+      | l = e^{2 w}
+      | r = \frac {z + 1} {z - 1}
+      | c = [[Complex Exponential Function has Imaginary Period]]
+}}
+{{eqn | lll= \implies
+      | l = z
+      | r = \frac {e^{2 w} + 1} {e^{2 w} - 1}
+      | c = 
+}}
+{{eqn | r = \frac {e^w + e^{-w} } {e^w - e^{-w} }
+      | c = multiplying top and bottom by $e^{-w}$
+}}
+{{eqn | lll= \implies
+      | l = z
+      | r = \coth w
+      | c = Definition of [[Definition:Hyperbolic Cotangent|Hyperbolic Cotangent]]
+}}
+{{eqn | lll= \implies
+      | l = w
+      | o = \in
+      | r = \left\{ {w \in \C: \coth \left({w}\right) = z}\right\}
+      | c = 
+}}
+{{end-eqn}}
+Thus by definition of [[Definition:Superset|superset]]:
+:$\left\{{w \in \C: \coth \left({w}\right) = z}\right\} \supseteq \left\{{\dfrac 1 2 \ln \left({\dfrac {z + 1} {z - 1} }\right) + k \pi i: k \in \Z}\right\}$
+{{qed|lemma}}
+Thus by definition of [[Definition:Set Equality|set equality]]:
+:$\left\{{w \in \C: \coth \left({w}\right) = z}\right\} = \left\{{\dfrac 1 2 \ln \left({\dfrac {z + 1} {z - 1} }\right) + k \pi i: k \in \Z}\right\}$
+{{qed}}
+[[Category:Inverse Hyperbolic Cotangent]]
+r1ftmi35rmzyxjelqoltdd60v3w8od5
+\end{proof}<|endoftext|>
+\section{Natural Numbers form Inductive Set}
+Tags: Natural Numbers, Inductive Sets, Natural Numbers form Inductive Set
+
+\begin{theorem}
+Let $\N$ denote the [[Definition:Natural Numbers in Real Numbers|natural numbers]] as [[Definition:Subset|subset]] of the [[Definition:Real Number|real numbers]] $\R$.
+Then $\N$ is an [[Definition:Inductive Set as Subset of Real Numbers|inductive set]].
+\end{theorem}
+
+\begin{proof}
+By definition of the [[Definition:Natural Numbers in Real Numbers|natural numbers]]:
+:$\N = \displaystyle \bigcap \II$
+where $\II$ is the collection of all [[Definition:Inductive Set as Subset of Real Numbers|inductive sets]].
+Suppose that $n \in \N$.
+Then by definition of [[Definition:Intersection of Set of Sets|intersection]]:
+:$\forall I \in \II: n \in I$
+Because all these $I$ are [[Definition:Inductive Set as Subset of Real Numbers|inductive]]:
+:$\forall I \in \II: n + 1 \in I$
+Again by definition of [[Definition:Intersection of Set of Sets|intersection]]:
+:$n + 1 \in \N$
+{{qed}}
+\end{proof}
+
+\begin{proof}
+By the given definition of the [[Definition:Natural Numbers in Real Numbers|natural numbers]]:
+:$\N = \bigcap \II$
+where $\II$ is the collection of all [[Definition:Inductive Set as Subset of Real Numbers|inductive sets]].
+The result is a direct application of [[Intersection of Inductive Set as Subset of Real Numbers is Inductive Set]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Square Root of Complex Number in Cartesian Form}
+Tags: Complex Analysis, Square Roots, Complex Numbers
+
+\begin{theorem}
+Let $z \in \C$ be a [[Definition:Complex Number|complex number]].
+Let $z = x + i y$ where $x, y \in \R$ are [[Definition:Real Number|real numbers]].
+Let $z$ not be [[Definition:Wholly Real|wholly real]], that is, such that $y \ne 0$.
+Then the [[Definition:Complex Square Root|square root]] of $z$ is given by:
+:$z^{1/2} = \pm \paren {a + i b}$
+where:
+{{begin-eqn}}
+{{eqn | l = a
+      | r = \sqrt {\frac {x + \sqrt {x^2 + y^2} } 2}
+      | c = 
+}}
+{{eqn | l = b
+      | r = \frac y {\cmod y} \sqrt {\frac {-x + \sqrt {x^2 + y^2} } 2}
+      | c = 
+}}
+{{end-eqn}}
+\end{theorem}
+
+\begin{proof}
+Let $a + i b \in z^{1/2}$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \paren {a + i b}^2
+      | r = x + i y
+      | c = {{Defof|Square Root|subdef = Complex Number|index = 4|Square Root of Complex Number}}
+}}
+{{eqn | n = 1
+      | ll= \leadsto
+      | l = a^2 + 2 i a b - b^2
+      | r = x + i y
+      | c = [[Square of Sum]] and $i^2 = -1$
+}}
+{{end-eqn}}
+Equating [[Definition:Imaginary Part|imaginary parts]] in $(1)$:
+{{begin-eqn}}
+{{eqn | l = 2 a b
+      | r = y
+      | c = 
+}}
+{{eqn | n = 2
+      | ll= \leadsto
+      | l = b
+      | r = \frac y {2 a}
+      | c = rearranging
+}}
+{{end-eqn}}
+Equating [[Definition:Real Part|real parts]] in $(1)$:
+{{begin-eqn}}
+{{eqn | l = a^2 - b^2
+      | r = x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = a^2 - \paren {\frac y {2 a} }^2
+      | r = x
+      | c = substituting for $b$ from $(2)$
+}}
+{{eqn | ll= \leadsto
+      | l = 4 a^4 - 4 a^2 x - y^2
+      | r = 0
+      | c = multiplying by $4 a^2$ and rearranging
+}}
+{{eqn | ll= \leadsto
+      | l = a^2
+      | r = \frac {4 x \pm \sqrt {16 x^2 + 16 y^2} } 8
+      | c = [[Quadratic Formula]]
+}}
+{{eqn | ll= \leadsto
+      | l = a^2
+      | r = \frac {x \pm \sqrt {x^2 + y^2} } 2
+      | c = dividing top and bottom by $4$
+}}
+{{eqn | n = 3
+      | ll= \leadsto
+      | l = a
+      | r = \pm \sqrt {\frac {x + \sqrt {x^2 + y^2} } 2}
+      | c = taking the [[Definition:Square Root|square root]]
+}}
+{{end-eqn}}
+Note that in $(3)$, only the [[Definition:Positive Square Root|positive square root]] of the [[Definition:Discriminant of Quadratic Equation|discriminant]] $x^2 + y^2$ is used.
+This is because the [[Definition:Negative Square Root|negative square root]] of $x^2 + y^2$ would yield $\dfrac {x - \sqrt {x^2 + y^2} } 2 < 0$.
+As $a \in \R$, it is necessary that $\dfrac {x + \sqrt {x^2 + y^2} } 2 > 0$.
+Hence $\sqrt {x^2 + y^2} > 0$.
+Then:
+{{begin-eqn}}
+{{eqn | l = b
+      | r = \frac y {2 a}
+      | c = from $(2)$
+}}
+{{eqn | r = \frac y {2 \paren {\pm \sqrt {\dfrac {x + \sqrt {x^2 + y^2} } 2} } }
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = b^2
+      | r = \frac {y^2} {2 \paren {x + \sqrt {x^2 + y^2} } }
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = b^2
+      | r = \frac {y^2 \paren {x - \sqrt {x^2 + y^2} } } {2 \paren {x + \sqrt {x^2 + y^2} } \paren {x - \sqrt {x^2 + y^2} } }
+      | c = multiplying top and bottom by $x - \sqrt {x^2 + y^2}$
+}}
+{{eqn | ll= \leadsto
+      | l = b^2
+      | r = \frac {y^2 \paren {x - \sqrt {x^2 + y^2} } } {2 \paren {x^2 - \paren {x^2 + y^2} } }
+      | c = [[Difference of Two Squares]]
+}}
+{{eqn | ll= \leadsto
+      | l = b^2
+      | r = \frac {y^2 \paren {x - \sqrt {x^2 + y^2} } } {- 2 y^2}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = b^2
+      | r = \frac {-x + \sqrt {x^2 + y^2} } 2
+      | c = 
+}}
+{{end-eqn}}
+But from $(2)$ we have:
+:$b = \dfrac y {2 a}$
+and so having picked either the [[Definition:Positive Square Root|positive square root]] or [[Definition:Negative Square Root|negative square root]] of either $a^2$ or $b^2$, the [[Definition:Square Root|root]] of the other is forced.
+So:
+:if $y > 0$, then $a$ and $b$ are both of the same [[Definition:Sign of Number|sign]].
+Thus:
+:$b = 1 \times \sqrt {\dfrac {-x + \sqrt {x^2 + y^2} } 2}$
+:if $y < 0$, then $a$ and $b$ are of opposite [[Definition:Sign of Number|sign]].
+Thus:
+:$b = \paren {-1} \times \sqrt {\dfrac {-x + \sqrt {x^2 + y^2} } 2}$
+Hence:
+:$b = \dfrac y {\cmod y} \sqrt {\dfrac {-x + \sqrt {x^2 + y^2} } 2}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Zero is Identity in Naturally Ordered Semigroup}
+Tags: Naturally Ordered Semigroup
+
+\begin{theorem}
+Let $\struct {S, \circ, \preceq}$ be a [[Definition:Naturally Ordered Semigroup|naturally ordered semigroup]].
+Let $0$ be the [[Definition:Zero of Naturally Ordered Semigroup|zero]] of $\struct {S, \circ, \preceq}$.
+Then $0$ is the [[Definition:Identity Element|identity]] for $\circ$.
+That is:
+:$\forall n \in S: n \circ 0 = n = 0 \circ n$
+\end{theorem}
+
+\begin{proof}
+By definition of an [[Definition:Ordering|ordering]]:
+:$0 \preceq 0$
+Thus from [[Definition:Naturally Ordered Semigroup Axioms|axiom $(\text {NO} 3)$]]:
+:$\exists p \in S: 0 \circ p = 0$
+By the definition of [[Definition:Zero of Naturally Ordered Semigroup|zero]]:
+:$0 \preceq 0 \circ 0$ and $0 \preceq p$
+Thus since $\preceq$ is [[Definition:Relation Compatible with Operation|compatible]] with $\circ$:
+:$0 \circ 0 \preceq 0 \circ p = 0$
+Thus:
+:$0 \circ 0 \preceq 0$ and $0 \preceq 0 \circ 0$ 
+Hence, as $\preceq$ is [[Definition:Antisymmetric Relation|antisymmetric]], it follows that:
+:$0 \circ 0 = 0$
+Because $\struct {S, \circ, \preceq}$ is a [[Definition:Semigroup|semigroup]], $\circ$ is [[Definition:Associative|associative]]. 
+So:
+:$\forall n \in S: \paren {n \circ 0} \circ 0 = n \circ \paren {0 \circ 0} = n \circ 0$
+Thus from [[Definition:Naturally Ordered Semigroup Axioms|axiom $(\text {NO} 2)$]]:
+:$\forall n \in S: n \circ 0 = n$
+Similarly:
+:$\forall n \in S: 0 \circ \paren {0 \circ n} = \paren {0 \circ 0} \circ n = 0 \circ n$
+meaning:
+:$\forall n \in S: 0 \circ n = n$
+Thus:
+:$\forall n \in S: n \circ 0 = n = 0 \circ n$
+and so $0$ is the [[Definition:Identity Element|identity]] for $\circ$.
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Reduction Formula for Integral of Power of Sine}
+Tags: Primitives involving Sine Function, Reduction Formula for Integral of Power of Sine
+
+\begin{theorem}
+Let $n \in \Z_{> 0}$ be a [[Definition:Strictly Positive Integer|(strictly) positive integer]].
+Then:
+:$\displaystyle \int \sin^n x \rd x = \dfrac {n - 1} n \int \sin^{n - 2} x \rd x - \dfrac {\sin^{n - 1} x \cos x} n$
+is a [[Definition:Reduction Formula|reduction formula]] for $\displaystyle \int \sin^n x \rd x$.
+\end{theorem}
+
+\begin{proof}
+Let $n \ge 2$.
+Let:
+: $\displaystyle I_n := \int \sin^n x \ \mathrm d x$
+Then:
+{{begin-eqn}}
+{{eqn | l = I_n
+      | r = \int \sin^n x \ \mathrm d x
+      | c = 
+}}
+{{eqn | r = \int \sin^{n-1} x \sin x \ \mathrm d x
+      | c = 
+}}
+{{eqn | r = \int \sin^{n-1} x \ \mathrm d \left({-\cos x}\right)
+      | c = [[Derivative of Cosine Function]]
+}}
+{{eqn | r = - \sin^{n-1} x \cos x - \int \left({-\cos x}\right) \mathrm d \left({\sin^{n-1} x }\right)
+      | c = [[Integration by Parts]]
+}}
+{{eqn | r = - \sin^{n-1} x \cos x - \int \left({-\cos x}\right) \left({n - 1}\right)\sin^{n-2} x \cos x \ \mathrm d x
+      | c = [[Derivative of Power]] and [[Chain Rule for Derivatives]]
+}}
+{{eqn | r = - \sin^{n-1} x \cos x + \left({n - 1}\right) \int \sin^{n-2} x \cos^2 x \ \mathrm d x
+      | c = [[Linear Combination of Integrals]] and rearranging
+}}
+{{eqn | r = - \sin^{n-1} x \cos x + \left({n - 1}\right) \int \sin^{n-2} x \left({1 - \sin^2 x}\right) \ \mathrm d x
+      | c = [[Sum of Squares of Sine and Cosine]]
+}}
+{{eqn | r = - \sin^{n-1} x \cos x + \left({n - 1}\right) \int \sin^{n-2} x \ \mathrm d x - \left({n - 1}\right) \int \sin^n x \ \mathrm d x
+      | c = 
+}}
+{{eqn | r = - \sin^{n-1} x \cos x + \left({n - 1}\right) I_{n-2} - \left({n - 1}\right) I_n
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = n I_n
+      | r = - \sin^{n-1} x \cos x + \left({n - 1}\right) I_{n-2}
+      | c = adding $\left({n - 1}\right) I_n$ to both sides
+}}
+{{eqn | ll= \implies
+      | l = I_n
+      | r = \dfrac {n - 1} n I_{n-2} - \dfrac {\sin^{n-1} x \cos x} n
+      | c = dividing by $n$ and rearranging
+}}
+{{end-eqn}}
+thus demonstrating the identity for all $n \ge 2$.
+When $n = 1$ this degenerates to:
+{{begin-eqn}}
+{{eqn | l = \int \sin x \ \mathrm d x
+      | r = \dfrac 0 1 \int \frac 1 {\sin x} \ \mathrm d x - \dfrac {\sin^0 x \cos x} 1
+      | c = 
+}}
+{{eqn | r = 0 - 1 \cdot \cos x
+      | c = 
+}}
+{{eqn | r = -\cos x
+      | c =
+}}
+{{end-eqn}}
+From [[Primitive of Sine Function]] this shows that the identity still holds for $n = 1$.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+With a view to expressing the problem in the form:
+:$\displaystyle \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$
+let:
+{{begin-eqn}}
+{{eqn | l = u
+      | r = \sin^{n - 1} x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d u} {\d x}
+      | r = \paren {n - 1} \sin ^{n - 2} x \cos x
+      | c = [[Chain Rule for Derivatives]], [[Derivative of Sine Function]], [[Derivative of Power]]
+}}
+{{end-eqn}}
+and let:
+{{begin-eqn}}
+{{eqn | l = \frac {\d v} {\d x}
+      | r = \sin x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = v
+      | r = -\cos x
+      | c = [[Primitive of Sine Function]]
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = \int \sin^n x \rd x
+      | r = \int \sin^{n - 1} x \sin x \rd x
+      | c = 
+}}
+{{eqn | r = \sin^{n - 1} x \paren {-\cos x} - \int \paren {-\cos x} \paren {\paren {n - 1} \sin^{n - 2} x \cos x} \rd x
+      | c = [[Integration by Parts]]
+}}
+{{eqn | r = \int \paren {n - 1} \sin^{n - 2} x \cos^2 x \rd x - \sin^{n - 1} x \cos x
+      | c = rearranging
+}}
+{{eqn | r = \int \paren {n - 1} \sin^{n - 2} x \paren {1 - \sin^2 x} \rd x - \sin^{n - 1} x \cos x
+      | c = [[Sum of Squares of Sine and Cosine]]
+}}
+{{eqn | r = \paren {n - 1} \int \sin^{n - 2} x \rd x - \paren {n - 1} \int \sin^n x \rd x - \sin^{n - 1} x \cos x
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | ll= \leadsto
+      | l = n \int \sin^n x \rd x
+      | r = \paren {n - 1} \int \sin^{n - 2} x \rd x - \sin^{n - 1} x \cos x
+      | c = rearranging
+}}
+{{eqn | ll= \leadsto
+      | l = \int \sin^n x \rd x
+      | r = \dfrac {n - 1} n \int \sin^{n - 2} x \rd x - \dfrac {\sin^{n - 1} x \cos x} n
+      | c = dividing both sides by $n$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Reduction Formula for Definite Integral of Power of Sine}
+Tags: Definite Integrals involving Sine Function
+
+\begin{theorem}
+Let $n \in \Z_{> 0}$ be a [[Definition:Positive Integer|positive integer]].
+Let $I_n$ be defined as:
+:$\displaystyle I_n = \int_0^{\frac \pi 2} \sin^n x \rd x$
+Then $\left\langle{I_n}\right\rangle$ is a [[Definition:Decreasing Real Sequence|decreasing sequence of real numbers]] which satisfies:
+:$n I_n = \left({n - 1}\right) I_{n - 2}$
+Thus:
+:$I_n = \dfrac {n - 1} n  I_{n - 2}$
+is a [[Definition:Reduction Formula|reduction formula]] for $I_n$.
+\end{theorem}
+
+\begin{proof}
+From [[Shape of Sine Function]]:
+:$\forall x \in \left[{0 \,.\,.\, \dfrac \pi 2}\right]: 0 \le \sin x \le 1$
+and so on the same [[Definition:Open Real Interval|interval]]:
+:$0 \le \sin^{n + 1} x \le \sin^n x$
+therefore:
+:$\forall n \in \N: 0 < I_{n + 1} < I_n$
+From [[Reduction Formula for Integral of Power of Sine]]:
+:$\displaystyle \int \sin^n x \rd x = \dfrac {n - 1} n \int \sin^{n - 2} x \rd x - \dfrac {\sin^{n - 1} x \cos x} n$
+Thus:
+{{begin-eqn}}
+{{eqn | l = I_n
+      | r = \int_0^{\frac \pi 2} \sin^n x \rd x
+      | c = 
+}}
+{{eqn | r = \dfrac {n - 1} n \int_0^{\frac \pi 2} \sin^{n - 2} x \rd x - \left[{\dfrac {\sin^{n - 1} x \cos x} n}\right]_0^{\pi / 2}
+      | c = 
+}}
+{{eqn | r = \dfrac {n - 1} n I_{n - 2} - \frac {\left({\sin^{n - 1} \frac \pi 2 \cos \frac \pi 2 - \sin^{n - 1} 0 \cos 0}\right)} n
+      | c = 
+}}
+{{eqn | r = \dfrac {n - 1} n I_{n - 2} - \frac {\left({0 - \sin^{n - 1} 0 \cos 0}\right)} n
+      | c = [[Cosine of Right Angle]]
+}}
+{{eqn | r = \dfrac {n - 1} n I_{n - 2} - \frac {\left({0 - 0}\right)} n
+      | c = [[Sine of Zero is Zero]]
+}}
+{{eqn | ll= \implies
+      | l = n I_n
+      | r = \left({n - 1}\right) I_{n - 2}
+      | c = multiplying both sides by $n$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Stirling's Formula/Proof 2/Lemma 4}
+Tags: Sine Function, Stirling's Formula
+
+\begin{theorem}
+Let $I_n$ be defined as:
+:$\displaystyle I_n = \int_0^{\frac \pi 2} \sin^n x \rd x$
+Then:
+:$\displaystyle \lim_{n \mathop \to \infty} \frac {I_{2 n} } {I_{2 n + 1} } = 1$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = n I_n
+      | r = \left({n - 1}\right) I_{n - 2}
+      | cc= [[Reduction Formula for Definite Integral of Power of Sine]]
+}}
+{{eqn | ll= \implies
+      | l = 1
+      | o = <
+      | r = \frac {I_{ 2n} } {I_{2 n + 1} }
+      | cc= 
+}}
+{{eqn | o = <
+      | r = \frac {I_{2 n - 1} } {I_{2 n + 1} }
+      | cc= [[Reduction Formula for Definite Integral of Power of Sine]]
+}}
+{{eqn | r = \frac {2 n + 1}  {2 n}
+      | cc= [[Reduction Formula for Definite Integral of Power of Sine]]
+}}
+{{eqn | r = \to 0
+      | c = as $n \to \infty$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Definite Integral from 0 to Half Pi of Even Power of Sine x}
+Tags: Definite Integrals involving Sine Function, Definite Integral from 0 to Half Pi of Even Power of Sine x
+
+\begin{theorem}
+:$\displaystyle \int_0^{\frac \pi 2} \sin^{2 n} x \rd x = \dfrac {\paren {2 n}!} {\paren {2^n n!}^2} \dfrac \pi 2$
+\end{theorem}
+
+\begin{proof}
+{{improve|This would be more rigorous if implemented as a formal induction proof.}}
+Let $I_n = \displaystyle \int_0^{\frac \pi 2} \sin^n x \rd x$.
+Then:
+{{begin-eqn}}
+{{eqn | l = I_{2 n}
+      | r = \frac {2 n - 1} {2 n} I_{2 n - 2}
+      | c = [[Reduction Formula for Definite Integral of Power of Sine]]
+}}
+{{eqn | r = \frac {\paren {2 n - 1} \paren {2 n - 3} } {2 n \paren {2 n - 2} } I_{2 n - 4}
+      | c = [[Reduction Formula for Definite Integral of Power of Sine]] again
+}}
+{{eqn | r = \frac {\paren {2 n - 1} \paren {2 n - 3} \cdots 1} {2 n \paren {2 n - 2} \cdots 2} I_0
+      | c = [[Reduction Formula for Definite Integral of Power of Sine]] until the end
+}}
+{{eqn | r = \frac {\paren {2 n - 1} \paren {2 n - 3} \cdots 1} {2 n \paren {2 n - 2} \cdots 2} \int_0^{\pi / 2} \rd x
+      | c = Definition of $I_n$
+}}
+{{eqn | r = \frac {\paren {2 n - 1} \paren {2 n - 3} \cdots 1} {2 n \paren {2 n - 2} \cdots 2} \bigintlimits x 0 {\pi / 2}
+      | c = [[Integral of Constant]]
+}}
+{{eqn | r = \frac {\paren {2 n - 1} \paren {2 n - 3} \cdots 1} {2 n \paren {2 n - 2} \cdots 2} \frac \pi 2
+      | c = [[Integral of Constant]]
+}}
+{{eqn | r = \frac {2n \paren {2 n - 1} \paren {2 n - 2} \paren {2 n - 3} \cdots 2 \cdot 1} {\paren {2 n}^2 \paren {2 n - 2}^2 \cdots 2^2} \frac \pi 2
+      | c = multiplying [[Definition:Numerator|top]] and [[Definition:Denominator|bottom]] by [[Definition:Denominator|bottom]]
+}}
+{{eqn | r = \frac {2n \paren {2 n - 1} \paren {2 n - 2} \paren {2 n - 3} \cdots 2 \cdot 1} {\paren {2^n}^2 n^2 \paren {n - 1}^2 \cdots 1^2} \frac \pi 2
+      | c = extracting factor of $\paren {2^n}^2$ from the [[Definition:Denominator|bottom]]
+}}
+{{eqn | r = \frac {\paren {2 n}!} {\paren {2^n}^2 \paren {n!}^2} \frac \pi 2
+      | c = {{Defof|Factorial}}
+}}
+{{eqn | r = \frac {\paren {2 n}!} {\paren {2^n n!}^2} \frac \pi 2
+      | c = rearranging 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Definite Integral from 0 to Half Pi of Odd Power of Sine x}
+Tags: Definite Integrals involving Sine Function, Definite Integral from 0 to Half Pi of Odd Power of Sine x
+
+\begin{theorem}
+:$\displaystyle \int_0^{\frac \pi 2} \sin^{2 n + 1} x \rd x = \dfrac {\paren {2^n n!}^2} {\paren {2 n + 1}!}$
+\end{theorem}
+
+\begin{proof}
+{{improve|This would be more rigorous if implemented as a formal induction proof.}}
+Let $I_n = \displaystyle \int_0^{\frac \pi 2} \sin^n x \rd x$.
+Then:
+{{begin-eqn}}
+{{eqn | l = I_{2 n + 1}
+      | r = \frac {2 n} {2 n + 1} I_{2 n - 1}
+      | c = [[Reduction Formula for Definite Integral of Power of Sine]]
+}}
+{{eqn | r = \frac {2 n \paren {2 n - 2} } {\paren {2 n + 1} \paren {2 n - 1} } I_{2 n - 3}
+      | c = [[Reduction Formula for Definite Integral of Power of Sine]] again
+}}
+{{eqn | r = \frac {2 n \paren {2 n - 2} \dotsm 2} {\paren {2 n + 1} \paren {2 n - 1} \dotsm 3} I_1
+      | c = [[Reduction Formula for Definite Integral of Power of Sine]] until the end
+}}
+{{eqn | r = \frac {2 n \paren {2 n - 2} \dotsm 2} {\paren {2 n + 1} \paren {2 n - 1} \dotsm 3} \int_0^{\pi / 2} \sin x \rd x
+      | c = Definition of $I_n$
+}}
+{{eqn | r = \frac {2 n \paren {2 n - 2} \dotsm 2} {\paren {2 n + 1} \paren {2 n - 1} \dotsm 3} \intlimits {-\cos x} 0 {\pi / 2}
+      | c = [[Primitive of Sine Function]]
+}}
+{{eqn | r = \frac {2 n \paren {2 n - 2} \dotsm 2} {\paren {2 n + 1} \paren {2 n - 1} \dotsm 3} \paren {0 - \paren {-1} }
+      | c = [[Cosine of Right Angle]] and [[Cosine of Zero is One]]
+}}
+{{eqn | r = \frac {2 n \paren {2 n - 2} \dotsm 2} {\paren {2 n + 1} \paren {2 n - 1} \dotsm 3}
+      | c = 
+}}
+{{eqn | r = \frac {\paren {2 n}^2 \paren {2 n - 2}^2 \dotsm 2^2} {\paren {2 n + 1} \paren {2 n} \paren {2 n - 1} \paren {2 n - 2} \paren {2 n - 3} \dotsm 3 \cdot 2}
+      | c = multiplying [[Definition:Numerator|top]] and [[Definition:Denominator|bottom]] by [[Definition:Numerator|top]]
+}}
+{{eqn | r = \frac {\paren {2 n}^2 n^2 \paren {n - 1}^2 \dotsm 1^2} {\paren {2 n + 1} \paren {2 n} \paren {2 n - 1} \paren {2 n - 2} \paren {2 n - 3} \dotsm 3 \cdot 2}
+      | c = extracting factor of $\left({2^n}\right)^2$ from the [[Definition:Numerator|top]]
+}}
+{{eqn | r = \frac {\paren {2^n n!}^2} {\paren {2 n + 1}!}
+      | c = {{Defof|Factorial}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Zero Complement is Not Empty}
+Tags: Naturally Ordered Semigroup
+
+\begin{theorem}
+Let $\struct {S, \circ, \preceq}$ be a [[Definition:Naturally Ordered Semigroup|naturally ordered semigroup]].
+Let $S^*$ be the [[Definition:Zero Complement|zero complement]] of $S$.
+Then $S^*$ is not [[Definition:Empty Set|empty]].
+\end{theorem}
+
+\begin{proof}
+From [[Definition:Naturally Ordered Semigroup Axioms|axiom $(\text {NO} 4)$]], we have:
+:$\exists m, n \in S: m \ne n$
+That is, there are at least two [[Definition:Distinct Elements|distinct]] [[Definition:Element|elements]] in $S$.
+Therefore, there must be at least one [[Definition:Element|element]] in $S^* = S \setminus \set 0$.
+So:
+:$S^* = S \setminus \set 0 \ne \O$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Stirling's Formula/Proof 2/Lemma 5}
+Tags: Stirling's Formula
+
+\begin{theorem}
+:$\dfrac {n!} {n^n \sqrt n e^{-n} } \to \sqrt {2 \pi}$ as $n \to \infty$
+\end{theorem}
+
+\begin{proof}
+By previous work done in [[Stirling's Formula/Proof 2|Stirling's Formula: Proof 2]] it is noted that $\left\langle{\dfrac {n!} {n^n \sqrt n e^{-n} } }\right\rangle_{n \mathop \in \N}$ is a [[Definition:Convergent Sequence (Analysis)|convergent sequence]].
+Let $\dfrac {n!} {n^n \sqrt n e^{-n} } \to C$ as $n \to \infty$.
+Let $I_n = \displaystyle \int_0^{\frac \pi 2} \sin^n x \rd x$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \frac {I_{2 n} } {I_{2 n + 1} }
+      | r = \frac {\left({2n}\right)!} {\left({2^n n!}\right)^2} \frac \pi 2 \cdot \frac {\left({2 n + 1}\right)!} {\left({2^n n!}\right)^2}
+      | c = [[Definite Integral from 0 to Half Pi of Even Power of Sine x|Definite Integral from $0$ to $\dfrac \pi 2$ of Even Power of $\sin x$]] and [[Reduction Formula for Definite Integral of Power of Sine/Corollary 2|Corollary 2]]
+}}
+{{eqn | r = \frac \pi 2 \left({2 n + 1}\right) \frac {\left({\left({2n}\right)!}\right)^2} {\left({2^n n!}\right)^4}
+      | c = extracting $\left({2 n + 1}\right)$ as a factor and rearranging
+}}
+{{eqn | o = \sim
+      | r = \frac \pi 2 \left({2 n + 1}\right) \frac {\left({C \left({2n}\right)^{2n + 1/2} e^{-2 n} }\right)^2} {\left({2^n C n^{n + 1/2} e^{-n} }\right)^4}
+      | c = entering $C$ into top and bottom
+}}
+{{eqn | r = \frac \pi 2 \frac {\left({2 n + 1}\right)} {C^2} \frac {2^{4 n} 2 \left({n^{4 n + 1} }\right) e^{-4 n} } {2^{4 n} \left({n^{4 n + 2} }\right) e^{-4 n} }
+      | c = rearranging
+}}
+{{eqn | r = \frac {\pi \left({2 n + 1}\right)} {n C^2}
+      | c = much simplification
+}}
+{{eqn | r = \frac {2 \pi} {C^2} \text{ as } n \to \infty
+      | c = 
+}}
+{{end-eqn}}
+From [[Stirling's Formula/Proof 2/Lemma 4|Lemma 4]], we have that:
+:$\displaystyle \lim_{n \to \infty} \frac {I_{2 n} } {I_{2 n + 1} } = 1$
+from which
+{{begin-eqn}}
+{{eqn | l = \frac {2 \pi} {C^2}
+      | r = 1
+}}
+{{eqn | ll= \implies
+      | l = C^2
+      | r = 2 \pi
+}}
+{{end-eqn}}
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Limit of Error in Stirling's Formula}
+Tags: Stirling's Formula
+
+\begin{theorem}
+Consider [[Stirling's Formula]]:
+:$n! \sim \sqrt {2 \pi n} \left({\dfrac n e}\right)^n$
+The ratio of $n!$ to its approximation $\sqrt {2 \pi n} \left({\dfrac n e}\right)^n$ is [[Definition:Bounded Real Sequence|bounded]] as follows:
+:$e^{1 / \left({12 n + 1}\right)} \le \dfrac {n!} {\sqrt {2 \pi n} n^n e^{-n} } \le e^{1 / 12 n}$
+\end{theorem}
+
+\begin{proof}
+Let $d_n = \ln n! - \left({n + \dfrac 1 2}\right) \ln n + n$.
+From the argument in [[Stirling's Formula/Proof 2/Lemma 3|Stirling's Formula: Proof 2: Lemma 3]] we have that $\left\langle{d_n - \dfrac 1 {12 n} }\right\rangle$ is an [[Definition:Increasing Real Sequence|increasing sequence]].
+Then:
+{{begin-eqn}}
+{{eqn | l = d_n - d_{n+1}
+      | r = \ln \left({n!}\right) - \left({n + \frac 1 2}\right) \ln n + n
+      | c = 
+}}
+{{eqn | o = 
+      | ro= -
+      | r = \left({\ln \left({\left({n+1}\right)!}\right) - \left({n + 1 + \frac 1 2}\right) \ln \left({n + 1}\right) + n + 1}\right)
+      | c = 
+}}
+{{eqn | r = - \ln \left({n+1}\right) - \left({n + \frac 1 2}\right) \ln n + \left({n + \frac 3 2}\right) \ln \left({n+1}\right) - 1
+      | c = (as $\ln \left({\left({n+1}\right)!}\right) = \ln \left({n+1}\right) + \ln \left({n!}\right)$)
+}}
+{{eqn | r = \left({n + \frac 1 2}\right) \ln \left({\frac {n+1} n}\right) - 1
+      | c =
+}}
+{{eqn | n = 1
+      | r = \frac {2n + 1} 2 \ln \left({\frac {1 + \left({2n + 1}\right)^{-1} } {1 - \left({2n + 1}\right)^{-1} } }\right) - 1
+      | c =
+}}
+{{end-eqn}}
+Let:
+: $f \left({x}\right) := \dfrac 1 {2 x} \ln \left({\dfrac {1 + x} {1 - x} }\right) - 1$
+for $\left\vert{x}\right\vert < 1$.
+By [[Stirling's Formula/Proof 2/Lemma 1|Stirling's Formula: Proof 2: Lemma 1]]:
+{{begin-eqn}}
+{{eqn | l = f \left({x}\right)
+      | r = \sum_{k \mathop = 1}^\infty \frac {x^{2n} } {2n + 1}
+      | c = 
+}}
+{{eqn | r = \frac {x^2} 3 + \frac {x^4} 5 + \frac {x^6} 7 + \cdots
+      | c = 
+}}
+{{eqn | o = >
+      | r = \frac {x^2} 3
+      | c = 
+}}
+{{end-eqn}}
+:$\displaystyle f \left({x}\right) = \sum_{k \mathop = 1}^\infty \frac {x^{2n} } {2n + 1}$
+As $-1 < \dfrac 1 {2n + 1} < 1$ it can be substituted for $x$ from $(1)$:
+{{begin-eqn}}
+{{eqn | l = d_n - d_{n+1}
+      | o = >
+      | r = \frac 1 3 \frac 1 {\left({2n + 1}\right)^2}
+      | c = 
+}}
+{{eqn | r = \frac 1 {12 n^2 + 12 n + 3}
+      | c = 
+}}
+{{end-eqn}}
+Next:
+{{begin-eqn}}
+{{eqn | l = \frac 1 {12 n + 1} - \frac 1 {12 \left({n + 1}\right) + 1}
+      | r = \frac {12 n + 13 - 12 n - 1} {\left({12 n + 1}\right) \left({12 n + 13}\right)}
+      | c = 
+}}
+{{eqn | r = \frac {12} {\left({12 n + 1}\right) \left({12 n + 13}\right)}
+      | c = 
+}}
+{{end-eqn}}
+It follows that:
+{{begin-eqn}}
+{{eqn | o =
+      | r = \left({d_n - \frac 1 {12 n + 1} }\right) - \left({\frac 1 {12 \left({n + 1}\right) + 1} }\right)
+      | c = 
+}}
+{{eqn | o = >
+      | r = \frac {\left({12 n + 1}\right) \left({12 n + 13}\right) - 12 \left({12 n^2 + 12 n + 3}\right)} {3 \left({12 n + 1}\right) \left({12 n + 13}\right) \left({2n + 1}\right)^2}
+      | c = 
+}}
+{{end-eqn}}
+The [[Definition:Numerator|numerator]] equals:
+:$12 n \left({14 - 12}\right) + \left({13 - 36}\right) = 24 n - 23 > 0$
+for $n = 1, 2, 3, \ldots$
+Therefore the [[Definition:Real Sequence|sequence]] $\left\langle{d_n - \dfrac 1 {12 n + 1} }\right\rangle$ is [[Definition:Decreasing Real Sequence|decreasing]].
+From [[Stirling's Formula]], we have that:
+: $\displaystyle \lim_{n \to \infty} d_n = d$
+where $d = \ln \left({\sqrt{2 \pi} }\right)$
+and so:
+:$d_n - \dfrac 1 {12 n} < d < d_n - \dfrac 1 {12 n + 1}$
+for $n = 1, 2, 3, \ldots$
+Taking [[Definition:Real Exponential Function|exponentials]], and again from [[Stirling's Formula]]:
+:$e^{-1/12n} < \dfrac {\left({\sqrt{2n} }\right) n^{n + 1/2} e^{-n} } {n!} < e^{-1/\left({12n + 1}\right)}$
+from whence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Zero Strictly Precedes One}
+Tags: Naturally Ordered Semigroup
+
+\begin{theorem}
+Let $\left({S, \circ, \preceq}\right)$ be a [[Definition:Naturally Ordered Semigroup|naturally ordered semigroup]].
+Let $0$ be the [[Definition:Zero of Naturally Ordered Semigroup|zero]] of $S$.
+Let $1$ be the [[Definition:One of Naturally Ordered Semigroup|one]] of $S$.
+Then:
+:$0 \prec 1$
+\end{theorem}
+
+\begin{proof}
+This follows directly from the [[Definition:Strictly Precede|definition of $\prec$]].
+First note that:
+:$\forall n \in S: 0 \preceq n$
+from the definition of [[Definition:Zero of Naturally Ordered Semigroup|zero]].
+Next, from the definition of [[Definition:One of Naturally Ordered Semigroup|one]]:
+:$0 \ne 1$
+Thus:
+{{begin-eqn}}
+{{eqn | o = 
+      | r = 0 \preceq 1 \land 0 \ne 1
+}}
+{{eqn | o = \implies
+      | r = 0 \prec 1
+      | c = Definition of [[Definition:Strictly Precede|Strictly Precede]]
+}}
+{{end-eqn}}
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Stirling's Formula/Proof 2/Lemma 2}
+Tags: Stirling's Formula
+
+\begin{theorem}
+The [[Definition:Real Sequence|sequence]] $\left \langle {d_n} \right \rangle$ defined as:
+: $\displaystyle d_n = \ln \left({n!}\right) - \left({n + \frac 1 2}\right) \ln n + n$
+is [[Definition:Decreasing Real Sequence|decreasing]].
+\end{theorem}
+
+\begin{proof}
+The proof strategy is to demonstrate that the sign of $d_n - d_{n+1}$ is [[Definition:Positive Real Number|positive]].
+{{begin-eqn}}
+{{eqn | l = d_n - d_{n+1}
+      | r = \ln \left({n!}\right) - \left({n + \frac 1 2}\right) \ln n + n
+      | c = 
+}}
+{{eqn | o = 
+      | ro= -
+      | r = \left({\ln \left({\left({n+1}\right)!}\right) - \left({n + 1 + \frac 1 2}\right) \ln \left({n + 1}\right) + n + 1}\right)
+      | c = 
+}}
+{{eqn | r = - \ln \left({n+1}\right) - \left({n + \frac 1 2}\right) \ln n + \left({n + \frac 3 2}\right) \ln \left({n+1}\right) - 1
+      | c = (as $\ln \left({\left({n+1}\right)!}\right) = \ln \left({n+1}\right) + \ln \left({n!}\right)$)
+}}
+{{eqn | r = \left({n + \frac 1 2}\right) \ln \left({\frac {n+1} n}\right) - 1
+      | c =
+}}
+{{eqn | n = 1
+      | r = \frac {2n + 1} 2 \ln \left({\frac {1 + \left({2n + 1}\right)^{-1} } {1 - \left({2n + 1}\right)^{-1} } }\right) - 1
+      | c =
+}}
+{{end-eqn}}
+Let:
+: $f \left({x}\right) := \dfrac 1 {2 x} \ln \left({\dfrac {1 + x} {1 - x} }\right) - 1$
+for $\left\vert{x}\right\vert < 1$.
+Then from [[Stirling's Formula/Proof 2/Lemma 1|Lemma 1]]:
+: $(2): \quad \displaystyle f \left({x}\right) = \sum_{k \mathop = 1}^\infty \frac {x^{2n} } {2n + 1}$
+Thus $f \left({x}\right) > 0$ for $\left\vert{x}\right\vert < 1$.
+Putting $x = \dfrac 1 {2n + 1}$ it can be seen that $(1)$ is $f \left({\dfrac 1 {2n + 1} }\right)$.
+As $-1 < \dfrac 1 {2n + 1} < 1$ it can be seen that $(2)$ can be applied and so:
+:$\forall n \in \N: d_n - d_{n+1} \ge 0$
+Thus $\left\langle{d_n}\right\rangle$ is a [[Definition:Decreasing Real Sequence|decreasing sequence]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Power Series Expansion for Real Arctangent Function}
+Tags: Examples of Power Series, Arctangent Function
+
+\begin{theorem}
+The [[Definition:Inverse Tangent|arctangent function]] has a [[Definition:Taylor Series|Taylor series expansion]]:
+:$\arctan x = \begin {cases} \displaystyle \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n + 1} } {2 n + 1} & : -1 \le x \le 1
+\\ \displaystyle \frac \pi 2 - \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac 1 {\paren {2 n + 1} x^{2 n + 1} } & : x \ge 1
+\\ \displaystyle -\frac \pi 2 - \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac 1 {\paren {2 n + 1} x^{2 n + 1} } & : x \le -1
+\end {cases}$
+That is:
+:$\arctan x = \begin {cases} \displaystyle x - \frac {x^3} 3 + \frac {x^5} 5 - \frac {x^7} 7 + \cdots & : -1 \le x \le 1
+\\ \displaystyle \frac \pi 2 - \frac 1 x + \frac 1 {3 x^3} - \frac 1 {5 x^5} + \cdots & : x \ge 1
+\\ \displaystyle -\frac \pi 2 - \frac 1 x + \frac 1 {3 x^3} - \frac 1 {5 x^5} + \cdots & : x \le -1
+\end {cases}$
+\end{theorem}
+
+\begin{proof}
+From [[Sum of Infinite Geometric Sequence]]:
+:$(1): \quad \displaystyle \sum_{n \mathop = 0}^\infty \paren {-x^2}^n = \frac 1 {1 + x^2}$
+for $-1 < x < 1$.
+From [[Power Series is Termwise Integrable within Radius of Convergence]], $(1)$ can be [[Definition:Integration|integrated]] term by term:
+{{begin-eqn}}
+{{eqn | l = \int_0^x \frac 1 {1 + t^2} \rd t
+      | r = \sum_{n \mathop = 0}^\infty \int_0^x \paren {-t^2}^n \rd t
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \arctan x 
+      | r = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n + 1} } {2 n + 1}
+      | c = [[Primitive of Reciprocal of x squared plus a squared/Arctangent Form|Primitive of Reciprocal of $\dfrac 1 {1 + t^2}$]], [[Integral of Power]]
+}}
+{{end-eqn}}
+For $-1 \le x \le 1$, the [[Definition:Real Sequence|sequence]] $\sequence {\dfrac {x^{2 n + 1}} {2 n + 1} }$ is [[Definition:Decreasing Real Sequence|decreasing]] and [[Definition:Convergent Sequence (Analysis)|converges]] to zero.
+Therefore the series converges in the given range by the [[Alternating Series Test]].
+{{qed|lemma}}
+Now consider the case $x \ge 1$:
+{{begin-eqn}}
+{{eqn	| l = \arctan x 
+	| r = \frac \pi 2 - \map \arctan {\frac 1 x}
+	| c = [[Sum of Arctangent and Arccotangent]]
+}}
+{{eqn	| r = \frac \pi 2 - \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac 1 {2 n + 1} \paren {\frac 1 x}^{2 n + 1}
+	| c = as $x \ge 1$, $0 < \dfrac 1 x \le 1$, so may apply the above result
+}}
+{{eqn	| r = \frac \pi 2 - \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac 1 {\paren {2 n + 1} x^{2 n + 1} }
+}}
+{{end-eqn}}
+{{qed|lemma}}
+We also have: 
+{{begin-eqn}}
+{{eqn	| l = \map \arctan {-x}
+	| r = -\arctan x
+	| c = [[Arctangent Function is Odd]]
+}}
+{{eqn	| r = -\frac \pi 2 - \sum_{n \mathop = 0}^\infty \paren {-1}^{n + 1} \frac 1 {\paren {2 n + 1} x^{2 n + 1} }
+	| c = per above result
+}}
+{{end-eqn}}
+Substituting $x$ for $-x$ gives us the expansion for $x \le -1$: 
+{{begin-eqn}}
+{{eqn	| l = \arctan x
+	| r = -\frac \pi 2 - \sum_{n \mathop = 0}^\infty \paren {-1}^{n + 1} \frac 1 {\paren {2 n + 1} \paren {-x}^{2 n + 1} }
+}}
+{{eqn	| r = -\frac \pi 2 - \sum_{n \mathop = 0}^\infty \paren {-1}^{n + 1} \frac 1 {-\paren {2 n + 1} x^{2 n + 1} }
+}}
+{{eqn	| r = -\frac \pi 2 - \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac 1 {\paren {2 n + 1} x^{2 n + 1} }
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Ordering in terms of Addition}
+Tags: Naturally Ordered Semigroup
+
+\begin{theorem}
+Let $\left({S, \circ, \preceq}\right)$ be a [[Definition:Naturally Ordered Semigroup|naturally ordered semigroup]].
+Then $\forall m, n \in S$: 
+:$m \preceq n \iff \exists p \in S: m \circ p = n$
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+From [[Definition:Naturally Ordered Semigroup Axioms|axiom $(NO3)$]], we have:
+:$\forall m, n \in S: m \preceq n \implies \exists p \in S: m \circ p = n$
+{{qed|lemma}}
+=== Sufficient Condition ===
+Suppose that $m \circ p = n$.
+{{begin-eqn}}
+{{eqn | l = 0
+      | o = \preceq
+      | r = p
+      | c = Definition of [[Definition:Zero of Naturally Ordered Semigroup|Zero]]
+}}
+{{eqn | ll= \implies
+      | l = m \circ 0
+      | o = \preceq
+      | r = m \circ p
+      | c = $\preceq$ is [[Definition:Relation Compatible with Operation|compatible]] with $\circ$
+}}
+{{eqn | ll= \implies
+      | l = m
+      | o = \preceq
+      | r = n
+      | c = [[Zero is Identity in Naturally Ordered Semigroup]]
+}}
+{{end-eqn}}
+{{qed|lemma}}
+So $\forall m, n \in S$: 
+:$m \preceq n \iff \exists p \in S: m \circ p = n$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Difference in Naturally Ordered Semigroup is Unique}
+Tags: Naturally Ordered Semigroup
+
+\begin{theorem}
+Let $\struct {S, \circ, \preceq}$ be a [[Definition:Naturally Ordered Semigroup|naturally ordered semigroup]].
+Let $n, m \in S$ such that $m \preceq n$.
+Then there exists a [[Definition:Unique|unique]] [[Definition:Difference (Naturally Ordered Semigroup)|difference]] $n - m$ of $m$ and $n$.
+\end{theorem}
+
+\begin{proof}
+Since $m \preceq n$, by [[Definition:Naturally Ordered Semigroup Axioms|axiom $(NO3)$]]:
+:$\exists p \in S: m + p = n$
+Now suppose that $p, q \in S$ are such that:
+:$m + p = m + q = n$
+Then it follows from [[Definition:Naturally Ordered Semigroup Axioms|axiom $(NO2)$]] that:
+:$p = q$
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Beta Function}
+Tags: Beta Function
+
+\begin{theorem}
+{{TFAE|def = Beta Function}}
+For $\map \Re x, \map \Re y > 0$:
+\end{theorem}
+
+\begin{proof}
+=== [[Definition:Beta Function/Definition 1|Definition 1]] is equivalent to [[Definition:Beta Function/Definition 2|Definition 2]] ===
+{{begin-eqn}}
+{{eqn | l = \map \Beta {x, y}
+      | r = \int_0^1 t^{x - 1} \paren {1 - t}^{y - 1} \rd t
+      | c = {{Defof|Beta Function|index = 1}}
+}}
+{{eqn | r = \int_0^{\pi / 2} \paren {\sin \theta}^{2 x - 2} \paren {\cos \theta}^{2 y - 2} 2 \sin \theta \cos \theta \rd \theta
+      | c = [[Integration by Substitution|Substitute]] $t = \sin^2 \theta$
+}}
+{{eqn | r = 2 \int_0^{\pi / 2} \paren {\sin \theta}^{2 x - 1} \paren {\cos \theta}^{2 y - 1} \rd \theta
+      | c = {{Defof|Beta Function|index = 2}}
+}}
+{{end-eqn}}
+{{qed|lemma}}
+=== [[Definition:Beta Function/Definition 2|Definition 2]] is equivalent to [[Definition:Beta Function/Definition 3|Definition 3]] ===
+By definition of [[Definition:Integral Form of Gamma Function|Gamma function]]:
+:$\displaystyle \map \Gamma x \, \map \Gamma y = \int_0^\infty t^{x - 1} e^{-t} \rd t \int_0^\infty s^{y - 1} e^{-s} \rd s$
+Substitute $t = u^2$ and $s = v^2$:
+{{begin-eqn}}
+{{eqn | l = \map \Gamma x \, \map \Gamma y
+      | r = 4 \int_0^\infty u^{2 x - 1} e^{-u^2} \rd u \int_0^\infty v^{2 y - 1} e^{-v^2} \rd v
+      | c =
+}}
+{{eqn | r = 4 \int_0^\infty \int_0^\infty u^{2 x - 1} v^{2 y - 1} e^{-\paren {u^2 + v^2} } \rd u \rd v
+      | c = [[Fubini's Theorem]]
+}}
+{{eqn | r = 4 \int_0^\infty \int_0^{\pi / 2} r^{2 x - 1} \paren {\sin \theta}^{2 x - 1} r^{2 y - 1} \paren {\cos \theta}^{2 y - 1} e^{-r^2} r \rd \theta \rd r
+      | c = Substitute $v = r \cos \theta$ and $u = r \sin \theta$
+}}
+{{eqn | r = 4 \int_0^\infty r^{2 x + 2 y - 2} e^{-r^2} r \rd r \int_0^{\pi / 2} \paren {\sin \theta}^{2 x - 1} \paren {\cos \theta}^{2 y - 1} \rd \theta
+      | c = [[Fubini's Theorem]]
+}}
+{{eqn | r = 2 \int_0^\infty u^{x + y - 1} e^{-u} \rd u \int_0^{\pi / 2} \paren {\sin \theta}^{2 x - 1} \paren {\cos \theta}^{2 y - 1} \rd \theta
+      | c = Substitute $u = r^2$
+}}
+{{eqn | r = 2 \, \map \Gamma {x + y} \int_0^{\pi / 2} \paren {\sin \theta}^{2 x - 1} \paren {\cos \theta}^{2 y - 1} \rd \theta
+      | c = {{Defof|Gamma Function}}
+}}
+{{eqn | ll= \leadsto
+      | o = 
+      | r = \frac {\map \Gamma x \, \map \Gamma y} {\map \Gamma {x + y} }
+      | c = {{Defof|Beta Function|index = 3}}
+}}
+{{eqn | r = 2 \int_0^{\pi / 2} \paren {\sin \theta}^{2 x - 1} \paren {\cos \theta}^{2 y - 1} \rd \theta
+      | c = {{Defof|Beta Function|index = 2}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Ordering of Naturally Ordered Semigroup is Strongly Compatible}
+Tags: Naturally Ordered Semigroup
+
+\begin{theorem}
+:$\forall m, n, p \in S: m \preceq n \iff m \circ p \preceq n \circ p$
+\end{theorem}
+
+\begin{proof}
+The forward implication is immediate from $\preceq$ being [[Definition:Relation Compatible with Operation|compatible]] with $\circ$:
+:$\forall m, n, p \in S: m \preceq n \implies m \circ p \preceq n \circ p$
+Conversely, suppose that $m \circ p \preceq n \circ p$.
+Suppose that $n \prec m$.
+Then as $\preceq$ is [[Definition:Relation Compatible with Operation|compatible]] with $\circ$:
+:$n \circ p \preceq m \circ p$
+Since $\preceq$ is an [[Definition:Ordering|ordering]], this implies:
+:$n \circ p = m \circ p$
+By [[Definition:Naturally Ordered Semigroup Axioms|axiom $(\text {NO} 2)$]], it follows that:
+:$n = m$
+contradicting our assumption that $n \prec m$.
+Hence, since $\preceq$ is a [[Definition:Total Ordering|total ordering]]:
+:$m \preceq n$
+as desired.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Strict Ordering of Naturally Ordered Semigroup is Strongly Compatible}
+Tags: Naturally Ordered Semigroup
+
+\begin{theorem}
+:$\forall m, n, p \in S: m \prec n \iff m \circ p \prec n \circ p$
+\end{theorem}
+
+\begin{proof}
+By [[Definition:Naturally Ordered Semigroup Axioms|axiom $(NO2)$]], all $n \in S$ are [[Definition:Cancellable Element|cancellable]].
+Hence from [[Strict Ordering Preserved under Product with Cancellable Element]]:
+:$\forall m, n, p \in S: m \prec n \implies m \circ p \prec n \circ p$
+By [[Definition:Naturally Ordered Semigroup Axioms|axiom $(NO1)$]], $\preceq$ is a [[Definition:Total Ordering|total ordering]].
+Therefore, the [[Definition:Contrapositive Statement|contrapositive]] of:
+:$\forall m, n, p \in S: m \circ p \prec n \circ p \implies m \prec n$
+is:
+:$\forall m, n, p \in S: m \preceq n \implies m \circ p \preceq n \circ p$
+which we know to be true by virtue of [[Definition:Naturally Ordered Semigroup Axioms|axiom $(NO2)$]].
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sum with One is Immediate Successor in Naturally Ordered Semigroup}
+Tags: Naturally Ordered Semigroup
+
+\begin{theorem}
+Let $\left({S, \circ, \preceq}\right)$ be a [[Definition:Naturally Ordered Semigroup|naturally ordered semigroup]].
+Let $1$ be the [[Definition:One of Naturally Ordered Semigroup|one]] of $S$.
+Let $n \in S$.
+Then $n \circ 1$ is the [[Definition:Immediate Successor Element|immediate successor]] of $n$.
+That is, for all $m \in S$:
+:$n \prec m \iff n \circ 1 \preceq m$
+\end{theorem}
+
+\begin{proof}
+By [[Zero Strictly Precedes One]], $0 \prec 1$, where $0$ is the [[Definition:Zero of Naturally Ordered Semigroup|zero]] of $S$.
+Hence from [[Strict Ordering of Naturally Ordered Semigroup is Strongly Compatible]]:
+:$n \circ 0 \prec n \circ 1$
+and by [[Zero is Identity in Naturally Ordered Semigroup]], $n \circ 0 = n$.
+Now suppose that $n \prec m$.
+Then by [[Definition:Naturally Ordered Semigroup Axioms|axiom $(NO3)$]], there exists $p \in S$ such that:
+:$n \circ p = m$
+Moreover, since $n \ne m$, it follows that $p \ne 0$.
+Hence $0 \prec p$ by definition of [[Definition:Zero of Naturally Ordered Semigroup|zero]].
+Therefore, by definition of [[Definition:One of Naturally Ordered Semigroup|one]]:
+:$1 \preceq p$
+Now by [[Definition:Relation Compatible with Operation|compatibility]] of $\preceq$ with $\circ$:
+:$n \circ 1 \preceq n \circ p = m$
+as desired.
+Conversely, if $n \circ 1 \preceq m$, it is immediate from:
+:$n \prec n \circ 1$
+that $n \prec m$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Riemann Zeta Function at Even Integers}
+Tags: Riemann Zeta Function, Analytic Number Theory, Riemann Zeta Function at Even Integers, Bernoulli Numbers
+
+\begin{theorem}
+The [[Definition:Riemann Zeta Function|Riemann $\zeta$ function]] can be calculated for [[Definition:Even Integer|even integers]] as follows:
+{{begin-eqn}}
+{{eqn | l = \map \zeta {2 n}
+      | r = \paren {-1}^{n + 1} \dfrac {B_{2 n} 2^{2 n - 1} \pi^{2 n} } {\paren {2 n}!}
+      | c = 
+}}
+{{eqn | r = \frac 1 {1^{2 n} } + \frac 1 {2^{2 n} } + \frac 1 {3^{2 n} } + \frac 1 {4^{2 n} } + \cdots
+      | c = 
+}}
+{{end-eqn}}
+where:
+:$B_n$ are the [[Definition:Bernoulli Numbers|Bernoulli numbers]]
+:$n$ is a [[Definition:Positive Integer|positive integer]].
+\end{theorem}
+
+\begin{proof}
+Let $k \in \N$.
+Let $\map S x$ be equal to $x^{2 k}$ on $\closedint {-\pi} \pi$ and be [[Definition:Periodic Real Function|periodic]] with [[Definition:Period of Function|period]] $2 \pi$.
+Let $\displaystyle \map I {2 m, n} = \int_0^\pi x^{2 m} \cos n x \rd x$.
+Let $\map A {2 m, n} = \dfrac {\pi^{2 m - 1} \paren {-1}^n 2 m} {n^2}$.
+Let $\map B {2 m, n} = -\dfrac {2 m \paren {2 m - 1} } {n^2}$.
+By [[Fourier Series for Even Function over Symmetric Range]]:
+{{begin-eqn}}
+{{eqn | l = \map S x
+      | r = \frac {\frac 2 \pi \int_0^\pi \map S x  \rd x} 2 + \sum_{n \mathop = 1}^\infty \cos n x \frac 2 \pi \int_0^\pi \map S x \cos n x \rd x
+}}
+{{eqn | r = \frac {\int_0^\pi x^{2 k} \rd x} \pi + \frac 2 \pi \sum_{n \mathop = 1}^\infty \cos n x \int_0^\pi x^{2 k} \cos n x \rd x
+}}
+{{eqn | r = \frac {\pi^{2 k} } {2 k + 1} + \frac 2 \pi \sum_{n \mathop = 1}^\infty \cos n x \map I {2 k, n}
+}}
+{{eqn | ll= \leadsto
+      | l = \map S \pi
+      | r = \frac {\pi^{2 k} } {2 k + 1} + \frac 2 \pi \sum_{n \mathop = 1}^\infty \cos n \pi \map I {2 k, n}
+}}
+{{eqn | ll= \leadsto
+      | l = \pi^{2 k}
+      | r = \frac {\pi^{2 k} } {2 k + 1} + \frac 2 \pi \sum_{n \mathop = 1}^\infty \paren {-1}^n \map I {2 k, n}
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {2 k \pi^{2 k} } {2 k + 1}
+      | r = \frac 2 \pi \sum_{n \mathop = 1}^\infty \paren {-1}^n \map I {2 k, n}
+}}
+{{end-eqn}}
+We have:
+{{begin-eqn}}
+{{eqn | l = \map I {0, n}
+      | r = \int_0^\pi \cos n x \rd x
+}}
+{{eqn | r = \bigintlimits {\frac {\sin n x} n} 0 \pi
+}}
+{{eqn | r = \frac {\sin n \pi} n - \frac {\sin 0} n
+}}
+{{eqn | r = 0 - 0
+}}
+{{eqn | r = 0 
+}}
+{{eqn | l = \map I {2 k, n}
+      | r = \int_0^\pi x^{2 k} \cos n x \rd x
+}}
+{{eqn | r = \intlimits {\frac {x^{2 k} \sin n x} n} 0 \pi - \frac {2 k} n \int_0^\pi x^{2 k - 1} \cos n x \rd x
+      | c = [[Integration by Parts]]
+}}
+{{eqn | r = 0 - 0 - \frac {2 k} n \int_0^\pi x^{2 k - 1} \sin n x \rd x
+}}
+{{eqn | r = -\frac {2 k} n \int_0^\pi x^{2 k - 1} \sin n x \rd x
+}}
+{{eqn | r = -\frac {2 k} n \paren {\intlimits {\frac {-x^{2 k - 1} \cos n x} n} 0 \pi + \frac {2 k - 1} n \int_0^\pi x^{2 k - 2} \cos n x \rd x}
+      | c = [[Integration by Parts]]
+}}
+{{eqn | r = -\frac {2 k} n \paren {\frac {\pi^{2 k - 1} \paren {-1}^{n + 1} } n + \frac{2 k - 1} n \map I {2 k - 2, n} }
+}}
+{{eqn | r = \frac {\pi^{2 k - 1} \paren {-1}^n 2 k} {n^2} - \frac {2 k \paren {2 k - 1} } {n^2} \map I {2 k - 2, n} 
+}}
+{{eqn | r = \map A {2 k, n} + \map B {2 k, n} \map I {2 k - 2, n}
+}}
+{{eqn | r = \map A {2 k, n} + \map B {2 k, n} \map A {2 k - 2, n} + \map B {2 k, n} \map B {2 k - 2, n} \map I {2 k - 4, n} 
+}}
+{{eqn | r = \sum_{m \mathop = 0}^{k - 1} \map A {2 k - 2 m, n} \prod_{j \mathop = 0}^{m - 1} \map B {2 k - 2 j, n}
+}}
+{{eqn | r = \sum_{m \mathop = 0}^{k - 1} \frac {\pi^{2 k - 2 m - 1} \paren {-1}^n \paren {2 k - 2 m} } {n^2} \prod_{j \mathop = 0}^{m - 1} -\frac {\paren {2 k - 2 j} \paren {2 k - 2 j - 1} } {n^2}
+}}
+{{eqn | r = \sum_{m \mathop = 0}^{k - 1} \frac {\pi^{2 k - 2 m - 1} \paren {-1}^{n + m} \paren {2 k - 2 m} } {n^{2 \paren {m + 1} } } \prod_{j \mathop = 0}^{m - 1} \paren {2 k - 2 j} \paren {2 k - 2 j - 1}
+}}
+{{eqn | r = \sum_{m \mathop = 0}^{k - 1} \frac {\pi^{2 k - 2 m - 1} \paren {-1}^{n + m} \paren {2 k}!} {n^{2 \paren {m + 1} } \paren {2 k - 2 m - 1}!} 
+}}
+{{eqn | r = \sum_{m \mathop = 1}^k \frac {\pi^{2 k - 2m + 1} \paren {-1}^{n + m - 1} \paren {2 k}!} {n^{2 \paren m} \paren {2 k - 2 m + 1}!} 
+}}
+{{end-eqn}}
+Thus:
+{{begin-eqn}}
+{{eqn | l = \frac 2 \pi \sum_{n \mathop = 1}^\infty \paren {-1}^n \map I {2 k, n}
+      | r = \frac 2 \pi \sum_{n \mathop = 1}^\infty \paren {-1}^n \sum_{m \mathop = 1}^k \frac {\pi^{2 k - 2 m + 1} \paren {-1}^{n+ m - 1} \paren {2 k}!} {n^{2 \paren m} \paren {2 k - 2 m + 1}!} 
+}}
+{{eqn | r = 2 \sum_{n \mathop = 1}^\infty \sum_{m \mathop = 1}^k \frac {\pi^{2 k - 2 m} \paren {-1}^{m - 1} \paren {2 k}!} {n^{2 \paren m} \paren {2 k - 2 m + 1}!} 
+}}
+{{eqn | r = 2 \sum_{m \mathop = 1}^k \sum_{n \mathop = 1}^\infty \frac {\pi^{2 k - 2 m} \paren {-1}^{m - 1} \paren {2 k}!} {n^{2 \paren m} \paren {2 k - 2 m + 1}!} 
+}}
+{{eqn | r = 2 \sum_{m \mathop = 1}^k \frac{\pi^{2 k - 2 m} \paren {-1}^{m - 1} \paren {2 k}!} {\paren {2 k - 2 m + 1}!} \sum_{n \mathop = 1}^\infty \frac 1 {n^{2 \paren m} } 
+}}
+{{eqn | r = 2 \sum_{m \mathop = 1}^k \frac{\pi^{2 k - 2 m} \paren {-1}^{m - 1} \paren {2 k}!} {\paren {2 k - 2 m + 1}!} \map \zeta {2 m}
+      | c = {{Defof|Riemann Zeta Function}}
+}}
+{{eqn | ll= \leadsto
+      | l = \frac{ 2 k \pi^{2 k} } {2 k + 1}
+      | r = 2  \sum_{m \mathop = 1}^k \frac{\pi^{2 k - 2 m} \paren {-1}^{m - 1} \paren {2 k}!} {\paren {2 k - 2 m + 1}!} \map \zeta {2 m}
+}}
+{{eqn | ll= \leadsto
+      | l = 2 \paren {2 k}! \paren {-1}^{k - 1} \map \zeta {2 k}
+      | r = \frac{ 2 k \pi^{2 k} } {2 k + 1} - 2  \sum_{m \mathop = 1}^{k - 1} \frac{\pi^{2 k - 2 m} \paren {-1}^{m - 1} \paren {2 k}!} {\paren {2 k - 2 m + 1}!} \map \zeta {2 m}
+}}
+{{eqn | ll= \leadsto
+      | l = \map \zeta {2 k}
+      | r = \frac {\paren {-1}^{k - 1} k \pi^{2 k} } {\paren {2 k + 1}!} -  \sum_{m \mathop = 1}^{k - 1} \frac{\pi^{2 k - 2 m} \paren {-1}^{k - m} } {\paren {2 k - 2 m + 1}!} \map \zeta {2 m}
+}}
+{{end-eqn}}
+From the above: 
+:$\map \zeta 2 = \dfrac {\pi^2} 6$
+which serves as our [[Definition:Basis for the Induction|base case]].
+Assume the [[Definition:Induction Hypothesis|induction hypothesis]] that, for $1 \le m \le k - 1$:
+:$\map \zeta {2 m} = \paren {-1}^{m + 1} \dfrac {B_{2 m} 2^{2 m - 1} \pi^{2 m} } {\paren {2 m}!}$
+Then: 
+{{begin-eqn}}
+{{eqn | l = \map \zeta {2 k}
+      | r = \frac {\paren {-1}^{k - 1} k \pi^{2 k} } {\paren {2 k + 1}!} -  \sum_{m \mathop = 1}^{k - 1} \frac{\pi^{2 k - 2 m} \paren {-1}^{k - m} } {\paren {2 k - 2 m + 1}!} \map \zeta {2 m}
+}}
+{{eqn | r = \frac {\paren {-1}^{k - 1} k \pi^{2 k} } {\paren {2 k + 1}!} -  \sum_{m \mathop = 1}^{k - 1} \frac{\pi^{2 k - 2 m} \paren {-1}^{k - m} } {\paren {2 k - 2 m + 1}!} \paren {-1}^{m + 1} \dfrac {B_{2 m} 2^{2 m - 1} \pi^{2 m} } {\paren {2 m}!}
+}} 
+{{eqn | r = \frac {\paren {-1}^{k - 1} k \pi^{2 k} } {\paren {2 k + 1}!} +  \sum_{m \mathop = 1}^{k - 1} \frac{\pi^{2 k} \paren {-1}^k B_{2 m} 2^{2 m - 1} } {\paren {2 k - 2 m + 1}! \paren {2 m}!}  
+}} 
+{{eqn | r = \frac {\paren {-1}^{k - 1} k \pi^{2 k} } {\paren {2 k + 1}!} +  \frac{\pi^{2 k} \paren {-1}^k} {2 \paren {2 k + 1}!} \sum_{m \mathop = 1}^{k - 1} \binom {2 k + 1}{2m} B_{2 m} 2^{2 m}
+}} 
+{{eqn | r = \frac {\paren {-1}^{k - 1} k \pi^{2 k} } {\paren {2 k + 1}!} +  \frac{\pi^{2 k} \paren {-1}^k} {2 \paren {2 k + 1}!} \paren {2 k - 2^{2 k} B_{2 k} \paren {2 k + 1} }
+      | c = [[Sum of Bernoulli Numbers by Power of Two and Binomial Coefficient]]
+}} 
+{{eqn | r = \frac {\paren {-1}^{k - 1} k \pi^{2 k} } {\paren {2 k + 1}!} +  \frac{k \pi^{2 k} \paren {-1}^k} {\paren {2 k + 1}!} + \frac{B_{2 k} 2^{2 k - 1}\pi^{2 k} \paren {-1}^{k + 1} } {\paren {2 k}!} 
+}} 
+{{eqn | r = \frac {B_{2 k} 2^{2 k - 1}\pi^{2 k} \paren {-1}^{k + 1} } {\paren {2 k}!} 
+}} 
+{{end-eqn}}
+which completes the [[Definition:Induction Step|induction step]].
+Thus by [[Proof by Mathematical Induction]], for all $n \ge 1$:
+:$\map \zeta {2 n} = \paren {-1}^{n + 1} \dfrac {B_{2 n} 2^{2 n - 1} \pi^{2 n} } {\paren {2 n}!}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Tangent Inequality}
+Tags: Tangent Function, Inequalities
+
+\begin{theorem}
+:$x < \tan x$
+for all $x$ in the [[Definition:Open Real Interval|interval]] $\left({0 \,.\,.\, \dfrac {\pi} 2}\right)$.
+\end{theorem}
+
+\begin{proof}
+Let $f \left({x}\right) = \tan x - x$.
+By [[Derivative of Tangent Function]], $f' \left({x}\right) = \sec^2 x - 1$.
+By [[Shape of Secant Function]], $\sec^2 x > 1$ for $x \in \left({0 \,.\,.\, \dfrac {\pi} 2}\right)$.
+Hence $f' \left({x}\right) > 0$.
+From [[Derivative of Monotone Function]], $f \left({x}\right)$ is [[Definition:Strictly Increasing|strictly increasing]] in this [[Definition:Open Real Interval|interval]].
+Since $f \left({0}\right) = 0$, it follows that $f \left({x}\right) > 0$ for all $x$ in $x \in \left({0 \,.\,.\, \dfrac {\pi} 2}\right)$.
+{{qed}}
+[[Category:Tangent Function]]
+[[Category:Inequalities]]
+jyj6srbdnfe2spqq4ob5afa9mk69130
+\end{proof}<|endoftext|>
+\section{Integral Representation of Riemann Zeta Function in terms of Gamma Function}
+Tags: Definite Integrals involving Exponential Function, Riemann Zeta Function, Analytic Number Theory
+
+\begin{theorem}
+For $\Re \paren s > 1$, the [[Definition:Riemann Zeta Function|Riemann Zeta function]] is given by:
+:$\displaystyle \map \zeta s = \frac 1 {\map \Gamma s} \int_0^\infty \frac {t^{s - 1}} {e^t - 1} \rd t$
+where $\Gamma$ is the [[Definition:Gamma Function|Gamma function]].
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int_0^\infty \frac {t^{s - 1} } {e^t - 1} \rd t
+      | r = \int_0^\infty \frac {t^{s - 1} e^{-t} } {1 - e^{-t} } \rd t
+      | c = 
+}}
+{{eqn | r = \int_0^\infty t^{s - 1} e^{-t} \sum_{n \mathop = 0}^\infty e^{-n t} \rd t
+      | c = [[Sum of Infinite Geometric Sequence]]
+}}
+{{eqn | r = \sum_{n \mathop = 1}^\infty \int_0^\infty t^{s - 1} e^{-n t} \rd t
+      | c = interchange of sum and integral is valid by [[Fubini's Theorem]]
+}}
+{{eqn | r = \sum_{n \mathop = 1}^\infty \frac 1 {n^s} \int_0^\infty t^{s - 1} e^{-t} \rd t
+      | c = [[Integration by Substitution|substituting]] $n t \to t$
+}}
+{{eqn | r = \map \zeta s \, \map \Gamma s
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Gamma Function is Unique Extension of Factorial}
+Tags: Gamma Function
+
+\begin{theorem}
+Let $f: \R_{>0} \to \R$ be a [[Definition:Real Function|real function]] which is [[Definition:Positive Real Number|positive]] and [[Definition:Continuous Real Function|continuous]].
+Let $\ln \mathop \circ f$ be [[Definition:Convex Real Function|convex]] on $\R_{>0}$.
+Let $f$ satisfy the conditions:
+:$f \left({x + 1}\right) = \begin{cases}
+1 & : x = 0 \\
+x f \left({x}\right) & : x > 0 \end{cases}$
+Then:
+: $\forall x \in \R_{>0}: f \left({x}\right) = \Gamma \left({x}\right)$
+where $\Gamma \left({x}\right)$ is the [[Definition:Gamma Function|Gamma function]].
+\end{theorem}
+
+\begin{proof}
+From the [[Gamma Function Extends Factorial]]:
+:$f \left({x + 1}\right) = \begin{cases}
+1 & : x = 0 \\
+x f \left({x}\right) & : x > 0 \end{cases}$
+From [[Gamma Function is Continuous on Positive Reals]], $\Gamma$ is [[Definition:Positive Real Number|positive]] and [[Definition:Continuous Real Function|continuous]] on $\R_{>0}$.
+From [[Log of Gamma Function is Convex on Positive Reals]], $\ln \mathop \circ \Gamma$ is [[Definition:Convex Real Function|convex]] on $\R_{>0}$.
+It remains to be shown that $\Gamma$ is the ''only'' [[Definition:Real Function|real function]] which satisfies these conditions.
+Let $s, t \in \R_{>0}$ such that $s \le t \le s + 1$.
+Let $t = \alpha s + \beta \left({s + 1}\right)$ where $\alpha, \beta \in \R_{>0}, \alpha + \beta = 1$.
+Then:
+:$t = \left({\alpha + \beta}\right) s + \beta = s + \beta$
+and so:
+:$\beta = t - s$
+Thus:
+{{begin-eqn}}
+{{eqn | l = \ln \left({f \left({t}\right)}\right)
+      | o = \le
+      | r = \alpha \ln \left({f \left({s}\right)}\right) + \beta \ln \left({f \left({s + 1}\right)}\right)
+      | c = $\ln \mathop \circ f$ is [[Definition:Convex Real Function|convex]] on $\R_{>0}$
+}}
+{{eqn | ll= \leadsto
+      | l = f \left({t}\right)
+      | o = \le
+      | r = \left({f \left({s}\right)}\right)^\alpha \left({f \left({s + 1}\right)}\right)^\beta
+      | c = 
+}}
+{{eqn | r = \left({f \left({s}\right)}\right)^\alpha \left({s f \left({s}\right)}\right)^\beta
+      | c = 
+}}
+{{eqn | r = s^\beta f \left({s}\right)
+      | c = as $\alpha + \beta = 1$
+}}
+{{eqn | n = 1
+      | r = s^{t - s} f \left({s}\right)
+      | c = as $\beta = t - s$
+}}
+{{end-eqn}}
+Because $s \le t \le s + 1$, it follows that $t - 1 \le s \le t$.
+Substituting as appropriate in $(1)$:
+{{begin-eqn}}
+{{eqn | l = f \left({s}\right)
+      | o = \le
+      | r = \left({t - 1}\right)^{s - t + 1} f \left({t - 1}\right)
+      | c = 
+}}
+{{eqn | n = 2
+      | r = \left({t - 1}\right)^{s - t} f \left({t}\right)
+      | c = 
+}}
+{{end-eqn}}
+Combining $(1)$ and $(2)$:
+{{begin-eqn}}
+{{eqn | l = \left({t - 1}\right)^{t - s} f \left({s}\right)
+      | o = \le
+      | r = f \left({t}\right)
+      | c = 
+}}
+{{eqn | n = 3
+      | o = \le
+      | r = s^{t - s} f \left({s}\right)
+      | c = 
+}}
+{{end-eqn}}
+Let $x \in \R$ such that $0 < x \le 1$.
+Let $n \in \N$.
+Let:
+:$s = n + 1$
+:$t = x + n + 1$
+Then substituting in $(3)$:
+{{begin-eqn}}
+{{eqn | l = \left({x + n}\right)^x f \left({n + 1}\right)
+      | o = \le
+      | r = f \left({x + n + 1}\right)
+      | c = 
+}}
+{{eqn | o = \le
+      | r = \left({n + 1}\right)^x f \left({n + 1}\right)
+      | c = 
+}}
+{{end-eqn}}
+Hence:
+{{begin-eqn}}
+{{eqn | l = \left({x + n}\right)^x n!
+      | o = \le
+      | r = \left({x + n}\right) \left({x + n - 1}\right) \cdots x f \left({x}\right)
+      | c = 
+}}
+{{eqn | o = \le
+      | r = \left({n + 1}\right)^x n!
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \left({1 + \frac x n}\right)^x
+      | o = \le
+      | r = \frac {\left({x + n}\right) \left({x + n - 1}\right) \cdots x f \left({x}\right)} {n^x n!}
+      | c = 
+}}
+{{eqn | o = \le
+      | r = \left({1 + \frac 1 n}\right)^x
+      | c = 
+}}
+{{end-eqn}}
+Thus when $0 < x \le 1$:
+: $\displaystyle f \left({x}\right) = \lim_{n \mathop \to \infty} \frac {x \left({x + 1}\right) \cdots \left({x + n}\right)} {n^x n!}$
+which is the [[Definition:Euler Form of Gamma Function|Euler form of the Gamma function]].
+The general case is deduced by the use of:
+:$f \left({x + 1}\right) = x f \left({x}\right)$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Convex Real Function}
+Tags: Convex Real Functions
+
+\begin{theorem}
+Let $f$ be a [[Definition:Real Function|real function]] which is defined on a [[Definition:Real Interval|real interval]] $I$.
+{{TFAE|def = Convex Real Function}}
+\end{theorem}
+
+\begin{proof}
+Let $f$ be [[Definition:Convex Real Function/Definition 1|convex real function on $I$ according to definition 1]].
+That is:
+:$\forall x, y \in I: \forall \alpha, \beta \in \R_{>0}, \alpha + \beta = 1: \map f {\alpha x + \beta y} \le \alpha \map f x + \beta \map f y$
+{{WLOG}}, assume $x \le y$.
+Make the substitutions $x_1 = x, x_2 = \alpha x + \beta y, x_3 = y$.
+As $\alpha + \beta = 1$, we have $x_2 = \alpha x_1 + \paren {1 - \alpha} x_3$.
+Thus:
+:$\alpha = \dfrac {x_3 - x_2} {x_3 - x_1}, \beta = \dfrac {x_2 - x_1} {x_3 - x_1}$
+So:
+{{begin-eqn}}
+{{eqn | n = 1
+      | l = \map f {x_2}
+      | o = \le
+      | r = \frac {x_3 - x_2} {x_3 - x_1} \map f {x_1} + \frac {x_2 - x_1} {x_3 - x_1} \map f {x_3}
+      | c = substituting for $\alpha$ and $\beta$
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \paren {x_3 - x_1} \map f {x_2}
+      | o = \le
+      | r = \paren {x_3 - x_2} \map f {x_1} + \paren {x_2 - x_1} \map f {x_3}
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \paren {x_3 - x_1} \map f {x_2}
+      | o = \le
+      | r = \paren {x_3 - x_1 - x_2 + x_1} \map f {x_1} + \paren {x_2 - x_1} \map f {x_3}
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \paren {x_3 - x_1} \map f {x_2} - \paren {x_3 - x_1} \map f {x_1}
+      | o = \le
+      | r = \paren {x_2 - x_1} \map f {x_3} - \paren {x_2 - x_1} \map f {x_1}
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \frac {\map f {x_2} - \map f {x_1} } {x_2 - x_1}
+      | o = \le
+      | r = \frac {\map f {x_3} - \map f {x_1} } {x_3 - x_1}
+      | c = 
+}}
+{{end-eqn}}
+Again from $(1)$:
+{{begin-eqn}}
+{{eqn | n = 1
+      | l = \map f {x_2}
+      | o = \le
+      | r = \frac {x_3 - x_2} {x_3 - x_1} \map f {x_1} + \frac {x_2 - x_1} {x_3 - x_1} \map f {x_3}
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \paren {x_3 - x_1} \map f {x_2}
+      | o = \le
+      | r = \paren {x_3 - x_2} \map f {x_1} + \paren {x_2 - x_1} \map f {x_3}
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \paren {x_3 - x_1} \map f {x_2}
+      | o = \le
+      | r = \paren {x_3 - x_1 - x_3 + x_2} \map f {x_3} + \paren {x_3 - x_2} \map f {x_1}
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \paren {x_3 - x_2} \map f {x_3} - \paren {x_3 - x_2} \map f {x_1}
+      | o = \le
+      | r = \paren {x_3 - x_1} \map f {x_3} - \paren {x_3 - x_1} \map f {x_2}
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \frac {\map f {x_3} - \map f {x_1} } {x_3 - x_1}
+      | o = \le
+      | r = \frac {\map f {x_3} - \map f {x_2} } {x_3 - x_2}
+      | c = 
+}}
+{{end-eqn}}
+Thus:
+:$\dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1} \le \dfrac {\map f {x_3} - \map f {x_1} } {x_3 - x_1} \le \dfrac {\map f {x_3} - \map f {x_2} } {x_3 - x_2}$
+So:
+:$\dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1} \le \dfrac {\map f {x_3} - \map f {x_1} } {x_3 - x_1}$
+demonstrating that $f$ is [[Definition:Convex Real Function/Definition 3|convex on $I$ according to definition 3]], and:
+:$\dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1} \le \dfrac {\map f {x_3} - \map f {x_2} } {x_3 - x_2}$
+demonstrating that $f$ is [[Definition:Convex Real Function/Definition 2|convex on $I$ according to definition 2]].
+As each step is an equivalence, the argument reverses throughout.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Concave Real Function}
+Tags: Concave Real Functions
+
+\begin{theorem}
+Let $f$ be a [[Definition:Real Function|real function]] which is defined on a [[Definition:Real Interval|real interval]] $I$.
+{{TFAE|def = Concave Real Function}}
+\end{theorem}
+
+\begin{proof}
+Let $f$ be [[Definition:Concave Real Function/Definition 1|concave real function on $I$ according to definition 1]].
+That is:
+:$\forall x, y \in I: \forall \alpha, \beta \in \R_{>0}, \alpha + \beta = 1: \map f {\alpha x + \beta y} \ge \alpha \map f x + \beta \map f y$
+Make the substitutions $x_1 = x, x_2 = \alpha x + \beta y, x_3 = y$.
+As $\alpha + \beta = 1$, we have $x_2 = \alpha x_1 + \paren {1 - \alpha} x_3$.
+Thus:
+:$\alpha = \dfrac {x_3 - x_2} {x_3 - x_1}, \beta = \dfrac {x_2 - x_1} {x_3 - x_1}$
+So:
+{{begin-eqn}}
+{{eqn | n = 1
+      | l = \map f {x_2}
+      | o = \ge
+      | r = \frac {x_3 - x_2} {x_3 - x_1} \map f {x_1} + \frac {x_2 - x_1} {x_3 - x_1} \map f {x_3}
+      | c = substituting for $\alpha$ and $\beta$
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \paren {x_3 - x_1} \map f {x_2}
+      | o = \ge
+      | r = \paren {x_3 - x_2} \map f {x_1} + \paren {x_2 - x_1} \map f {x_3}
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \paren {x_3 - x_1} \map f {x_2}
+      | o = \ge
+      | r = \paren {x_3 - x_1 - x_2 + x_1} \map f {x_1} + \paren {x_2 - x_1} \map f {x_3}
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \paren {x_3 - x_1} \map f {x_2} - \paren {x_3 - x_1} \map f {x_1}
+      | o = \ge
+      | r = \paren {x_2 - x_1} \map f {x_3} - \paren {x_2 - x_1} \map f {x_1}
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \frac {\map f {x_2} - \map f {x_1} } {x_2 - x_1}
+      | o = \ge
+      | r = \frac {\map f {x_3} - \map f {x_1} } {x_3 - x_1}
+      | c = 
+}}
+{{end-eqn}}
+Again from $(1)$:
+{{begin-eqn}}
+{{eqn | n = 1
+      | l = \map f {x_2}
+      | o = \ge
+      | r = \frac {x_3 - x_2} {x_3 - x_1} \map f {x_1} + \frac {x_2 - x_1} {x_3 - x_1} \map f {x_3}
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \paren {x_3 - x_1} \map f {x_2}
+      | o = \ge
+      | r = \paren {x_3 - x_2} \map f {x_1} + \paren {x_2 - x_1} \map f {x_3}
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \paren {x_3 - x_1} \map f {x_2}
+      | o = \ge
+      | r = \paren {x_3 - x_1 - x_3 + x_2} \map f {x_3} + \paren {x_3 - x_2} \map f {x_1}
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \paren {x_3 - x_2} \map f {x_3} - \paren {x_3 - x_2} \map f {x_1}
+      | o = \ge
+      | r = \paren {x_3 - x_1} \map f {x_3} - \paren {x_3 - x_1} \map f {x_2}
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \frac {\map f {x_3} - \map f {x_1} } {x_3 - x_1}
+      | o = \ge
+      | r = \frac {\map f {x_3} - \map f {x_2} } {x_3 - x_2}
+      | c = 
+}}
+{{end-eqn}}
+Thus:
+:$\dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1} \ge \dfrac {\map f {x_3} - \map f {x_1} } {x_3 - x_1} \ge \dfrac {\map f {x_3} - \map f {x_2} } {x_3 - x_2}$
+So:
+:$\dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1} \ge \dfrac {\map f {x_3} - \map f {x_1} } {x_3 - x_1}$
+demonstrating that $f$ is [[Definition:Concave Real Function/Definition 3|concave on $I$ according to definition 3]], and:
+:$\dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1} \ge \dfrac {\map f {x_3} - \map f {x_2} } {x_3 - x_2}$
+demonstrating that $f$ is [[Definition:Concave Real Function/Definition 2|concave on $I$ according to definition 2]].
+As each step is an equivalence, the argument reverses throughout.
+{{qed}}
+[[Category:Concave Real Functions]]
+1w5eny3dousu088aarfj8vxup3b675l
+\end{proof}<|endoftext|>
+\section{Power Series Expansion for Real Arcsine Function}
+Tags: Examples of Power Series, Arcsine Function
+
+\begin{theorem}
+The [[Definition:Real Arcsine|(real) arcsine]] function has a [[Definition:Taylor Series|Taylor series expansion]]:
+{{begin-eqn}}
+{{eqn | l = \arcsin x
+      | r = \sum_{n \mathop = 0}^\infty \frac {\paren {2 n}!} {2^{2 n} \paren {n!}^2} \frac {x^{2 n + 1} } {2 n + 1}
+      | c = 
+}}
+{{eqn | r = x + \frac 1 2 \frac {x^3} 3 + \frac {1 \times 3} {2 \times 4} \frac {x^5} 5 + \frac {1 \times 3 \times 5} {2 \times 4 \times 6} \frac {x^7} 7 + \cdots
+      | c = 
+}}
+{{end-eqn}}
+which [[Definition:Convergent Series|converges]] for $-1 \le x \le 1$.
+\end{theorem}
+
+\begin{proof}
+From the [[General Binomial Theorem]]:
+{{begin-eqn}}
+{{eqn | l = \paren {1 - x^2}^{-1/2} 
+      | r = 1 + \frac 1 2 x^2 + \frac {1 \times 3} {2 \times 4} x^4 + \frac {1 \times 3 \times 5} {2 \times 4 \times 6} x^6 + \cdots
+      | c =
+}}      
+{{eqn | n = 1
+      | r = \sum_{n \mathop = 0}^\infty \frac {\paren {2 n}!} {2^{2 n} \paren {n!}^2} x^{2 n}
+      | c = 
+}}
+{{end-eqn}}
+for $-1 < x < 1$.
+From [[Power Series is Termwise Integrable within Radius of Convergence]], $(1)$ can be [[Definition:Integration|integrated]] term by term:
+{{begin-eqn}}
+{{eqn | l = \int_0^x \frac 1 {\sqrt{1 - t^2} } \rd t
+      | r = \sum_{n \mathop = 0}^\infty \int_0^x \frac {\paren {2 n}!} {2^{2 n} \paren {n!}^2} t^{2 n} \rd t
+      | c =
+}}
+{{eqn | ll= \leadsto
+      | l = \arcsin x
+      | r = \sum_{n \mathop = 0}^\infty \frac {\paren {2 n}!} {2^{2 n} \paren {n!}^2} \frac {x^{2 n + 1} } {2 n + 1}
+      | c = [[Derivative of Arcsine Function]]
+}}
+{{end-eqn}}
+We will now prove that the series [[Definition:Convergent Series|converges]] for $-1 \le x \le 1$.
+By [[Stirling's Formula]]:
+{{begin-eqn}}
+{{eqn | l = \frac {\paren {2 n}!} {2^{2 n} \paren {n!}^2} \frac {x^{2 n + 1} } {2 n + 1} 
+      | o = \sim
+      | r = \frac {\paren {2 n}^{2 n} e^{-2 n} \sqrt {4 \pi n} } {2^{2 n} n^{2 n} e^{-2 n} 2 \pi n} \frac {x^{2 n + 1} } {2 n + 1}
+      | c =
+}}
+{{eqn | r = \frac 1 {\sqrt {\pi n} } \frac {x^{2 n + 1} } {2 n + 1}
+      | c =
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = \size {\frac 1 {\sqrt {\pi n} } \frac {x^{2 n + 1} } {2 n + 1} }
+      | o = <
+      | r = \size {\frac {x^{2 n + 1} } {n^{3/2} } }
+      | c = 
+}}
+{{eqn | o = \le
+      | r = \frac 1 {n^{3/2} }
+      | c = 
+}}
+{{end-eqn}}
+Hence by [[Convergence of P-Series]]:
+:$\displaystyle \sum_{n \mathop = 1}^\infty \frac 1 {n^{3/2} }$
+is [[Definition:Convergent Series|convergent]].
+So by the [[Comparison Test]], the [[Definition:Taylor Series|Taylor series]] is [[Definition:Convergent Series|convergent]] for $-1 \le x \le 1$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Power Series Expansion for Real Arccosine Function}
+Tags: Examples of Power Series, Arccosine Function
+
+\begin{theorem}
+The [[Definition:Inverse Cosine|arccosine]] function has a [[Definition:Taylor Series|Taylor Series]] expansion:
+{{begin-eqn}}
+{{eqn | l = \arccos x
+      | r = \frac \pi 2 - \sum_{n \mathop = 0}^\infty \frac {\paren {2 n}!} {2^{2 n} \paren {n!}^2} \frac {x^{2 n + 1} } {2 n + 1}
+      | c = 
+}}
+{{eqn | r = \frac \pi 2 - \paren {x + \frac 1 2 \frac {x^3} 3 + \frac {1 \times 3} {2 \times 4} \frac {x^5} 5 + \frac {1 \times 3 \times 5} {2 \times 4 \times 6} \frac {x^7} 7 + \cdots}
+      | c = 
+}}
+{{end-eqn}}
+which [[Definition:Convergent Series|converges]] for $-1 \le x \le 1$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \arccos x 
+      | r = \frac {\pi} 2 - \arcsin x
+      | c = [[Sum of Arcsine and Arccosine]]
+}}
+{{eqn | r = \frac {\pi} 2 - \sum_{n \mathop = 0}^\infty \frac {\paren {2 n}!} {2^{2 n} \paren {n!}^2} \frac {x^{2 n + 1} } {2 n + 1}
+      | c = [[Power Series Expansion for Real Arcsine Function]]
+}}
+{{end-eqn}}
+It follows from [[Power Series Expansion for Real Arcsine Function]] that the series is convergent for $-1 \le x \le 1$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Real Function is Concave iff its Negative is Convex}
+Tags: Convex Real Functions, Concave Real Functions
+
+\begin{theorem}
+Let $f$ be a [[Definition:Real Function|real function]].
+Let $I \subseteq \R$ be an [[Definition:Real Interval|interval of $\R$]].
+Then $f$ is [[Definition:Concave Real Function|concave on $I$]] {{iff}} $-f$ is [[Definition:Convex Real Function|convex]] on $\R$.
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Let $f$ be [[Definition:Concave Real Function|concave on $I$]].
+Let $\alpha, \beta \in \R_{>0}$ such that $\alpha + \beta = 1$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \map f {\alpha x + \beta y}
+      | o = \ge
+      | r = \alpha \map f x + \beta \map f y
+      | c = {{Defof|Concave Real Function}}
+}}
+{{eqn | ll= \leadsto
+      | l = -\map f {\alpha x + \beta y}
+      | o = \le
+      | r = -\paren {\alpha \map f x + \beta \map f y}
+      | c = 
+}}
+{{eqn | r = \alpha \paren {-\map f x} + \beta \paren {-\map f y}
+      | c = 
+}}
+{{end-eqn}}
+and so $-f$ is [[Definition:Convex Real Function|convex]] by definition.
+{{qed|lemma}}
+=== Sufficient Condition ===
+Let $-f$ be [[Definition:Convex Real Function|convex on $I$]].
+Let $\alpha, \beta \in \R_{>0}$ such that $\alpha + \beta = 1$.
+Then:
+{{begin-eqn}}
+{{eqn | l = -\map f {\alpha x + \beta y}
+      | o = \le
+      | r = \alpha \paren {-\map f x} + \beta \paren {-\map f y}
+      | c = {{Defof|Convex Real Function}}
+}}
+{{eqn | ll= \leadsto
+      | l = \map f {\alpha x + \beta y}
+      | o = \ge
+      | r = - \paren {\alpha \paren {-\map f x} + \beta \paren {-\map f y} }
+      | c = 
+}}
+{{eqn | r = \alpha \map f x + \beta \map f y
+      | c = 
+}}
+{{end-eqn}}
+and so $f$ is [[Definition:Concave Real Function|concave]] by definition.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Strictly Concave Real Function}
+Tags: Concave Real Functions
+
+\begin{theorem}
+Let $f$ be a [[Definition:Real Function|real function]] which is defined on a [[Definition:Real Interval|real interval]] $I$.
+{{TFAE|def = Strictly Concave Real Function}}
+\end{theorem}
+
+\begin{proof}
+Let $f$ be [[Definition:Concave Real Function/Definition 1/Strictly|strictly concave real function on $I$ according to definition 1]].
+That is:
+:$\forall x, y \in I: \forall \alpha, \beta \in \R_{>0}, \alpha + \beta = 1: f \left({\alpha x + \beta y}\right) > \alpha f \left({x}\right) + \beta f \left({y}\right)$
+Make the substitutions $x_1 = x, x_2 = \alpha x + \beta y, x_3 = y$.
+As $\alpha + \beta = 1$, we have $x_2 = \alpha x_1 + \left({1 - \alpha}\right) x_3$.
+Thus:
+:$\alpha = \dfrac {x_3 - x_2} {x_3 - x_1}, \beta = \dfrac {x_2 - x_1} {x_3 - x_1}$
+So:
+{{begin-eqn}}
+{{eqn | n = 1
+      | l = f \left({x_2}\right)
+      | o = >
+      | r = \frac {x_3 - x_2} {x_3 - x_1} f \left({x_1}\right) + \frac {x_2 - x_1} {x_3 - x_1} f \left({x_3}\right)
+      | c = substituting for $\alpha$ and $\beta$
+}}
+{{eqn | ll= \iff
+      | l = \left({x_3 - x_1}\right) f \left({x_2}\right)
+      | o = >
+      | r = \left({x_3 - x_2}\right) f \left({x_1}\right) + \left({x_2 - x_1}\right) f \left({x_3}\right)
+      | c = 
+}}
+{{eqn | ll= \iff
+      | l = \left({x_3 - x_1}\right) f \left({x_2}\right)
+      | o = >
+      | r = \left({x_3 - x_1 - x_2 + x_1}\right) f \left({x_1}\right) + \left({x_2 - x_1}\right) f \left({x_3}\right)
+      | c = 
+}}
+{{eqn | ll= \iff
+      | l = \left({x_3 - x_1}\right) f \left({x_2}\right) - \left({x_3 - x_1}\right) f \left({x_1}\right)
+      | o = >
+      | r = \left({x_2 - x_1}\right) f \left({x_3}\right) - \left({x_2 - x_1}\right) f \left({x_1}\right)
+      | c = 
+}}
+{{eqn | ll= \iff
+      | l = \frac {f \left({x_2}\right) - f \left({x_1}\right)} {x_2 - x_1}
+      | o = >
+      | r = \frac {f \left({x_3}\right) - f \left({x_1}\right)} {x_3 - x_1}
+      | c = 
+}}
+{{end-eqn}}
+Again from $(1)$:
+{{begin-eqn}}
+{{eqn | n = 1
+      | l = f \left({x_2}\right)
+      | o = >
+      | r = \frac {x_3 - x_2} {x_3 - x_1} f \left({x_1}\right) + \frac {x_2 - x_1} {x_3 - x_1} f \left({x_3}\right)
+      | c = 
+}}
+{{eqn | ll= \iff
+      | l = \left({x_3 - x_1}\right) f \left({x_2}\right)
+      | o = >
+      | r = \left({x_3 - x_2}\right) f \left({x_1}\right) + \left({x_2 - x_1}\right) f \left({x_3}\right)
+      | c = 
+}}
+{{eqn | ll= \iff
+      | l = \left({x_3 - x_1}\right) f \left({x_2}\right)
+      | o = >
+      | r = \left({x_3 - x_1 - x_3 + x_2}\right) f \left({x_3}\right) + \left({x_3 - x_2}\right) f \left({x_1}\right)
+      | c = 
+}}
+{{eqn | ll= \iff
+      | l = \left({x_3 - x_2}\right) f \left({x_3}\right) - \left({x_3 - x_2}\right) f \left({x_1}\right)
+      | o = >
+      | r = \left({x_3 - x_1}\right) f \left({x_3}\right) - \left({x_3 - x_1}\right) f \left({x_2}\right)
+      | c = 
+}}
+{{eqn | ll= \iff
+      | l = \frac {f \left({x_3}\right) - f \left({x_1}\right)} {x_3 - x_1}
+      | o = >
+      | r = \frac {f \left({x_3}\right) - f \left({x_2}\right)} {x_3 - x_2}
+      | c = 
+}}
+{{end-eqn}}
+Thus:
+:$\dfrac {f \left({x_2}\right) - f \left({x_1}\right)} {x_2 - x_1} > \dfrac {f \left({x_3}\right) - f \left({x_1}\right)} {x_3 - x_1} > \dfrac {f \left({x_3}\right) - f \left({x_2}\right)} {x_3 - x_2}$
+So:
+:$\dfrac {f \left({x_2}\right) - f \left({x_1}\right)} {x_2 - x_1} > \dfrac {f \left({x_3}\right) - f \left({x_1}\right)} {x_3 - x_1}$
+demonstrating that $f$ is [[Definition:Concave Real Function/Definition 3/Strictly|strictly concave on $I$ according to definition 3]], and:
+:$\dfrac {f \left({x_2}\right) - f \left({x_1}\right)} {x_2 - x_1} > \dfrac {f \left({x_3}\right) - f \left({x_2}\right)} {x_3 - x_2}$
+demonstrating that $f$ is [[Definition:Concave Real Function/Definition 2/Strictly|strictly concave on $I$ according to definition 2]].
+As each step is an equivalence, the argument reverses throughout.
+{{qed}}
+[[Category:Concave Real Functions]]
+ilxskpfcipiah0tt4dl464e0w4e2lrp
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Strictly Convex Real Function}
+Tags: Convex Real Functions
+
+\begin{theorem}
+Let $f$ be a [[Definition:Real Function|real function]] which is defined on a [[Definition:Real Interval|real interval]] $I$.
+{{TFAE|def = Strictly Convex Real Function}}
+\end{theorem}
+
+\begin{proof}
+Let $f$ be [[Definition:Convex Real Function/Definition 1/Strictly|strictly convex real function on $I$ according to definition 1]].
+That is:
+:$\forall x, y \in I: \forall \alpha, \beta \in \R_{>0}, \alpha + \beta = 1: \map f {\alpha x + \beta y} < \alpha \, \map f x + \beta \, \map f y$
+Make the substitutions $x_1 = x, x_2 = \alpha x + \beta y, x_3 = y$.
+As $\alpha + \beta = 1$, we have:
+:$x_2 = \alpha x_1 + \paren {1 - \alpha} x_3$
+Thus:
+:$\alpha = \dfrac {x_3 - x_2} {x_3 - x_1}$
+:$\beta = \dfrac {x_2 - x_1} {x_3 - x_1}$
+So:
+{{begin-eqn}}
+{{eqn | n = 1
+      | l = \map f {x_2}
+      | o = <
+      | r = \frac {x_3 - x_2} {x_3 - x_1} \, \map f {x_1} + \frac {x_2 - x_1} {x_3 - x_1} \map f {x_3}
+      | c = substituting for $\alpha$ and $\beta$
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \paren {x_3 - x_1} \map f {x_2}
+      | o = <
+      | r = \paren {x_3 - x_2} \map f {x_1} + \paren {x_2 - x_1} \map f {x_3}
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \paren {x_3 - x_1} \map f {x_2}
+      | o = <
+      | r = \paren {x_3 - x_1 - x_2 + x_1} \map f {x_1} + \paren {x_2 - x_1} \map f {x_3}
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \paren {x_3 - x_1} \map f {x_2} - \paren {x_3 - x_1} \map f {x_1}
+      | o = <
+      | r = \paren {x_2 - x_1} \map f {x_3} - \paren {x_2 - x_1} \map f {x_1}
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \frac {\map f {x_2} - \map f {x_1} } {x_2 - x_1}
+      | o = <
+      | r = \frac {\map f {x_3} - \map f {x_1} } {x_3 - x_1}
+      | c = 
+}}
+{{end-eqn}}
+Again from $(1)$:
+{{begin-eqn}}
+{{eqn | n = 1
+      | l = \map f {x_2}
+      | o = <
+      | r = \frac {x_3 - x_2} {x_3 - x_1} \map f {x_1} + \frac {x_2 - x_1} {x_3 - x_1} \map f {x_3}
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \paren {x_3 - x_1} \map f {x_2}
+      | o = <
+      | r = \paren {x_3 - x_2} \map f {x_1} + \paren {x_2 - x_1} \map f {x_3}
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \paren {x_3 - x_1} \map f {x_2}
+      | o = <
+      | r = \paren {x_3 - x_1 - x_3 + x_2} \map f {x_3} + \paren {x_3 - x_2} \map f {x_1}
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \paren {x_3 - x_2} \map f {x_3} - \paren {x_3 - x_2} \map f {x_1}
+      | o = <
+      | r = \paren {x_3 - x_1} \map f {x_3} - \paren {x_3 - x_1} \map f {x_2}
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \frac {\map f {x_3} - \map f {x_1} } {x_3 - x_1}
+      | o = <
+      | r = \frac {\map f {x_3} - \map f {x_2} } {x_3 - x_2}
+      | c = 
+}}
+{{end-eqn}}
+Thus:
+:$\dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1} < \dfrac {\map f {x_3} - \map f {x_1} } {x_3 - x_1} < \dfrac {\map f {x_3} - \map f {x_2} } {x_3 - x_2}$
+So:
+:$\dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1} < \dfrac {\map f {x_3} - \map f {x_1} } {x_3 - x_1}$
+demonstrating that $f$ is [[Definition:Convex Real Function/Definition 3/Strictly|strictly convex on $I$ according to definition 3]], and:
+:$\dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1} < \dfrac {\map f {x_3} - \map f {x_2} } {x_3 - x_2}$
+demonstrating that $f$ is [[Definition:Convex Real Function/Definition 2/Strictly|strictly convex on $I$ according to definition 2]].
+As each step is an equivalence, the argument reverses throughout.
+{{qed}}
+[[Category:Convex Real Functions]]
+iv7r4z0j5dr5r9p1shuq2fus954kxnz
+\end{proof}<|endoftext|>
+\section{Concave Real Function is Left-Hand and Right-Hand Differentiable}
+Tags: Concave Real Functions
+
+\begin{theorem}
+Let $f$ be a [[Definition:Real Function|real function]] which is [[Definition:Concave Real Function|concave]] on the [[Definition:Open Real Interval|open interval]] $\left({a \,.\,.\, b}\right)$.
+Then the [[Definition:Left-Hand Derivative|left-hand derivative]] $f'_- \left({x}\right)$ and [[Definition:Right-Hand Derivative|right-hand derivative]] $f'_+ \left({x}\right)$ both exist for all $x \in \left({a \,.\,.\, b}\right)$.
+\end{theorem}
+
+\begin{proof}
+Let $f$ be [[Definition:Concave Real Function|concave]] on $\left({a \,.\,.\, b}\right)$.
+Then by definition of [[Definition:Concave Real Function|concavity]]:
+:$\forall x_1, x_2, x_3 \in \left({a \,.\,.\, b}\right): x_1 < x_2 < x_3: \dfrac {f \left({x_2}\right) - f \left({x_1}\right)} {x_2 - x_1} \ge \dfrac {f \left({x_3}\right) - f \left({x_1}\right)} {x_3 - x_1}$
+Let $0 < h_1 < h_2$.
+Substitute $x_1 = x$, $x_2 = x + h_1$, $x_3 = x + h_2$. Then:
+:$\dfrac {f \left({x + h_1}\right) - f \left({x}\right)} {h_1} \ge \dfrac {f \left({x + h_2}\right) - f \left({x}\right)} {h_2}$
+Hence the function $F \left({h}\right) = \dfrac {f \left({x + h}\right) - f \left({x}\right)} h$ [[Definition:Decreasing Real Function|decreases]] in some $\left({0 \,.\,.\, \delta}\right)$.
+Thus from [[Limit of Decreasing Function]] it follows that both $\displaystyle \lim_{h \to 0^+} F \left({h}\right) = f'_+ \left({x}\right)$ and $\displaystyle \lim_{h \to 0^-} F \left({h}\right) = f'_-\left({x}\right)$ exist.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Concave Real Function is Continuous}
+Tags: Concave Real Functions, Continuous Functions
+
+\begin{theorem}
+Let $f$ be a [[Definition:Real Function|real function]] which is [[Definition:Concave Real Function|concave]] on the [[Definition:Open Real Interval|open interval]] $\left({a \,.\,.\, b}\right)$.
+Then $f$ is [[Definition:Continuous on Interval|continuous]] on $\left({a \,.\,.\, b}\right)$.
+\end{theorem}
+
+\begin{proof}
+From [[Concave Real Function is Left-Hand and Right-Hand Differentiable]], we have:
+:$\displaystyle \lim_{h \to 0^-} f \left({x + h}\right) - f \left({x}\right) = \left({\lim_{h \to 0^-} \frac {f \left({x + h}\right) - f \left({x}\right)} h}\right) \left({\lim_{h \to 0^-} h}\right) = 0$
+and similarly:
+:$\displaystyle \lim_{h \to 0^+} f \left({x + h}\right) - f \left({x}\right) = \left({\lim_{h \to 0^+} \frac {f \left({x + h}\right) - f \left({x}\right)} h}\right) \left({\lim_{h \to 0^+} h}\right) = 0$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Absolute Value Function is Convex}
+Tags: Absolute Value Function, Convex Real Functions, Absolute Value Function is Convex
+
+\begin{theorem}
+Let $f: \R \to \R$ be the [[Definition:Absolute Value|absolute value function]] on the [[Definition:Real Number|real numbers]].
+Then $f$ is [[Definition:Convex Real Function|convex]].
+\end{theorem}
+
+\begin{proof}
+Let $x_1, x_2, x_3 \in \R$ such that $x_1 < x_2 < x_3$.
+Consider the expressions:
+:$\dfrac {f \left({x_2}\right) - f \left({x_1}\right)} {x_2 - x_1}$
+:$\dfrac {f \left({x_3}\right) - f \left({x_2}\right)} {x_3 - x_2}$
+The following cases are investigated:
+:$(1): \quad x_1, x_2, x_3 < 0$:
+Then:
+{{begin-eqn}}
+{{eqn | l = \frac {f \left({x_2}\right) - f \left({x_1}\right)} {x_2 - x_1}
+      | r = \frac {-\left({x_2 - x_1}\right)} {x_2 - x_1}
+      | c = {{Defof|Absolute Value}}
+}}
+{{eqn | r = -1
+      | c = 
+}}
+{{eqn | l = \frac {f \left({x_3}\right) - f \left({x_2}\right)} {x_3 - x_2}
+      | r = \frac {-\left({x_3 - x_2}\right)} {x_3 - x_2}
+      | c = {{Defof|Absolute Value}}
+}}
+{{eqn | r = -1
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \frac {f \left({x_2}\right) - f \left({x_1}\right)} {x_2 - x_1}
+      | o = \le
+      | r = \frac {f \left({x_3}\right) - f \left({x_2}\right)} {x_3 - x_2}
+      | c = {{Defof|Convex Real Function}}
+}}
+{{end-eqn}}
+:$(2): \quad x_1, x_2, x_3 > 0$:
+Then:
+{{begin-eqn}}
+{{eqn | l = \frac {f \left({x_2}\right) - f \left({x_1}\right)} {x_2 - x_1}
+      | r = \frac {x_2 - x_1} {x_2 - x_1}
+      | c = {{Defof|Absolute Value}}
+}}
+{{eqn | r = 1
+      | c = 
+}}
+{{eqn | l = \frac {f \left({x_3}\right) - f \left({x_2}\right)} {x_3 - x_2}
+      | r = \frac {x_3 - x_2} {x_3 - x_2}
+      | c = {{Defof|Absolute Value}}
+}}
+{{eqn | r = 1
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \frac {f \left({x_2}\right) - f \left({x_1}\right)} {x_2 - x_1}
+      | o = \le
+      | r = \frac {f \left({x_3}\right) - f \left({x_2}\right)} {x_3 - x_2}
+      | c = {{Defof|Convex Real Function}}
+}}
+{{end-eqn}}
+:$(3): \quad x_1 < 0, x_2, x_3 > 0$:
+{{begin-eqn}}
+{{eqn | l = \frac {f \left({x_2}\right) - f \left({x_1}\right)} {x_2 - x_1}
+      | r = \frac {x_2 - \left({-x_1}\right)} {x_2 - x_1}
+      | c = {{Defof|Absolute Value}}
+}}
+{{eqn | r = \frac {\left({x_2 + x_1}\right) + \left({x_1 - x_1}\right)} {x_2 - x_1} 
+      | c = 
+}}
+{{eqn | r = \frac {x_2 - x_1} {x_2 - x_1} + \frac {2 x_1} {x_2 - x_1}
+      | c = 
+}}
+{{eqn | r = 1 + \frac {2 x_1} {x_2 - x_1}
+      | c = 
+}}
+{{eqn | o = <
+      | r = 1
+      | c = as $2 x_1 < 0$
+}}
+{{eqn | l = \frac {f \left({x_3}\right) - f \left({x_2}\right)} {x_3 - x_2}
+      | r = \frac {x_3 - x_2} {x_3 - x_2}
+      | c = {{Defof|Absolute Value}}
+}}
+{{eqn | r = 1
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \frac {f \left({x_2}\right) - f \left({x_1}\right)} {x_2 - x_1}
+      | o = \le
+      | r = \frac {f \left({x_3}\right) - f \left({x_2}\right)} {x_3 - x_2}
+      | c = Definition of [[Definition:Convex Real Function|Convex Real Function]]
+}}
+{{end-eqn}}
+:$(4): \quad x_1, x_2 < 0, x_3 > 0$:
+{{begin-eqn}}
+{{eqn | l = \frac {f \left({x_2}\right) - f \left({x_1}\right)} {x_2 - x_1}
+      | r = \frac {-\left({x_2 - x_1}\right)} {x_2 - x_1}
+      | c = {{Defof|Absolute Value}}
+}}
+{{eqn | r = -1
+      | c = 
+}}
+{{eqn | l = \frac {f \left({x_3}\right) - f \left({x_2}\right)} {x_3 - x_2}
+      | r = \frac {x_3 - \left({-x_2}\right)} {x_3 - x_3}
+      | c = Definition of [[Definition:Absolute Value|Absolute Value]]
+}}
+{{eqn | r = \frac {x_3 + x_2 + \left({x_3 - x_3}\right)} {x_3 - x_2} 
+      | c = 
+}}
+{{eqn | r = \frac {-x_3 + x_2 + x_3 + x_3} {x_3 - x_2} 
+      | c = 
+}}
+{{eqn | r = \frac {-\left({x_3 - x_2}\right)} {x_3 - x_2} + \frac {2 x_3} {x_3 - x_2}
+      | c = 
+}}
+{{eqn | r = -1 + \frac {2 x_3} {x_3 - x_2}
+      | c = 
+}}
+{{eqn | o = >
+      | r = -1
+      | c = as $2 x_3 > 0$
+}}
+{{eqn | ll= \implies
+      | l = \frac {f \left({x_2}\right) - f \left({x_1}\right)} {x_2 - x_1}
+      | o = \le
+      | r = \frac {f \left({x_3}\right) - f \left({x_2}\right)} {x_3 - x_2}
+      | c = {{Defof|Convex Real Function}}
+}}
+{{end-eqn}}
+Thus for all cases, the condition for $f$ to be [[Definition:Convex Real Function|convex]] is fulfilled.
+Hence the result.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $x, y \in \R$.
+Let $\alpha, \beta \in \R_{\ge 0}$ where $\alpha + \beta = 1$.
+{{begin-eqn}}
+{{eqn | l = \map f {\alpha x + \beta y}
+      | r = \size {\alpha x + \beta y}
+      | c = Definition of $f$
+}}
+{{eqn | o = \le
+      | r = \size {\alpha x} + \size {\beta y}
+      | c = [[Triangle Inequality for Real Numbers]]
+}}
+{{eqn | r = \size \alpha \size x + \size \beta \size y
+      | c = [[Absolute Value Function is Completely Multiplicative]]
+}}
+{{eqn | r = \alpha \size x + \beta \size y
+      | c = {{Defof|Absolute Value}}
+}}
+{{eqn | r = \alpha \, \map f x + \beta \, \map f y
+      | c = Definition of $f$
+}}
+{{end-eqn}}
+Hence the result by definition of [[Definition:Convex Real Function/Definition 1|convex real function]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Real Function is Concave iff Derivative is Decreasing}
+Tags: Differential Calculus, Concave Real Functions
+
+\begin{theorem}
+Let $f$ be a [[Definition:Real Function|real function]] which is [[Definition:Differentiable on Interval|differentiable]] on the [[Definition:Open Real Interval|open interval]] $\openint a b$.
+Then $f$ is [[Definition:Concave Real Function|concave]] on $\openint a b$ {{iff}} its [[Definition:Derivative on Interval|derivative]] $f'$ is [[Definition:Decreasing Real Function|decreasing]] on $\openint a b$.
+Thus the intuitive result that a [[Definition:Concave Real Function|concave function]] "gets less steep".
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Let $f$ be [[Definition:Concave Real Function|concave]] on $\openint a b$.
+Let $x_1, x_2, x_3, x_4 \in \openint a b$ such that:
+:$x_1 < x_2 < x_3 < x_4$
+By the definition of [[Definition:Concave Real Function|concave]]:
+:$\dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1} \ge \dfrac {\map f {x_3} - \map f {x_2} } {x_3 - x_2} \ge \dfrac {\map f {x_4} - \map f {x_3} } {x_4 - x_3}$
+Ignore the middle term and let $x_2 \to x_1^+$ and $x_3 \to x_4^-$.
+Thus:
+:$\map {f'} {x_1} \ge \map {f'} {x_4}$
+Hence $f'$ is [[Definition:Decreasing Real Function|decreasing]] on $\openint a b$.
+{{qed|lemma}}
+=== Sufficient Condition ===
+Let $f'$ be [[Definition:Decreasing Real Function|decreasing]] on $\openint a b$.
+Let $x_1, x_2, x_3 \in \openint a b: x_1 < x_2 < x_3$.
+By the [[Mean Value Theorem]]:
+:$\exists \xi: \dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1} = \map {f'} \xi$
+:$\exists \eta: \dfrac {\map f {x_3} - \map f {x_2} } {x_3 - x_2} = \map {f'} \eta$
+where $x_1 < \xi < x_2 < \eta < x_3$.
+Since $f'$ is [[Definition:Decreasing Real Function|decreasing]]:
+:$\map {f'} \xi \ge \map {f'} \eta$
+Thus:
+:$\dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1} \ge \dfrac {\map f {x_3} - \map f {x_2} } {x_3 - x_2}$
+Hence $f$ is [[Definition:Concave Real Function|concave]] by definition.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Real Function is Strictly Convex iff Derivative is Strictly Increasing}
+Tags: Differential Calculus, Convex Real Functions
+
+\begin{theorem}
+Let $f$ be a [[Definition:Real Function|real function]] which is [[Definition:Differentiable on Interval|differentiable]] on the [[Definition:Open Real Interval|open interval]] $\openint a b$.
+Then $f$ is [[Definition:Strictly Convex Real Function|strictly convex]] on $\openint a b$ {{iff}} its [[Definition:Derivative on Interval|derivative]] $f'$ is [[Definition:Strictly Increasing Real Function|strictly increasing]] on $\openint a b$.
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Let $f$ be [[Definition:Strictly Convex Real Function|strictly convex]] on $\openint a b$.
+Let $r, s \in \openint a b$ be arbitrarily selected such that $r < s$.
+We are to show that:
+:$\map {f'} r < \map {f'} s$
+Let $x_1, x_2, x_3 \in \openint a b$ be chosen such that:
+:$r < x_1 < x_2 < x_3 < s$
+By the definition of [[Definition:Strictly Convex Real Function|strictly convex]]:
+:$(1): \quad \dfrac {\map f {x_1} - \map f r} {x_1 - r} < \dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1} < \dfrac {\map f {x_3} - \map f {x_2} } {x_3 - x_2} < \dfrac {\map f s - \map f {x_3} } {s - x_3}$
+and:
+:$(2): \quad \dfrac {\map f {x_2} - \map f r} {x_2 - r} < \dfrac {\map f s - \map f {x_2} } {s - x_2}$
+Let $x_1 \to r^+$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \dfrac {\map f {x_1} - \map f r} {x_1 - r}
+      | o = <
+      | r = \dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \lim_{x_1 \mathop \to r^+} \dfrac {\map f {x_1} - \map f r} {x_1 - r}
+      | o = \le
+      | r = \lim_{x_1 \mathop \to r^+} \dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1}
+      | c = [[Limits Preserve Inequalities]]
+}}
+{{eqn | n = 3
+      | ll= \leadsto
+      | l = \map {f'} r
+      | o = \le
+      | r = \dfrac {\map f {x_2} - \map f r} {x_2 - r}
+      | c = {{Defof|Derivative of Real Function at Point}}
+}}
+{{end-eqn}}
+Similarly, let $x_3 \to s^-$.
+We have:
+{{begin-eqn}}
+{{eqn | l = \dfrac {\map f {x_3} - \map f {x_2} } {x_3 - x_2}
+      | o = <
+      | r = \dfrac {\map f s - \map f {x_3} } {s - x_3}
+      | c = from $(1)$
+}}
+{{eqn | ll= \leadsto
+      | l = \lim_{x_3 \mathop \to s^-} \dfrac {\map f {x_3} - \map f {x_2} } {x_3 - x_2}
+      | o = \le
+      | r = \lim_{x_3 \mathop \to s^-} \dfrac {\map f s - \map f {x_3} } {s - x_3}
+      | c = [[Limits Preserve Inequalities]]
+}}
+{{eqn | n = 4
+      | ll= \leadsto
+      | l = \dfrac {\map f s - \map f {x_2} } {s - x_2}
+      | o = \le
+      | r = \map {f'} s
+      | c = {{Defof|Derivative of Real Function at Point}}
+}}
+{{end-eqn}}
+Thus we have:
+{{begin-eqn}}
+{{eqn | l = \map {f'} r
+      | o = \le
+      | r = \dfrac {\map f {x_2} - \map f r} {x_2 - r}
+      | c = from $(3)$
+}}
+{{eqn | o = <
+      | r = \dfrac {\map f s - \map f {x_2} } {s - x_2}
+      | c = from $(2)$
+}}
+{{eqn | o = \le
+      | r = \map {f'} s
+      | c = from $(4)$
+}}
+{{end-eqn}}
+Thus:
+:$\map {f'} r < \map {f'} s$
+Hence $f'$ is [[Definition:Strictly Decreasing Real Function|strictly decreasing]] on $\openint a b$.
+{{Proofread}}
+{{qed|lemma}}
+=== Sufficient Condition ===
+Let $f'$ be [[Definition:Strictly Increasing Real Function|strictly increasing]] on $\openint a b$.
+Let $x_1, x_2, x_3 \in \openint a b: x_1 < x_2 < x_3$.
+By the [[Mean Value Theorem]]:
+:$\exists \xi: \dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1} = \map {f'} \xi$
+:$\exists \eta: \dfrac {\map f {x_3} - \map f {x_2} } {x_3 - x_2} = \map {f'} \eta$
+where $x_1 < \xi < x_2 < \eta < x_3$.
+Since $f'$ is [[Definition:Strictly Increasing Real Function|strictly increasing]]:
+:$\map {f'} \xi < \map {f'} \eta$
+Thus:
+:$\dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1} < \dfrac {\map f {x_3} - \map f {x_2} } {x_3 - x_2}$
+Hence $f$ is [[Definition:Strictly Convex Real Function|strictly convex]] by definition.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Real Function is Strictly Concave iff Derivative is Strictly Decreasing}
+Tags: Differential Calculus, Concave Real Functions
+
+\begin{theorem}
+Let $f$ be a [[Definition:Real Function|real function]] which is [[Definition:Differentiable on Interval|differentiable]] on the [[Definition:Open Real Interval|open interval]] $\openint a b$.
+Then $f$ is [[Definition:Strictly Concave Real Function|strictly concave]] on $\openint a b$ {{iff}} its [[Definition:Derivative on Interval|derivative]] $f'$ is [[Definition:Strictly Decreasing Real Function|strictly decreasing]] on $\openint a b$.
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Let $f$ be [[Definition:Strictly Concave Real Function|strictly concave]] on $\openint a b$.
+Let $r, s \in \openint a b$ be arbitrarily selected such that $r < s$.
+We are to show that:
+:$\map {f'} r > \map {f'} s$
+Let $x_1, x_2, x_3 \in \openint a b$ be chosen such that:
+:$r < x_1 < x_2 < x_3 < s$
+By the definition of [[Definition:Strictly Concave Real Function|strictly concave]]:
+:$(1): \quad \dfrac {\map f {x_1} - \map f r} {x_1 - r} > \dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1} > \dfrac {\map f {x_3} - \map f {x_2} } {x_3 - x_2} > \dfrac {\map f s - \map f {x_3} } {s - x_3}$
+and:
+:$(2): \quad \dfrac {\map f {x_2} - \map f r} {x_2 - r} > \dfrac {\map f s - \map f {x_2} } {s - x_2}$
+Let $x_1 \to r^+$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \dfrac {\map f {x_1} - \map f r} {x_1 - r}
+      | o = >
+      | r = \dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1}
+      | c = from $(1)$
+}}
+{{eqn | ll= \leadsto
+      | l = \lim_{x_1 \mathop \to r^+} \dfrac {\map f {x_1} - \map f r} {x_1 - r}
+      | o = \ge
+      | r = \lim_{x_1 \mathop \to r^+} \dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1}
+      | c = [[Limits Preserve Inequalities]]
+}}
+{{eqn | n = 3
+      | ll= \leadsto
+      | l = \map {f'} r
+      | o = \ge
+      | r = \dfrac {\map f {x_2} - \map f r} {x_2 - r}
+      | c = {{Defof|Derivative of Real Function at Point}}
+}}
+{{end-eqn}}
+Similarly, let $x_3 \to s^-$.
+We have:
+{{begin-eqn}}
+{{eqn | l = \dfrac {\map f {x_3} - \map f {x_2} } {x_3 - x_2}
+      | o = >
+      | r = \dfrac {\map f s - \map f {x_3} } {s - x_3}
+      | c = from $(1)$
+}}
+{{eqn | ll= \leadsto
+      | l = \lim_{x_3 \mathop \to s^-} \dfrac {\map f {x_3} - \map f {x_2} } {x_3 - x_2}
+      | o = \ge
+      | r = \lim_{x_3 \mathop \to s^-} \dfrac {\map f s - \map f {x_3} } {s - x_3}
+      | c = [[Limits Preserve Inequalities]]
+}}
+{{eqn | n = 4
+      | ll= \leadsto
+      | l = \dfrac {\map f s - \map f {x_2} } {s - x_2}
+      | o = \ge
+      | r = \map {f'} s
+      | c = {{Defof|Derivative of Real Function at Point}}
+}}
+{{end-eqn}}
+Thus we have:
+{{begin-eqn}}
+{{eqn | l = \map {f'} r
+      | o = \ge
+      | r = \dfrac {\map f {x_2} - \map f r} {x_2 - r}
+      | c = from $(3)$
+}}
+{{eqn | o = >
+      | r = \dfrac {\map f s - \map f {x_2} } {s - x_2}
+      | c = from $(2)$
+}}
+{{eqn | o = \ge
+      | r = \map {f'} s
+      | c = from $(4)$
+}}
+{{end-eqn}}
+Thus:
+:$\map {f'} r > \map {f'} s$
+Hence $f'$ is [[Definition:Strictly Decreasing Real Function|strictly decreasing]] on $\openint a b$.
+{{Proofread}}
+{{qed|lemma}}
+=== Sufficient Condition ===
+Let $f'$ be [[Definition:Strictly Decreasing Real Function|strictly decreasing]] on $\openint a b$.
+Let $x_1, x_2, x_3 \in \openint a b: x_1 < x_2 < x_3$.
+By the [[Mean Value Theorem]]:
+:$\exists \xi: \dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1} = \map {f'} \xi$
+:$\exists \eta: \dfrac {\map f {x_3} - \map f {x_2} } {x_3 - x_2} = \map {f'} \eta$
+where $x_1 < \xi < x_2 < \eta < x_3$.
+Since $f'$ is [[Definition:Strictly Decreasing Real Function|strictly decreasing]]:
+:$\map {f'} \xi > \map {f'} \eta$
+Thus:
+:$\dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1} > \dfrac {\map f {x_3} - \map f {x_2} } {x_3 - x_2}$
+Hence $f$ is [[Definition:Strictly Concave Real Function|strictly concave]] by definition.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Inverse of Strictly Decreasing Convex Real Function is Convex}
+Tags: Convex Real Functions
+
+\begin{theorem}
+Let $f$ be a [[Definition:Real Function|real function]] which is [[Definition:Convex Real Function|convex]] on the [[Definition:Open Real Interval|open interval]] $I$.
+Let $J = f \left[{I}\right]$.
+If $f$ be [[Definition:Strictly Decreasing Real Function|strictly decreasing]] on $I$, then $f^{-1}$ is [[Definition:Convex Real Function|convex]] on $J$.
+\end{theorem}
+
+\begin{proof}
+Let:
+:$X = f \left({x}\right) \in J$
+:$Y = f \left({y}\right) \in J$.
+From the definition of [[Definition:Convex Real Function|convex]]:
+: $\forall \alpha, \beta \in \R_{>0}, \alpha + \beta = 1: f \left({\alpha x + \beta y}\right) \le \alpha f \left({x}\right) + \beta f \left({y}\right)$
+Let $f$ be [[Definition:Strictly Decreasing Real Function|strictly decreasing]] on $I$.
+Then from [[Inverse of Strictly Monotone Function]] it follows that $f^{-1}$ is [[Definition:Strictly Decreasing Real Function|strictly decreasing]] on $J$.
+Thus:
+: $\alpha f^{-1} \left({X}\right) + \beta f^{-1} \left({Y}\right) = \alpha x + \beta y \le f^{-1} \left({\alpha X + \beta Y}\right)$
+Hence $f^{-1}$ is [[Definition:Convex Real Function|convex]] on $J$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Inverse of Strictly Increasing Concave Real Function is Convex}
+Tags: Convex Real Functions, Concave Real Functions
+
+\begin{theorem}
+Let $f$ be a [[Definition:Real Function|real function]] which is [[Definition:Concave Real Function|concave]] on the [[Definition:Open Real Interval|open interval]] $I$.
+Let $J = f \sqbrk I$.
+If $f$ is [[Definition:Strictly Increasing Real Function|strictly increasing]] on $I$, then $f^{-1}$ is [[Definition:Concave Real Function|convex]] on $J$.
+\end{theorem}
+
+\begin{proof}
+Let $X = \map f x \in J, Y = \map f y \in J$.
+From the definition of [[Definition:Concave Real Function|concave]]:
+:$\forall \alpha, \beta \in \R_{>0}, \alpha + \beta = 1: \map f {\alpha x + \beta y} \ge \alpha \map f x + \beta \map f y$
+Let $f$ be [[Definition:Strictly Increasing Real Function|strictly increasing]] on $I$.
+From [[Inverse of Strictly Monotone Function]], it follows that $f^{-1}$ is [[Definition:Strictly Increasing Real Function|strictly increasing]] on $J$.
+Thus:
+:$\alpha \map {f^{-1} } X + \beta \map {f^{-1} } Y = \alpha x + \beta y \ge \map {f^{-1} } {\alpha X + \beta Y}$
+Hence $f^{-1}$ is [[Definition:Convex Real Function|convex]] on $J$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Inverse of Strictly Decreasing Concave Real Function is Concave}
+Tags: Concave Real Functions
+
+\begin{theorem}
+Let $f$ be a [[Definition:Real Function|real function]] which is [[Definition:Concave Real Function|concave]] on the [[Definition:Open Real Interval|open interval]] $I$.
+Let $J = f \left[{I}\right]$.
+If $f$ be [[Definition:Strictly Decreasing Real Function|strictly decreasing]] on $I$, then $f^{-1}$ is [[Definition:Concave Real Function|concave]] on $J$.
+\end{theorem}
+
+\begin{proof}
+Let:
+:$X = f \left({x}\right) \in J$
+:$Y = f \left({y}\right) \in J$.
+From the definition of [[Definition:Concave Real Function|concave]]:
+: $\forall \alpha, \beta \in \R_{>0}, \alpha + \beta = 1: f \left({\alpha x + \beta y}\right) \ge \alpha f \left({x}\right) + \beta f \left({y}\right)$
+Let $f$ be [[Definition:Strictly Decreasing Real Function|strictly decreasing]] on $I$.
+From [[Inverse of Strictly Monotone Function]], it follows that $f^{-1}$ is [[Definition:Strictly Decreasing Real Function|strictly decreasing]] on $J$.
+Thus:
+: $\alpha f^{-1} \left({X}\right) + \beta f^{-1} \left({Y}\right) = \alpha x + \beta y \ge f^{-1} \left({\alpha X + \beta Y}\right)$
+Hence $f^{-1}$ is [[Definition:Concave Real Function|concave]] on $J$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Mean Value of Concave Real Function}
+Tags: Concave Real Functions, Differential Calculus
+
+\begin{theorem}
+Let $f$ be a [[Definition:Real Function|real function]] which is [[Definition:Continuous on Interval|continuous]] on the [[Definition:Closed Real Interval|closed interval]] $\left[{a \,.\,.\, b}\right]$ and [[Definition:Differentiable on Interval|differentiable]] on the [[Definition:Open Real Interval|open interval]] $\left({a \,.\,.\, b}\right)$.
+Let $f$ be [[Definition:Concave Real Function|concave]] on $\left({a \,.\,.\, b}\right)$.
+Then:
+: $\forall \xi \in \left({a \,.\,.\, b}\right): f \left({x}\right) - f \left({\xi}\right) \le f^{\prime} \left({\xi}\right) \left({x - \xi}\right)$
+\end{theorem}
+
+\begin{proof}
+By the [[Mean Value Theorem]]:
+: $\displaystyle \exists \eta \in \left({x \,.\,.\, \xi}\right): f^{\prime} \left({\eta}\right) = \dfrac {f \left({x}\right) - f \left({\xi}\right)} {x - \xi}$
+From [[Real Function is Concave iff Derivative is Decreasing]], the [[Definition:Derivative|derivative]] of $f$ is [[Definition:Decreasing Real Function|decreasing]].
+Thus:
+: $x > \xi \implies f' \left({\eta}\right) \le f' \left({\xi}\right)$
+: $x < \xi \implies f' \left({\eta}\right) \ge f' \left({\xi}\right)$
+Hence:
+: $f \left({x}\right) - f \left({\xi}\right) \le f' \left({\xi}\right) \left({x - \xi}\right)$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Differentiable Bounded Concave Real Function is Constant}
+Tags: Concave Real Functions, Differential Calculus
+
+\begin{theorem}
+Let $f$ be a [[Definition:Real Function|real function]] which is:
+:$(1): \quad$ [[Definition:Differentiable on Interval|Differentiable]] on $\R$
+:$(2): \quad$ [[Definition:Bounded Real-Valued Function|Bounded]] on $\R$
+:$(3): \quad$ [[Definition:Concave Real Function|Concave]] on $\R$.
+Then $f$ is [[Definition:Constant Mapping|constant]].
+\end{theorem}
+
+\begin{proof}
+Let $f$ be [[Definition:Differentiable on Interval|differentiable]] and [[Definition:Bounded Real-Valued Function|bounded]] on $\R$.
+Let $f$ be [[Definition:Concave Real Function|concave]] on $\R$.
+Let $\xi \in \R$.
+Aiming for a [[Proof by Contradiction|contradiction]], suppose $f' \left({\xi}\right) > 0$.
+Then by [[Mean Value of Concave Real Function]] it follows that:
+: $f \left({x}\right) \le f \left({\xi}\right) + f' \left({\xi}\right) \left({x - \xi}\right) \to -\infty$ as $x \to +\infty$
+and therefore is not [[Definition:Bounded Real-Valued Function|bounded]].
+Similarly, suppose $f' \left({\xi}\right) < 0$.
+Then by [[Mean Value of Concave Real Function]] it follows that:
+: $f \left({x}\right) \le f \left({\xi}\right) + f' \left({\xi}\right) \left({x - \xi}\right) \to -\infty$ as $x \to -\infty$
+and therefore is likewise not [[Definition:Bounded Real-Valued Function|bounded]].
+Hence $f' \left({\xi}\right) = 0$.
+From [[Zero Derivative implies Constant Function]], it follows that $f$ is [[Definition:Constant Mapping|constant]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Second Derivative of Strictly Concave Real Function is Strictly Negative}
+Tags: Differential Calculus, Concave Real Functions
+
+\begin{theorem}
+Let $f$ be a [[Definition:Real Function|real function]] which is [[Definition:Second Derivative|twice differentiable]] on the [[Definition:Open Real Interval|open interval]] $\left({a \,.\,.\, b}\right)$.
+Then $f$ is [[Definition:Strictly Concave Real Function|strictly concave]] on $\left({a \,.\,.\, b}\right)$ {{iff}} its [[Definition:Second Derivative|second derivative]] $f'' < 0$ on $\left({a \,.\,.\, b}\right)$.
+\end{theorem}
+
+\begin{proof}
+From [[Real Function is Strictly Concave iff Derivative is Strictly Decreasing]], $f$ is [[Definition:Strictly Concave Real Function|strictly concave]] {{iff}} $f'$ is [[Definition:Strictly Decreasing Real Function|strictly decreasing]].
+From [[Derivative of Monotone Function]], $f'$ is [[Definition:Strictly Decreasing Real Function|strictly decreasing]] {{iff}} its [[Definition:Second Derivative|second derivative]] $f'' < 0$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Second Derivative of Strictly Convex Real Function is Strictly Positive}
+Tags: Differential Calculus, Convex Real Functions
+
+\begin{theorem}
+Let $f$ be a [[Definition:Real Function|real function]] which is [[Definition:Second Derivative|twice differentiable]] on the [[Definition:Open Real Interval|open interval]] $\openint a b$.
+Then $f$ is [[Definition:Strictly Convex Real Function|strictly convex]] on $\openint a b$ {{iff}} its [[Definition:Second Derivative|second derivative]] $f'' > 0$ on $\openint a b$.
+\end{theorem}
+
+\begin{proof}
+From [[Real Function is Strictly Convex iff Derivative is Strictly Increasing]], $f$ is [[Definition:Strictly Convex Real Function|strictly convex]] {{iff}} $f'$ is [[Definition:Strictly Increasing Real Function|strictly increasing]].
+From [[Derivative of Monotone Function]], $f'$ is [[Definition:Strictly Increasing Real Function|strictly increasing]] {{iff}} its [[Definition:Second Derivative|second derivative]] $f'' > 0$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Inverse of Strictly Decreasing Strictly Convex Real Function is Strictly Convex}
+Tags: Convex Real Functions
+
+\begin{theorem}
+Let $f$ be a [[Definition:Real Function|real function]] which is [[Definition:Strictly Convex Real Function|strictly convex]] on the [[Definition:Open Real Interval|open interval]] $I$.
+Let $J = f \sqbrk I$.
+If $f$ be [[Definition:Strictly Decreasing Real Function|strictly decreasing]] on $I$, then $f^{-1}$ is [[Definition:Strictly Convex Real Function|strictly convex]] on $J$.
+\end{theorem}
+
+\begin{proof}
+Let:
+:$X = \map f x \in J$
+:$Y = \map f y \in J$.
+From the definition of [[Definition:Strictly Convex Real Function|strictly convex]]:
+:$\forall \alpha, \beta \in \R_{>0}, \alpha + \beta = 1: \map f {\alpha x + \beta y} < \alpha \map f x + \beta \map f y$
+Let $f$ be [[Definition:Strictly Decreasing Real Function|strictly decreasing]] on $I$.
+Then from [[Inverse of Strictly Monotone Function]] it follows that $f^{-1}$ is [[Definition:Strictly Decreasing Real Function|strictly decreasing]] on $J$.
+Thus:
+:$\alpha \map {f^{-1} } X + \beta \map {f^{-1} } Y = \alpha x + \beta y < \map {f^{-1} } {\alpha X + \beta Y}$
+Hence $f^{-1}$ is [[Definition:Strictly Convex Real Function|strictly convex]] on $J$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Inverse of Strictly Decreasing Strictly Concave Real Function is Strictly Concave}
+Tags: Concave Real Functions
+
+\begin{theorem}
+Let $f$ be a [[Definition:Real Function|real function]] which is [[Definition:Strictly Concave Real Function|strictly concave]] on the [[Definition:Open Real Interval|open interval]] $I$.
+Let $J = f \left[{I}\right]$.
+If $f$ be [[Definition:Strictly Decreasing Real Function|strictly decreasing]] on $I$, then $f^{-1}$ is [[Definition:Strictly Concave Real Function|strictly concave]] on $J$.
+\end{theorem}
+
+\begin{proof}
+Let:
+:$X = f \left({x}\right) \in J$
+:$Y = f \left({y}\right) \in J$.
+From the definition of [[Definition:Strictly Concave Real Function|strictly concave]]:
+: $\forall \alpha, \beta \in \R_{>0}, \alpha + \beta = 1: f \left({\alpha x + \beta y}\right) > \alpha f \left({x}\right) + \beta f \left({y}\right)$
+Let $f$ be [[Definition:Strictly Decreasing Real Function|strictly decreasing]] on $I$.
+From [[Inverse of Strictly Monotone Function]], it follows that $f^{-1}$ is [[Definition:Strictly Decreasing Real Function|strictly decreasing]] on $J$.
+Thus:
+: $\alpha f^{-1} \left({X}\right) + \beta f^{-1} \left({Y}\right) = \alpha x + \beta y > f^{-1} \left({\alpha X + \beta Y}\right)$
+Hence $f^{-1}$ is [[Definition:Strictly Concave Real Function|strictly concave]] on $J$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Inverse of Strictly Increasing Strictly Convex Real Function is Strictly Concave}
+Tags: Convex Real Functions, Concave Real Functions
+
+\begin{theorem}
+Let $f$ be a [[Definition:Real Function|real function]] which is [[Definition:Strictly Convex Real Function|strictly convex]] on the [[Definition:Open Real Interval|open interval]] $I$.
+Let $J = f \sqbrk I$.
+If $f$ be [[Definition:Strictly Increasing Real Function|strictly increasing]] on $I$, then $f^{-1}$ is [[Definition:Strictly Concave Real Function|strictly concave]] on $J$.
+\end{theorem}
+
+\begin{proof}
+Let:
+:$X = \map f x \in J$
+:$Y = \map f y \in J$
+From the definition of [[Definition:Strictly Convex Real Function|strictly convex]]:
+:$\forall \alpha, \beta \in \R_{>0}, \alpha + \beta = 1: \map f {\alpha x + \beta y} < \alpha \, \map f x + \beta \, \map f y$
+Let $f$ be [[Definition:Strictly Increasing Real Function|strictly increasing]] on $I$.
+From [[Inverse of Strictly Monotone Function]] it follows that, $f^{-1}$ is [[Definition:Strictly Increasing Real Function|strictly increasing]] on $J$.
+Thus:
+:$\alpha \, \map {f^{-1} } X + \beta \, \map {f^{-1} } Y = \alpha x + \beta y < \map {f^{-1} } {\alpha X + \beta Y}$
+Hence $f^{-1}$ is [[Definition:Strictly Concave Real Function|strictly concave]] on $J$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Inverse of Strictly Increasing Strictly Concave Real Function is Strictly Convex}
+Tags: Convex Real Functions, Concave Real Functions
+
+\begin{theorem}
+Let $f$ be a [[Definition:Real Function|real function]] which is [[Definition:Strictly Concave Real Function|strictly concave]] on the [[Definition:Open Real Interval|open interval]] $I$.
+Let $J = f \left[{I}\right]$.
+If $f$ be [[Definition:Strictly Increasing Real Function|strictly increasing]] on $I$, then $f^{-1}$ is [[Definition:Strictly Concave Real Function|strictly convex]] on $J$.
+\end{theorem}
+
+\begin{proof}
+Let $X = f \left({x}\right) \in J, Y = f \left({y}\right) \in J$.
+From the definition of [[Definition:Strictly Concave Real Function|Strictly concave]]:
+: $\forall \alpha, \beta \in \R_{>0}, \alpha + \beta = 1: f \left({\alpha x + \beta y}\right) > \alpha f \left({x}\right) + \beta f \left({y}\right)$
+Let $f$ be [[Definition:Strictly Increasing Real Function|strictly increasing]] on $I$.
+From [[Inverse of Strictly Monotone Function]], it follows that $f^{-1}$ is [[Definition:Strictly Increasing Real Function|strictly increasing]] on $J$.
+Thus:
+: $\alpha f^{-1} \left({X}\right) + \beta f^{-1} \left({Y}\right) = \alpha x + \beta y > f^{-1} \left({\alpha X + \beta Y}\right)$
+Hence $f^{-1}$ is [[Definition:Strictly Convex Real Function|strictly convex]] on $J$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Gamma Function as Integral of Natural Logarithm}
+Tags: Gamma Function
+
+\begin{theorem}
+Let $x \in \R_{>0}$ be a [[Definition:Strictly Positive Real Number|strictly positive real number]].
+Then:
+:$\displaystyle \map \Gamma x = \int_{\to 0}^1 \paren {\ln \frac 1 t}^{x - 1} \rd t$
+where $\Gamma$ denotes the [[Definition:Gamma Function|Gamma function]].
+\end{theorem}
+
+\begin{proof}
+By definition of the [[Definition:Gamma Function|Gamma function]]:
+:$\displaystyle \map \Gamma x = \int_0^{\to \infty} t^{x - 1} e^{-t} \rd t$
+In order to allow the [[Definition:Limits of Integration|limits]] to be evaluated, this is to be expressed as:
+:$\displaystyle \map \Gamma x = \lim_{\delta \mathop \to 0^+, \ \Delta \mathop \to +\infty} \int_\delta^\Delta t^{x - 1} e^{-t} \rd t$
+where $0 < \delta < \Delta$.
+Let $-t = \ln u$.
+Then by [[Derivative of Logarithm Function]] and the [[Chain Rule for Derivatives]]:
+:$-1 = \dfrac 1 u \dfrac {\rd u} {\rd t}$
+Also:
+{{begin-eqn}}
+{{eqn | l = -t
+      | r = \ln u
+}}
+{{eqn | ll= \leadsto
+      | l = e^{-t}
+      | r = u
+      | c = [[Exponential of Natural Logarithm]]
+}}
+{{end-eqn}}
+Hence:
+{{begin-eqn}}
+{{eqn | l = \int_\delta^\Delta t^{x - 1} e^{-t} \rd t
+      | r = \int_{e^{-\delta} }^{e^{-\Delta} } \paren {-\ln u}^{x - 1} e^{\ln u} \paren {-\frac 1 u \frac {\rd u} {\rd t} } \rd t
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \int_{e^{-\Delta} }^{e^{-\delta} } \paren {-\ln u}^{x - 1} e^{\ln u} \paren {\frac 1 u \frac {\rd u} {\rd t} } \rd t
+      | c = [[Reversal of Limits of Definite Integral]]
+}}
+{{eqn | r = \int_{e^{-\Delta} }^{e^{-\delta} } \paren {\ln \frac 1 u}^{x - 1} u \frac 1 u \rd u
+      | c = [[Exponential of Natural Logarithm]]
+}}
+{{eqn | n = 1
+      | r = \int_{e^{-\Delta} }^{e^{-\delta} } \paren {\ln \frac 1 u}^{x - 1} \rd u
+      | c = simplification
+}}
+{{end-eqn}}
+Next:
+{{begin-eqn}}
+{{eqn | l = \lim_{\delta \mathop \to 0^+} e^{-\delta}
+      | r = 1
+      | c = [[Exponential of Zero]]
+}}
+{{eqn | l = \lim_{\Delta \mathop \to +\infty} e^{-\Delta}
+      | r = 0
+      | c = [[Exponential Tends to Zero and Infinity]]
+}}
+{{end-eqn}}
+Thus:
+{{begin-eqn}}
+{{eqn | l = \lim_{\delta \mathop \to 0^+, \ \Delta \mathop \to +\infty} \int_{e^{-\Delta} }^{e^{-\delta} } \paren {\ln \frac 1 u}^{x - 1} \rd u
+      | r = \int_{e^{-\Delta} }^1 \paren {\ln \frac 1 u}^{x - 1} \rd u
+      | c = substituting for the top limit
+}}
+{{eqn | r = \int_{\to 0}^1 \paren {\ln \frac 1 u}^{x - 1} \rd u
+      | c = as $\ln 0$ is undefined
+}}
+{{end-eqn}}
+The result follows by renaming the [[Definition:Dummy Variable|dummy variable]] $u$ to $t$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Reversal of Limits of Definite Integral}
+Tags: Definite Integrals
+
+\begin{theorem}
+Let $a \le b$.
+Then:
+:$\displaystyle \int_a^b \map f x \rd x = - \int_b^a \map f x \rd x$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int_a^b \map f x \rd x + \int_b^a \map f x \rd x
+      | r = \int_a^a \map f x \rd x
+      | c = [[Sum of Integrals on Adjacent Intervals for Integrable Functions]]
+}}
+{{eqn | r = 0
+      | c = [[Definite Integral on Zero Interval]]
+}}
+{{eqn | ll = \leadsto
+      | l = \int_a^b \map f x \rd x
+      | r = - \int_b^a \map f x \rd x
+}}
+{{end-eqn}}
+{{CircularStructure|Careful with the above. Down at the bottom of [[Sum of Integrals on Adjacent Intervals for Integrable Functions]] use is made of this result. There has been a lot of activity on this area of mathematics, and several people have contributed, possibly without having completely followed through all the chains of implications. </br> My bad. Either way this has to be proven independently since the case $a {{=}} b$ was not considered in that theorem.}}
+\end{proof}<|endoftext|>
+\section{Beta Function is Defined for Positive Reals}
+Tags: Beta Function
+
+\begin{theorem}
+Let $x, y \in \R$ be [[Definition:Real Number|real numbers]].
+Let $\Beta \left({x, y}\right)$ be the [[Definition:Beta Function/Definition 1|Beta function]]:
+:$\displaystyle \Beta \left({x, y}\right) := \int_{\mathop \to 0}^{\mathop \to 1} t^{x - 1} \left({1 - t}\right)^{y - 1} \ \mathrm d t$
+Then $\Beta \left({x, y}\right)$ exists provided that $x, y > 0$.
+\end{theorem}
+
+\begin{proof}
+Consider the following inequalities, valid for $0 < t < 1$:
+{{begin-eqn}}
+{{eqn | l = t^{x - 1} \left({1 - t}\right)^{y - 1}
+      | o = <
+      | r = t^{x - 1}
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = t^{x - 1} \left({1 - t}\right)^{y - 1}
+      | o = <
+      | r = \left({1 - t}\right)^{y - 1}
+      | c = 
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = \int_{\mathop \to 0}^{\mathop \to 1} t^{x - 1} \ \mathrm d t
+      | r = \left.{\frac {t^x} x}\right\vert_{\mathop \to 0}^{\mathop \to 1}
+      | c = 
+}}
+{{eqn | r = x - 0
+      | c = 
+}}
+{{end-eqn}}
+and similarly:
+{{begin-eqn}}
+{{eqn | l = \int_{\mathop \to 0}^{\mathop \to 1} \left({1 - t}\right)^{y - 1} \ \mathrm d t
+      | r = -\left.{\frac {\left({1 - t}\right)^y} y}\right\vert_{\mathop \to 0}^{\mathop \to 1}
+      | c = 
+}}
+{{eqn | r = y - 0
+      | c = 
+}}
+{{end-eqn}}
+The result follows from the [[Comparison Test for Improper Integral]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Comparison Test for Improper Integral}
+Tags: Convergence Tests, Improper Integrals
+
+\begin{theorem}
+Let $I = \openint a b$ be an [[Definition:Open Real Interval|open real interval]].
+Let $\phi$ be a [[Definition:Real Function|real function]] which is [[Definition:Continuous Real Function on Open Interval|continuous on $I$]] and also [[Definition:Positive Real Function|non-negative]] on $I$.
+Let $f$ be a [[Definition:Real Function|real function]] which is [[Definition:Continuous Real Function on Open Interval|continuous on $I$]].
+Let $f$ satisfy:
+:$\forall x \in I: \size {\map f x} \le \map \phi x$
+If the [[Definition:Improper Integral over Open Interval|improper integral]] of $\phi$ over $I$ exists, then so does that of $f$.
+\end{theorem}
+
+\begin{proof}
+{{WLOG}}, we consider the case $I = \openint 0 \to$ such that $\displaystyle l = \int_0^{\mathop \to +\infty} \map \phi x \rd x$ exists.
+Let:
+:$\ds a_n = \int_{n - 1}^n \map f x \rd x$
+:$\ds b_n = \int_{n - 1}^n \map \phi x \rd x$
+for $n = 1, 2, \ldots$
+Then the [[Definition:Series|series]]:
+:$\ds \sum_{n \mathop = 1}^\infty b_n$
+is a [[Definition:Convergent Series|convergent series]] of [[Definition:Positive Real Number|positive]] terms whose sum is $l$.
+Also:
+$\size {a_n} \le b_n$
+for $n = 1, 2, \ldots$
+Hence $\ds \sum_{n \mathop = 1}^\infty a_n$ a [[Definition:Convergent Series|convergent series]] by the [[Comparison Test]].
+That is:
+:$\ds m = \lim_{N \mathop \to \infty} \int_0^N \map f x \rd x$
+exists.
+Given any $X > 0$, $N$ can be taken to be the smallest [[Definition:Natural Number|natural number]] which satisfies $N > X$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \size {\int_1^X \map f x \rd x - \int_1^{N - 1} \map f x \rd x}
+      | o = \le
+      | r = \int_{N - 1}^X \size {\map f x} \rd x
+      | c = 
+}}
+{{eqn | o = \le
+      | r = b_N
+      | c = 
+}}
+{{eqn | o = \to
+      | r = 0 \text{ as } N \to \infty
+      | c = 
+}}
+{{end-eqn}}
+It follows that:
+:$\ds \lim_{X \mathop \to \infty} \int_1^X \map f x \rd x = m$
+as required.
+The other cases are treated similarly.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Beta Function is Continuous and Positive on Positive Reals}
+Tags: Beta Function
+
+\begin{theorem}
+Let $x, y \in \R$ be [[Definition:Real Number|real numbers]].
+Let $\map \Beta {x, y}$ be the [[Definition:Beta Function/Definition 1|Beta function]]:
+:$\ds \map \Beta {x, y} := \int_{\mathop \to 0}^{\mathop \to 1} t^{x - 1} \paren {1 - t}^{y - 1} \rd t$
+Let $y \in \R_{>0}$ be given.
+Then $\map \Beta {x, y}$ is a [[Definition:Positive Real Function|positive]] and [[Definition:Continuous Real Function|continuous]] [[Definition:Real Function|function]] of $x$ on $\R_{>0}$.
+\end{theorem}
+
+\begin{proof}
+For each $x > 0$, we have for all $t$ with $0 < t < 1$ that:
+:$t^{x - 1} \paren {1 - t}^{y - 1} > 0$
+from which it is immediate that $\map \Beta {x, y} > 0$.
+{{ProofWanted|Continuity to be established.}}
+\end{proof}<|endoftext|>
+\section{Logarithm of Beta Function is Convex on Positive Reals}
+Tags: Beta Function
+
+\begin{theorem}
+Let $x, y \in \R$ be [[Definition:Real Number|real numbers]].
+Let $\Beta \left({x, y}\right)$ be the [[Definition:Beta Function/Definition 1|Beta function]]:
+:$\displaystyle \Beta \left({x, y}\right) := \int_{\mathop \to 0}^{\mathop \to 1} t^{x - 1} \left({1 - t}\right)^{y - 1} \ \mathrm d t$
+Let $y \in \R_{>0}$ be given.
+Then $\ln \left({\Beta \left({x, y}\right)}\right)$ is a [[Definition:Convex Real Function|convex function]] of $x$ on $\R_{>0}$.
+\end{theorem}<|endoftext|>
+\section{Gamma Function of One Half}
+Tags: Examples of Gamma Function Values, Gamma Function of One Half
+
+\begin{theorem}
+:$\map \Gamma {\dfrac 1 2} = \sqrt \pi$
+Its decimal expansion starts:
+:$\map \Gamma {\dfrac 1 2} = 1 \cdotp 77245 \, 38509 \, 05516 \, 02729 \, 81674 \, 83341 \, 14518 \, 27975 \ldots$
+\end{theorem}
+
+\begin{proof}
+From the definition of the [[Definition:Beta Function/Definition 3|Beta function]]:
+:$\Beta \left({x, y}\right) := \dfrac {\Gamma \left({x}\right) \Gamma \left({y}\right)} {\Gamma \left({x + y}\right)}$
+Setting $x = y = \dfrac 1 2$:
+{{begin-eqn}}
+{{eqn | l = \Beta \left({\dfrac 1 2, \dfrac 1 2}\right)
+      | r = \frac {\Gamma \left({\dfrac 1 2}\right) \Gamma \left({\dfrac 1 2}\right)} {\Gamma \left({\dfrac 1 2 + \dfrac 1 2}\right)}
+      | c = 
+}}
+{{eqn | r = \left({\Gamma \left({\dfrac 1 2}\right)}\right)^2
+      | c = 
+}}
+{{end-eqn}}
+Then from [[Beta Function of Half with Half]]:
+:$\Beta \left({\dfrac 1 2, \dfrac 1 2}\right) = \pi$
+Hence the result.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+From [[Euler's Reflection Formula]]:
+:$\forall z \notin \Z: \Gamma \left({z}\right) \Gamma \left({1 - z}\right) = \dfrac \pi {\sin \left({\pi z}\right)}$
+Setting $z = \dfrac 1 2$:
+{{begin-eqn}}
+{{eqn | l = \Gamma \left({\frac 1 2}\right) \Gamma \left({\frac 1 2}\right)
+      | r = \frac \pi {\sin \left({\frac \pi 2}\right)}
+      | c = 
+}}
+{{eqn | r = \frac \pi 1
+      | c = [[Sine of Right Angle]]
+}}
+{{eqn | ll= \implies
+      | l = \Gamma \left({\frac 1 2}\right)
+      | r = \pm \sqrt \pi
+      | c = 
+}}
+{{end-eqn}}
+By definition of the [[Definition:Gamma Function|$\Gamma$ function]]:
+:$\forall z \in \R_{\ge 0}: \Gamma \left({z}\right) > 0$
+and so the [[Definition:Negative Real Number|negative]] [[Definition:Square Root|square root]] can be discarded.
+Hence:
+:$\Gamma \left({\dfrac 1 2}\right) = \sqrt \pi$
+as required.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \Gamma \left({\dfrac 1 2}\right)
+      | r = \int_0^{\to \infty} t^{-\frac 1 2} e^{-t} \ \mathrm d t
+      | c = {{Defof|Integral Form of Gamma Function|Gamma Function}}
+}}
+{{eqn | r = \int_0^{\to \infty} u^{-1} e^{-u^2} 2 u \ \mathrm d u
+      | c = [[Integration by Substitution]], $\phi \left({u}\right) = u^2$
+}}
+{{eqn | r = 2 \int_0^{\to \infty} e^{-u^2} \ \mathrm d u
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \int_{\to -\infty}^{\to \infty} e^{-u^2} \ \mathrm d u
+      | c = [[Definite Integral of Even Function]]
+}}
+{{eqn | r = \sqrt {\pi}
+      | c = [[Gaussian Integral]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \Gamma \left({\frac 1 2}\right) 
+      | r = \frac {0!} {2^0 0!} \sqrt \pi
+      | c = [[Gamma Function of Positive Half-Integer]]
+}}
+{{eqn | r = \sqrt \pi
+      | c = [[Factorial of Zero]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map \Gamma 1 \, \map \Gamma {\frac 1 2} 
+      | r = 2^0 \sqrt \pi \ \map \Gamma 1 
+      | c = [[Legendre's Duplication Formula]] 
+}} 
+{{eqn | ll= \leadsto 
+      | l = 0! \, \map \Gamma {\frac 1 2} 
+      | r = 0! \, \sqrt \pi
+      | c = {{Defof|Gamma Function}}
+}}
+{{eqn | ll= \leadsto 
+      | l = \map \Gamma {\frac 1 2} 
+      | r = \sqrt \pi
+      | c = [[Factorial of Zero]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Integral of Exponent of Half Square over Reals}
+Tags: Gaussian Integral
+
+\begin{theorem}
+:$\displaystyle \int_{\mathop \to -\infty}^{\mathop \to +\infty} e^{- x^2 / 2} \rd x = \sqrt {2 \pi}$
+\end{theorem}
+
+\begin{proof}
+Let $t = \dfrac {x^2} 2$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \int_0^{\mathop \to +\infty} e^{- x^2 / 2} \rd x
+      | r = \int_0^{\mathop \to +\infty} \left({2 t}\right) e^{-t} \rd t
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \frac 1 {\sqrt 2} \Gamma \left({\frac 1 2}\right)
+      | c = {{Defof|Integral Form of Gamma Function|Gamma Function}}
+}}
+{{eqn | r = \frac 1 {\sqrt 2} \sqrt \pi
+      | c = [[Gamma Function of One Half]]
+}}
+{{eqn | r = \frac {\sqrt {2 \pi} } 2
+      | c = multiplying [[Definition:Numerator|top]] and [[Definition:Denominator|bottom]] by $\sqrt 2$
+}}
+{{end-eqn}}
+We have that $e^{- x^2 / 2}$ is an [[Definition:Even Function|even function]].
+From [[Definite Integral of Even Function/Corollary|Definite Integral of Even Function: Corollary]]:
+:$\displaystyle \int_{\mathop \to -\infty}^{\mathop \to +\infty} e^{- x^2 / 2} \rd x = 2 \int_0^{\mathop \to +\infty} e^{- x^2 / 2} \rd x$
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sum of Reciprocals of Primes is Divergent/Proof 2}
+Tags: Sum of Reciprocals of Primes is Divergent
+
+\begin{theorem}
+The [[Definition:Series|series]]:
+:$\displaystyle \sum_{p \mathop \in \Bbb P} \frac 1 p$
+where:
+:$\Bbb P$ is the [[Definition:Set|set]] of all [[Definition:Prime Number|prime numbers]]
+is [[Definition:Divergent Series|divergent]].
+\end{theorem}
+
+\begin{proof}
+Let $n \in \N$ be a [[Definition:Natural Number|natural number]].
+Let $p_n$ denote the $n$th [[Definition:Prime Number|prime number]].
+Consider the [[Definition:Product Notation (Algebra)|product]]:
+:$\displaystyle \prod_{k \mathop = 1}^n \frac 1 {1 - 1 / p_k}$
+By [[Sum of Geometric Sequence]], we have:
+{{begin-eqn}}
+{{eqn | l = \frac 1 {1 - \frac 1 2}
+      | r = 1 + \frac 1 2 + \frac 1 {2^2} + \cdots
+      | c = 
+}}
+{{eqn | l = \frac 1 {1 - \frac 1 3}
+      | r = 1 + \frac 1 3 + \frac 1 {3^2} + \cdots
+      | c = 
+}}
+{{eqn | l = \frac 1 {1 - \frac 1 5}
+      | r = 1 + \frac 1 5 + \frac 1 {5^2} + \cdots
+      | c = 
+}}
+{{eqn | o = \cdots
+      | c = 
+}}
+{{eqn | l = \frac 1 {1 - \frac 1 {p_n} }
+      | r = 1 + \frac 1 {p_n} + \frac 1 {p_n^2} + \cdots
+      | c = 
+}}
+{{end-eqn}}
+Consider what happens when all these [[Definition:Series|series]] are multiplied together.
+A new [[Definition:Series|series]] will be generated whose terms consist of all possible products of one term selected from each of the [[Definition:Series|series]] on the {{RHS}}.
+This new [[Definition:Series|series]] will [[Definition:Convergent Series|converge]] in any order to the product of the terms on the {{LHS}}.
+By the [[Fundamental Theorem of Arithmetic]], every [[Definition:Integer|integer]] greater than $1$ is uniquely expressible as a product of powers of different [[Definition:Prime Number|primes]].
+Hence the product of these [[Definition:Series|series]] is the [[Definition:Series|series]] of [[Definition:Reciprocal|reciprocals]] of all [[Definition:Strictly Positive Integer|(strictly) positive integers]] whose [[Definition:Prime Factor|prime factors]] are no greater than $p_n$.
+In particular, all [[Definition:Strictly Positive Integer|(strictly) positive integers]] up to $p_n$ have this property.
+So:
+:$\displaystyle \prod_{k \mathop = 1}^n \frac 1 {1 - 1 / p_k}$
+{{begin-eqn}}
+{{eqn | l = \prod_{k \mathop = 1}^n \frac 1 {1 - 1 / p_k}
+      | o = \ge
+      | r = \sum_{k \mathop = 1}^{p_n} \frac 1 k
+      | c = 
+}}
+{{eqn | o = >
+      | r = \int_1^{p_n + 1} \dfrac {\mathrm d x} x
+      | c = 
+}}
+{{eqn | r = \map \ln {p_n + 1}
+      | c = 
+}}
+{{eqn | r = \ln p_n
+      | c = 
+}}
+{{end-eqn}}
+It follows by taking [[Definition:Reciprocal|reciprocals]] that:
+:$\displaystyle \prod_{k \mathop = 1}^n \paren {1 - \frac 1 {p_k} } < \frac 1 {\ln p_n}$
+Taking [[Definition:Logarithm|logarithms]] of each side:
+:$(1): \quad \displaystyle \sum_{k \mathop = 1}^n \map \ln {1 - \frac 1 {p_k} } < - \ln \ln p_n$
+Next, note that the [[Definition:Straight Line|line]] $y = 2 x$ in the [[Definition:Cartesian Plane|cartesian plane]] lies below the curve $y = \map \ln {1 + x}$ on the [[Definition:Half-Open Real Interval|interval]] $\closedint {-\frac 1 2} 0$.
+Also note that all [[Definition:Prime Number|primes]] are greater than or equal to $2$.
+Thus it follows that:
+:$-\dfrac 2 {p_k} < \map \ln {1 - \dfrac 1 {p_k} }$
+Applying this to $(1)$ yields:
+:$\displaystyle -2 \sum_{k \mathop = 1}^n \dfrac 1 {p_k} < -\ln \ln p_n$
+and so:
+:$\displaystyle \sum_{k \mathop = 1}^n \dfrac 1 {p_k} > \dfrac 1 2 \ln \ln p_n$
+But:
+:$\displaystyle \lim_{n \mathop \to \infty} \ln \ln p_n \to \infty$
+and so the [[Definition:Series|series]]:
+:$\displaystyle \sum_{p \mathop \in \Bbb P} \frac 1 p$
+is [[Definition:Divergent Series|divergent]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sum of Reciprocals of Primes is Divergent/Lemma}
+Tags: Number Theory
+
+\begin{theorem}
+Let $C \in \R_{>0}$ be a [[Definition:Strictly Positive Real Number|(strictly) positive real number]].
+Then:
+:$\displaystyle \lim_{n \mathop \to \infty} \paren {\map \ln {\ln n} - C} = + \infty$
+\end{theorem}
+
+\begin{proof}
+Fix $c \in \R$.
+It is sufficient to show there exists $N \in \N$, such that:
+:$(1): \quad n \ge N \implies \map \ln {\ln n} - C > c$
+Proceed as follows:
+{{begin-eqn}}
+{{eqn | l = \map \ln {\ln n} - C
+      | o = >
+      | r = c
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \ln n
+      | o = >
+      | r = \map \exp {c + C}
+      | c = {{Defof|Exponential Function/Real|subdef = Sum of Series|Exponential}}
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = n
+      | o = >
+      | r = \map \exp {\map \exp {c + C} }
+      | c = {{Defof|Exponential Function/Real|subdef = Sum of Series|Exponential}}
+}}
+{{end-eqn}}
+Let $N \in \N$ such that $N > \map \exp {\map \exp {c + C} }$.
+By [[Logarithm is Strictly Increasing]] it follows that $N$ satisfies condition $(1)$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Strictly Increasing Infinite Sequence of Integers is Cofinal in Natural Numbers}
+Tags: Integers
+
+\begin{theorem}
+Let $S = \left\langle{x_n}\right\rangle$ be an [[Definition:Infinite Sequence|infinite sequence]] of [[Definition:Integer|integers]] which is [[Definition:Strictly Increasing Sequence|strictly increasing]].
+ 
+Then $S$ is a [[Definition:Cofinal Subset|cofinal subset of $\left({\Z, \le}\right)$]] where $\le$ is the [[Definition:Usual Ordering|usual ordering]] on the [[Definition:Integer|integers]].
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted}}
+[[Category:Integers]]
+4lesc476t30awzjfte4szdcdh1hi33j
+\end{proof}<|endoftext|>
+\section{Product of Even and Odd Functions}
+Tags: Real Analysis
+
+\begin{theorem}
+Let $\mathcal O$ be an [[Definition:Odd Function|odd]] [[Definition:Real Function|real function]] defined on some [[Definition:Symmetric Set|symmetric set]] $S$.
+Let $\mathcal E$ be an [[Definition:Even Function|even]] [[Definition:Real Function|real function]] defined on some [[Definition:Symmetric Set|symmetric set]] $S'$.
+Let $\mathcal O\mathcal E$ be their [[Definition:Pointwise Multiplication of Real-Valued Functions|pointwise product]], defined on the [[Definition:Set Intersection|intersection]] of the [[Definition:Domain (Set Theory)|domains]] of $\mathcal O$ and $\mathcal E$.
+Then $\mathcal O\mathcal E$ is odd. 
+That is:
+:$\forall x \in S \cap S': \left({\mathcal O\mathcal E}\right)\left({-x}\right) = - \left({\mathcal O\mathcal E}\right)\left({x}\right)$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \left({\mathcal O\mathcal E}\right)\left({-x}\right)
+      | r = \mathcal O\left({-x}\right)\mathcal E\left({-x}\right)
+      | c = Definition of [[Definition:Pointwise Multiplication of Real-Valued Functions|Pointwise Multiplication of Real-Valued Functions]]
+}}
+{{eqn | r = -\mathcal O\left({x}\right)\mathcal E\left({x}\right)
+      | c = as $\mathcal O$ is [[Definition:Odd Function|odd]] and $\mathcal E$ is [[Definition:Even Function|even]]
+}}
+{{eqn | r = -\left({\mathcal O\mathcal E}\right)\left({x}\right)
+}}
+{{end-eqn}}
+The result follows from the definition of an [[Definition:Odd Function|odd function]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Millin Series}
+Tags: Fibonacci Numbers
+
+\begin{theorem}
+The '''Millin series''' is the [[Definition:Series|series]] defined as:
+:$\displaystyle \sum_{n \mathop = 0}^\infty \frac 1 {F_{2^n}} = \frac {7 - \sqrt 5} 2$
+\end{theorem}
+
+\begin{proof}
+First we will prove that:
+:$\displaystyle \sum_{r \mathop = 0}^n \frac 1 {F_{2^r}} = 3 - \frac {F_{2^n - 1}} {F_{2^n}}$
+for $n \ge 1$.
+We see that:
+:$\dfrac 1 {F_1} + \dfrac 1 {F_2} = 2 = 3 - \dfrac {F_1} {F_2}$ 
+so the proposition holds for $n = 1$.
+Suppose the proposition is true for $n = k$.
+Then:
+{{begin-eqn}}
+{{eqn |l = \sum_{r \mathop = 0}^{k + 1}  \frac 1 {F_{2^r} } 
+      |r = 3 - \frac {F_{2^k - 1} } {F_{2^k} } + \frac 1 {F_{2^{k + 1} } }
+      |c = 
+}}
+{{eqn |r = 3 - \frac {F_{2^k - 1} F_{2^{k + 1} } - F_{2^k} } {F_{2^k} F_{2^{k + 1} } }
+      |c = 
+}}
+{{eqn |r = 3 - \frac {F_{2^k - 1} F_{2^k} \left({F_{2^k + 1} + F_{2^k - 1} }\right) - F_{2^k} } {F_{2^k} F_{2^{k + 1} } }
+      |c = by repeated [[Definition:Fibonacci Numbers|Fibonacci]] recurrence formula on $F_{2^{k + 1} }$ 
+}}
+{{eqn |r = 3 - \frac {F_{2^k - 1} F_{2^k + 1} + F_{2^k - 1}^2 - 1} {F_{2^{k + 1} } }
+      |c = 
+}}
+{{eqn |r = 3 - \frac {F_{2^k}^2 + F_{2^k - 1}^2} {F_{2^{k + 1} } }
+      |c = by [[Cassini's Identity]]
+}}
+{{eqn |r = 3 - \frac {F_{2^{k + 1} - 1} } {F_{2^{k + 1} } }
+      |c = by repeated [[Definition:Fibonacci Numbers|Fibonacci]] recurrence formula on $F_{2^{k + 1} - 1}$ 
+}}
+{{end-eqn}}
+Thus by [[Principle of Mathematical Induction]], the proof is complete. 
+{{qed|lemma}}
+Now taking the limit, we have:
+{{begin-eqn}}
+{{eqn |l = \sum_{r \mathop = 0}^\infty \frac 1 {F_{2^r} } 
+      |r = \lim_{n \mathop \to \infty} \sum_{r \mathop = 0}^n \frac 1 {F_{2^r} }
+      |c = 
+}}
+{{eqn |r = \lim_{n \mathop \to \infty} \left(3 - \frac {F_{2^n - 1} } {F_{2^n } }\right)
+      |c = 
+}}
+{{eqn |r = 3 - \frac 2 {1 + \sqrt 5}
+      |c = by [[Ratio of Consecutive Fibonacci Numbers]]
+}}
+{{eqn |r = 3 - \frac {\sqrt 5 - 1} 2
+      |c = 
+}}
+{{eqn |r = \frac {7 - \sqrt 5} 2 
+      |c = 
+}}
+{{end-eqn}}
+as required.
+{{qed}}
+{{Namedfor|D.A. Millin|cat=Millin}}
+\end{proof}<|endoftext|>
+\section{Odd Bernoulli Numbers Vanish}
+Tags: Bernoulli Numbers, Odd Bernoulli Numbers Vanish
+
+\begin{theorem}
+Let $B_n$ denote the $n$th [[Definition:Bernoulli Numbers|Bernoulli Number]].
+Then:
+:$B_{2n + 1} = 0$
+for $n \ge 1$.
+\end{theorem}
+
+\begin{proof}
+By definition of the [[Definition:Bernoulli Numbers|Bernoulli numbers]]:
+:$\displaystyle \frac x {e^x - 1} = \sum_{n \mathop = 0}^\infty \frac{B_n x^n} {n!}$
+or equivalently:
+:$\displaystyle \frac x {e^x - 1} + \frac x 2 = 1 + \sum_{n \mathop = 2}^\infty \frac{B_n x^n} {n!}$
+It remains to be shown that this is an [[Definition:Even Function|even function]].
+{{explain}}
+{{begin-eqn}}
+{{eqn | l = \frac x {e^x - 1} + \frac x 2
+      | r = \frac {x \left({1 + e^x}\right)} {2\left({e^x - 1}\right)}
+      | c = 
+}}
+{{eqn | r = \frac {-x \left({e^{-x} + 1}\right)} {2\left({e^{-x} - 1}\right)}
+      | c = 
+}}
+{{eqn | r = \frac {-x \left({2 + e^{-x} - 1}\right)} {2\left({e^{-x} - 1}\right)}
+      | c = 
+}}
+{{eqn | r = \frac {-x} {e^{-x} - 1} + \frac {-x} 2
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+By definition, the [[Definition:Bernoulli Numbers/Generating Function|Bernoulli numbers]] are given by:
+:$\displaystyle \frac x {e^x - 1} = \sum_{n \mathop = 0}^\infty \frac {B_n x^n} {n!}$
+We have:
+{{begin-eqn}}
+{{eqn | l = \frac x {e^x - 1}
+      | r = \frac x 2 \left({\frac 2 {e^x -  1} }\right)
+      | c = 
+}}
+{{eqn | r = \frac x 2 \left({\frac {e^x - e^x + 2} {e^x -  1} }\right)
+      | c = 
+}}
+{{eqn | r = \frac x 2 \left({\frac {\left({e^x + 1}\right) - \left({e^x - 1}\right)} {e^x -  1} }\right)
+      | c = 
+}}
+{{eqn | r = \frac x 2 \left({\frac {e^x + 1} {e^x - 1} - 1}\right)
+      | c = 
+}}
+{{eqn | r = -\frac x 2 + \frac x 2 \left({\frac {e^x + 1} {e^x - 1} }\right)
+      | c = 
+}}
+{{end-eqn}}
+Take $f \left({x}\right) := \dfrac x 2 \left({\dfrac {e^x + 1} {e^x - 1} }\right)$, and note that:
+{{begin-eqn}}
+{{eqn | l = f \left({-x}\right)
+      | r = \frac {-x} 2 \left({\dfrac {e^{-x} + 1} {e^{-x} - 1} }\right)
+      | c = 
+}}
+{{eqn | r = -\frac {-x} 2 \left({\dfrac {1 + e^x} {1 - e^x} }\right)
+      | c = multiplying [[Definition:Numerator|top]] and [[Definition:Denominator|bottom]] by $e^x$
+}}
+{{eqn | r = -\frac {-x} 2 \left({\dfrac {e^x + 1} {-\left({e^x - 1}\right)} }\right)
+      | c = 
+}}
+{{eqn | r = \frac x 2 \left({\frac {e^x + 1} {e^x - 1} }\right)
+      | c = 
+}}
+{{eqn | r = f \left({x}\right)
+      | c = 
+}}
+{{end-eqn}}
+and so $f \left({x}\right) := \dfrac x 2 \left({\dfrac {e^x + 1} {e^x - 1} }\right)$ is an [[Definition:Even Function|even function]].
+Thus $f \left({x}\right)$ can be seen to take the form:
+:$\displaystyle f \left({x}\right) = 1 + \sum_{n \mathop = 2}^\infty \frac {B_n} {n!} x^n$
+and so:
+:$\displaystyle f \left({x}\right) = 1 + \sum_{n \mathop = 2}^\infty \frac {B_n} {n!} x^{-n}$
+That is:
+:$\forall n \in \N: n > 1: \dfrac {B_n} {n!} x^n = \dfrac {B_n} {n!} x^{-n}$
+and so for [[Definition:Odd Number|odd]] $n$ where $n > 1$ it follows that $B_n = 0$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Gauss Multiplication Formula}
+Tags: Gamma Function
+
+\begin{theorem}
+Let $\Gamma$ denote the [[Definition:Gamma Function|Gamma Function]].
+Then:
+:$\displaystyle \forall z \notin \set {-\frac m n: m \in \N}: \prod_{k \mathop = 0}^{n - 1} \map \Gamma {z + \frac k n} = \paren {2 \pi}^{\paren {n - 1} / 2} n^{1/2 - n z} \map \Gamma {n z}$
+where $\N$ denotes the [[Definition:Natural Numbers|natural numbers]].
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map \Gamma {z + \frac k n}
+      | r = \paren {z + \frac k n - 1} \map \Gamma {z + \frac k n - 1}
+      | c = [[Gamma Difference Equation]]
+}}
+{{eqn | r = \lim_{m \mathop \to \infty} \frac {m! m^{z + k / n - 1} } {\paren {z + \frac k n} \paren {z + \frac k n + 1} \cdots \paren {z + \frac k n - 1 + m} } 
+      | c = {{Defof|Euler Form of Gamma Function}}
+}}
+{{eqn | r = \lim_{m \mathop \to \infty} \frac {\sqrt {2 \pi m} \paren {\frac m e}^m m^{z + k / n - 1} } {\paren {z + \frac k n} \paren {z + \frac k n + 1} \cdots \paren {z + \frac k n - 1 + m} }
+      | c = [[Stirling's Formula]]
+}}
+{{eqn | r = \lim_{m \mathop \to \infty} \frac {\sqrt {2 \pi} \paren {\frac {m n} e}^m m^{z + k / n - 1/2} } {\paren {n z + k} \paren {n z + k + n} \cdots \paren {n z + k - n + m n} }
+      | c = 
+}}
+{{end-eqn}}
+Taking the product for $k = 0$ to $n - 1$, we have:
+{{begin-eqn}}
+{{eqn | l = \prod_{k \mathop = 0}^{n - 1} \map \Gamma {z + \frac k n}
+      | r = \lim_{m \mathop \to \infty} \frac {\paren {\sqrt {2 \pi} }^n \paren {\frac {m n} e}^{m n} m^{n z  - n / 2} m^{\sum_{k \mathop = 0}^{n - 1} k / n} } {\paren {n z} \paren {n z + 1} \cdots \paren {n z - 1 + m n} }
+      | c = 
+}}
+{{eqn | r = \lim_{m \mathop \to \infty} \frac {\paren {\sqrt {2 \pi} }^n \paren {\frac {m n} e}^{m n} m^{n z  - 1 / 2} } {\paren {n z} \paren {n z + 1} \cdots \paren {n z - 1 + m n} }
+      | c = [[Sum of Arithmetic Sequence]]
+}}
+{{eqn | r = \lim_{m \mathop \to \infty} \frac {\paren {\sqrt {2 \pi} }^n \paren {\frac m e}^m m^{n z  - 1 / 2} n^{1 / 2 - n z} } {\paren {n z} \paren {n z + 1} \cdots \paren {n z - 1 + m} }
+      | c = by substituting $m n \mapsto m$
+}}
+{{eqn | r = \lim_{m \mathop \to \infty} \frac {\paren {\sqrt {2 \pi} }^{n - 1} m! m^{n z  - 1} n^{1 / 2 - n z} } {\paren {n z} \paren {n z + 1} \cdots \paren {n z - 1 + m} }
+      | c = [[Stirling's Formula]]
+}}
+{{eqn | r = \paren {2 \pi}^{\paren {n - 1} / 2} n^{1 / 2 - n z} \paren {n z - 1} \map \Gamma {n z - 1}
+      | c = {{Defof|Euler Form of Gamma Function}}
+}}
+{{eqn | r = \paren {2 \pi}^{\paren {n - 1} / 2} n^{1 / 2 - n z} \map \Gamma {n z}
+      | c = [[Gamma Difference Equation]]
+}}
+{{end-eqn}}
+{{qed}}
+{{Namedfor|Carl Friedrich Gauss|cat = Gauss}}
+\end{proof}<|endoftext|>
+\section{Dirichlet Integral}
+Tags: Integral Calculus, Dirichlet Integral, Definite Integrals involving Sine Function
+
+\begin{theorem}
+:$\displaystyle \int_0^\infty \frac {\sin x} x \rd x = \frac \pi 2$
+\end{theorem}
+
+\begin{proof}
+By [[Fubini's Theorem]]:
+:$\displaystyle \int_0^\infty \paren {\int_0^\infty e^{- x y} \sin x \rd y} \rd x = \int_0^\infty \paren {\int_0^\infty e^{- x y} \sin x \rd x} \rd y$
+Then:
+{{begin-eqn}}
+{{eqn | l = \int_0^\infty e^{- x y} \sin x \rd y
+      | r = \intlimits {-e^{- x y} \frac {\sin x} x} 0 \infty 
+      | c = [[Primitive of Exponential of a x|Primitive of $e^{a x}$]]
+}}
+{{eqn | r = \frac {\sin x} x
+      | c = 
+}}
+{{end-eqn}}
+and:
+{{begin-eqn}}
+{{eqn | l = \int_0^\infty e^{- x y} \sin x \rd x
+      | r = \intlimits {\frac {-e^{- x y} \paren {y \sin x + \cos x} } {y^2 + 1} } 0 \infty
+      | c = [[Primitive of Exponential of a x by Sine of b x|Primitive of $e^{a x} \sin b x$]]
+}}
+{{eqn | r = \frac 1 {y^2 + 1}
+      | c = 
+}}
+{{end-eqn}}
+Hence:
+{{begin-eqn}}
+{{eqn | l = \int_0^\infty \frac {\sin x} x \rd x
+      | r = \int_0^\infty \frac 1 {y^2 + 1} \rd y
+      | c =
+}}
+{{eqn | r = \bigintlimits {\arctan y} 0 \infty
+      | c = [[Primitive of Reciprocal of x squared plus a squared/Arctangent Form|Primitive of $\dfrac 1 {x^2 + a^2}$]]
+}}
+{{eqn | r = \frac \pi 2
+      | c = as $\displaystyle \lim_{y \mathop \to \infty} \arctan y = \frac \pi 2$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Define:
+:$\displaystyle \map I \alpha = \int_0^\infty \frac {e^{-\alpha x} \sin x} x \rd x$
+Using [[Definite Integral of Partial Derivative]]:
+{{begin-eqn}}
+{{eqn | l = \map {I'} \alpha
+      | r = \frac \partial {\partial \alpha} \int_0^\infty \frac {e^{-\alpha x} \sin x} x \rd x
+}}
+{{eqn | r = \int_0^\infty \frac \partial {\partial \alpha} \frac {e^{-\alpha x} \sin x} x \rd x
+      | c = [[Leibniz's Integral Rule]]
+}}
+{{eqn | r = -\int_0^\infty e^{-\alpha x} \sin x \rd x
+      | c = [[Derivative of Exponential Function]]
+}}
+{{eqn | r = \intlimits {-\frac {e^{-\alpha x} \paren {-\alpha \sin x + \cos x} } {\paren {-\alpha}^2 + 1} } 0 \infty 
+      | c = [[Primitive of Exponential of a x by Sine of b x|Primitive of $e^{\alpha x} \sin b x$]]
+}}
+{{eqn | r = -\frac 1 {\alpha^2 + 1} 
+}}
+{{end-eqn}}
+Integrating with respect to $\alpha$:
+{{begin-eqn}}
+{{eqn | l = \map I \alpha
+      | r = -\int \frac 1 {\alpha^2 + 1} \rd \alpha
+}}
+{{eqn | r = C - \arctan \alpha
+      | c = [[Primitive of Reciprocal of x squared plus a squared/Arctangent Form|Primitive of $\dfrac 1 {x^2 + a^2}$]]
+}}
+{{end-eqn}}
+To determine $C$, take $\alpha \to \infty$:
+{{begin-eqn}}
+{{eqn | l = \lim_{\alpha \mathop \to \infty} \int_0^\infty \frac {e^{-\alpha x} \sin x} x \rd x
+      | r = \lim_{\alpha \mathop \to \infty} \paren {C - \arctan \alpha}
+}}
+{{eqn | ll= \leadsto
+      | l = \int_0^\infty \frac {0 \times \sin x} x
+      | r = C - \frac \pi 2
+}}
+{{eqn | ll= \leadsto
+      | l = C
+      | r = \frac \pi 2
+}} 
+{{end-eqn}}
+Hence:
+:$\displaystyle \int_0^\infty \frac {e^{-\alpha x} \sin x} x \rd x = \frac \pi 2 - \arctan \alpha$
+Setting $\alpha = 0$ yields:
+:$\displaystyle \int_0^\infty \frac {\sin x} x \rd x = \frac \pi 2$
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let:
+:$\map f z = \dfrac {e^{i z} - 1} z$
+We have, by [[Euler's Formula]], on the [[Definition:real Number Line|real line]]:
+:$\map \Im {\map f z} = \dfrac {\sin z} z$
+So: 
+:$\displaystyle \map \Im {\int_0^\infty \dfrac {e^{i x} - 1} x \rd x} = \int_0^\infty \dfrac {\sin x} x \rd x$
+Let $C_R$ be the semicircular [[Definition:Contour (Complex Plane)|contour]] of radius $R$ situated on the upper [[Definition:Half-Plane|half plane]], centred at the origin, traversed anti-clockwise.
+Let $\Gamma_R = C_R \cup \closedint {-R} R$.
+Then, by [[Contour Integral of Concatenation of Contours]]:
+:$\displaystyle \oint_{\Gamma_R} \frac {e^{i z} - 1} z \rd z = \int_{C_R} \frac {e^{i z} - 1} z \rd z + \int_{-R}^R \frac {e^{i x} - 1} x \rd x$
+From [[Linear Combination of Contour Integrals]], we write: 
+:$\displaystyle \oint_{\Gamma_R} \frac {e^{i z} - 1} z \rd z = \int_{C_R} \frac {e^{i z} } z \rd z - \int_{C_R} \frac {\rd z} z + \int_{-R}^R \frac {e^{i x} - 1} x \rd x$
+The only [[Definition:Isolated Singularity|singularity]] of $f$ is at $z = 0$. However, we can show $\lim_{z \mathop \to 0} \map f z$ to be finite: 
+{{begin-eqn}}
+{{eqn	| l = \lim_{z \mathop \to 0} \frac {e^{i z} - 1} z
+	| r = \intlimits {\frac {\map \d {e^{i z} } } {\d z} } {z = 0} {}
+	| c = {{Defof|Derivative of Complex Function}}
+}}
+{{eqn	| r = i e^0
+	| c = [[Derivative of Exponential of a x]]
+}}
+{{eqn	| r = i
+}}
+{{end-eqn}}
+So, $z = 0$ is a [[Definition:Removable Singularity|removable singularity]] of $f$.
+Therefore, $f$ is [[Definition:Holomorphic Function|holomorphic]] inside our contour. 
+It then follows from the [[Cauchy-Goursat Theorem]], that:
+:$\displaystyle \oint_{\Gamma_R} \frac {e^{i z} - 1} z \rd z = 0$
+We also have: 
+{{begin-eqn}}
+{{eqn	| l = \size {\int_{C_R} \frac { e^{i z} } z \rd z}
+	| o = \le
+	| r = \pi \max_{0 \mathop \le \theta \mathop \le \pi} \size {\frac 1 {R e^{i \theta} } }
+	| c = [[Jordan's Lemma]]
+}}
+{{eqn	| r = \frac \pi R
+}}
+{{eqn	| o = \to
+	| r = 0
+	| c = as $R \to \infty$
+}}
+{{end-eqn}}
+Therefore: 
+:$\displaystyle \lim_{R \mathop \to \infty} \int_{C_R} \frac {\rd z} z = \lim_{R \mathop \to \infty} \int_{-R}^R \frac {e^{i x} - 1} x \rd x = \int_{-\infty}^\infty \frac {e^{i x} - 1} x \rd x$
+Evaluating the integral on the left hand side: 
+{{begin-eqn}}
+{{eqn	| l = \int_{C_R} \frac {\rd z} z
+	| r = \int_0^\pi \frac {i R e^{i \theta} } {R e^{i \theta} } \rd \theta
+	| c = {{Defof|Complex Contour Integral}}
+}}
+{{eqn	| r = i \int_0^\pi \rd \theta
+}}
+{{eqn	| r = \pi i
+}}
+{{end-eqn}}
+So:
+:$\displaystyle \int_{-\infty}^\infty \frac {e^{i x} - 1} x \rd x = \pi i$
+Taking the [[Definition:Imaginary Part|imaginary part]]:
+:$\displaystyle \int_{-\infty}^\infty \frac {\sin x} x \rd x = \pi$
+From [[Definite Integral of Even Function]]:
+:$\displaystyle \int_{-\infty}^\infty \frac {\sin x} x \rd x = 2 \int_0^\infty \frac {\sin x} x \rd x$
+Hence: 
+:$\displaystyle \int_0^\infty \frac {\sin x} x \rd x = \frac \pi 2$
+{{qed}}
+\end{proof}
+
+\begin{proof}
+From [[Integral to Infinity of Function over Argument]]:
+:$\displaystyle \int_0^\infty {\dfrac {\map f x} x} = \int_0^{\to \infty} \map F u \rd u$
+for a [[Definition:Real Function|real function]] $f$ and its [[Definition:Laplace Transform|Laplace transform]] $\laptrans f = F$, provided they exist.
+Let $\map f x := \sin x$.
+Then from [[Laplace Transform of Sine]]:
+:$\laptrans {\map f x} = \dfrac 1 {s^2 + 1}$
+Hence:
+{{begin-eqn}}
+{{eqn | l = \int_0^\infty \frac {\sin x} x \rd x
+      | r = \int_0^{\to \infty} \dfrac {\d u} {u^2 + 1}
+      | c = 
+}}
+{{eqn | r = \bigintlimits {\arctan u} 0 \infty
+      | c = [[Primitive of Reciprocal of x squared plus a squared/Arctangent Form|Primitive of $\dfrac 1 {x^2 + a^2}$]]
+}}
+{{eqn | r = \dfrac \pi 2
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Riemann Zeta Function and Prime Counting Function}
+Tags: Riemann Zeta Function, Number Theory
+
+\begin{theorem}
+For $\map \Re s > 1$:
+:$\displaystyle \log \map \zeta s = s \int_0^{\mathop \to \infty} \frac {\map \pi x} {x \paren {x^s - 1} } \rd x$
+where:
+:$\zeta$ denotes the [[Definition:Riemann Zeta Function|Riemann Zeta Function]]
+:$\pi$ denotes the [[Definition:Prime-Counting Function|Prime-Counting Function]].
+\end{theorem}
+
+\begin{proof}
+From the definition of the [[Definition:Riemann Zeta Function|Riemann Zeta Function]]: 
+{{begin-eqn}}
+{{eqn | l = \map \zeta s
+      | r = \prod_p \frac 1 {1 - p^{-s} }
+      | c =
+}}
+{{eqn | ll= \leadsto
+      | l = \log \map \zeta s
+      | r = \log \prod_p \frac 1 {1 - p^{-s} } 
+      | c = 
+}}
+{{eqn | r = \sum_p \map \log {\frac 1 {1 - p^{-s} } }
+      | c = [[Sum of Logarithms]]
+}}
+{{eqn | r = \sum_{n \mathop = 0}^\infty \paren {\map \pi n - \map \pi {n - 1} } \map \log {\frac 1 {1 - n^{-s} } }
+      | c = {{Defof|Prime-Counting Function}}
+}}
+{{eqn | r = \sum_{n \mathop = 0}^\infty \map \pi n \paren {\map \log {\frac 1 {1 - n^{-s} } } - \map \log {\frac 1 {1 - \paren {n + 1}^{-s} } } }
+      | c = as $\map \pi {-1} = 0$
+}}
+{{eqn | r = \sum_{n \mathop = 0}^\infty \map \pi n \paren {\map \log {1 - \paren {n + 1}^{-s} } - \map \log {1 - n^{-s} } }
+      | c = [[Logarithm of Power]]
+}}
+{{end-eqn}}
+By [[Derivative of Logarithm Function]] and the [[Chain Rule for Derivatives]]:
+{{begin-eqn}}
+{{eqn | l = \frac \d {\d x} \map \log {1 - x^{-s} }
+      | r = \frac {s x^{- s - 1} } {1 - x^{-s} }
+      | c = 
+}}
+{{eqn | r = \frac s {x \paren {x^s - 1} }
+      | c = 
+}}
+{{end-eqn}}
+Hence:
+{{begin-eqn}}
+{{eqn | l = \log \map \zeta s
+      | r = \sum_{n \mathop = 0}^\infty \map \pi n \int_n^{n + 1} \frac s {x \paren {x^s - 1} } \rd x
+      | c = [[Fundamental Theorem of Calculus/Second Part|Fundamental Theorem of Calculus]]
+}}
+{{eqn | r = \sum_{n \mathop = 0}^\infty s \int_n^{n + 1} \frac {\map \pi x} {x \paren {x^s - 1} } \rd x
+      | c = 
+}}
+{{eqn | r = s \int_0^{\mathop \to \infty} \frac {\map \pi x} {x \paren {x^s - 1} } \rd x
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Riemann Zeta Function]]
+[[Category:Number Theory]]
+gi6rrfeuq1n3xjizk0oausf9t4gexvi
+\end{proof}<|endoftext|>
+\section{Faulhaber's Formula}
+Tags: Number Theory, Bernoulli Numbers
+
+\begin{theorem}
+Let $n$ and $p$ be [[Definition:Strictly Positive Integer|positive integers]].
+Then:
+:$\displaystyle \sum_{k \mathop = 1}^n k^p = \frac 1 {p + 1} \sum_{i \mathop = 0}^p \paren {-1}^i \binom {p + 1} i B_i n^{p + 1 - i}$
+where $B_n$ denotes the $n$th [[Definition:Bernoulli Numbers|Bernoulli number]].
+\end{theorem}
+
+\begin{proof}
+Let $x \ge 0$.
+{{begin-eqn}}
+{{eqn | l = \sum_{k \mathop = 0}^{n - 1} e^{k x} 
+      | r = \sum_{k \mathop = 0}^{n - 1} \sum_{p \mathop = 0}^\infty \frac {\paren {k x}^p}
+ {p!}
+      | c = [[Power Series Expansion for Exponential Function]]
+}}
+{{eqn | r = \sum_{p \mathop = 0}^\infty \paren {\sum_{k \mathop = 0}^{n - 1} k^p} \frac {x^p} {p!} 
+      | c = rearrangement is valid by [[Tonelli's Theorem]]
+}}
+{{end-eqn}}
+We also have:
+{{begin-eqn}}
+{{eqn | l = \sum_{k \mathop = 0}^{n - 1} e^{k x} 
+      | r = \frac {1 - e^{n x} } {1 - e^x}
+      | c = [[Sum of Infinite Geometric Sequence]]
+}}
+{{eqn | r = \frac {e^{n x} - 1} x \frac x {e^x - 1}
+      | c =
+}}
+{{eqn | r = \sum_{p \mathop = 0}^\infty \frac {n^{p + 1} x^p} {\paren {p + 1}!} \sum_{p \mathop = 0}^\infty \frac {B_p x^p} {p!}
+      | c = {{Defof|Bernoulli Numbers}}
+}}
+{{eqn | r = \sum_{p \mathop = 0}^\infty \sum_{i \mathop = 0}^p \frac {n^{p + 1 - i} x^{p - i} } {\paren {p + 1 - i}!} \frac {B_i x^i} {i!}
+      | c = {{Defof|Cauchy Product}}
+}}
+{{eqn | r = \sum_{p \mathop = 0}^\infty \paren {\frac 1 {p + 1} \sum_{i \mathop = 0}^p \binom {p + 1} i B_i n^{p + 1 - i} } \frac {x^p} {p!}
+      | c = 
+}}
+{{end-eqn}}
+Equating coefficients:
+{{begin-eqn}}
+{{eqn | l = \sum_{k \mathop = 0}^{n - 1} k^p 
+      | r = \frac 1 {p + 1} \sum_{i \mathop = 0}^p \binom {p + 1} i B_i n^{p + 1 - i}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \sum_{k \mathop = 1}^n k^p 
+      | r = \frac 1 {p + 1} \sum_{i \mathop = 0}^p \paren {-1}^i \binom {p + 1} i B_i n^{p + 1 - i}
+      | c = as $B_1 = -\dfrac 1 2$ and [[Odd Bernoulli Numbers Vanish]]
+}}
+{{end-eqn}}
+{{qed}}
+{{Namedfor|Johann Faulhaber|cat = Faulhaber}}
+[[Category:Number Theory]]
+[[Category:Bernoulli Numbers]]
+rct8iugjp1o10ncpbqbea9nl00enaom
+\end{proof}<|endoftext|>
+\section{Complex Exponential Tends to Zero}
+Tags: Exponential Function
+
+\begin{theorem}
+Let $\exp z$ be the [[Definition:Complex Exponential Function|complex exponential function]].
+Then:
+:$\displaystyle \lim_{\map \Re z \mathop \to +\infty} e^{-z} = 0$
+where $\map \Re z$ denotes the [[Definition:Real Part|real part]] of $z$.
+\end{theorem}
+
+\begin{proof}
+Let $z = x + iy$.
+Let $\epsilon > 0$.
+By the definition of [[Definition:Limit at Infinity|limits at infinity]], we need to show that there is some $M > 0$ such that:
+:$x > M \implies \size {e^{-z} - 0} < \epsilon$
+But:
+{{begin-eqn}}
+{{eqn | l = \size {e^{-z} - 0}
+      | r = \size {e^{-z} }
+}}
+{{eqn | r = \size {e^{-x} }
+      | c = [[Modulus of Exponential is Modulus of Real Part]]
+}}
+{{eqn | r = \size {e^{-x} - 0}
+}}
+{{end-eqn}}
+so we need an $M$ such that:
+:$x > M \implies \size {e^{-x} - 0} < \epsilon$
+This is the definition of the [[Definition:Limit at Infinity|limits at infinity]] of the [[Definition:Real Exponential Function|''real'' exponential]].
+The result follows from [[Exponential Tends to Zero and Infinity]].
+{{qed}}
+[[Category:Exponential Function]]
+axmix7l39dppp8rend957tp5aykx8u2
+\end{proof}<|endoftext|>
+\section{Modulus of Exponential is Exponential of Real Part}
+Tags: Complex Modulus, Exponential Function
+
+\begin{theorem}
+Let $z \in \C$ be a [[Definition:Complex Number|complex number]].
+Let $\exp z$ denote the [[Definition:Complex Exponential Function|complex exponential function]].
+Let $\cmod {\, \cdot \,}$ denote the [[Definition:Complex Modulus|complex modulus]]
+Then:
+:$\cmod {\exp z} = \map \exp {\Re z}$
+where $\Re z$ denotes the [[Definition:Real Part|real part]] of $z$.
+\end{theorem}
+
+\begin{proof}
+Let $z = x + iy$.
+{{begin-eqn}}
+{{eqn | l = \cmod {\exp z}
+      | r = \cmod {\map \exp {x + iy} }
+}}
+{{eqn | r = \cmod {\paren {\exp x} \paren {\exp i y} }
+      | c = [[Exponential of Sum/Complex Numbers|Exponential of Sum]]
+}}
+{{eqn | r = \cmod {\exp x} \cmod {\exp i y}
+      | c = [[Complex Modulus of Product of Complex Numbers|Modulus of Product]]
+}}
+{{eqn | r = \cmod {\exp x}
+      | c = [[Modulus of Exponential of Imaginary Number is One]]
+}}
+{{eqn | r = e^x
+      | c = [[Exponential of Real Number is Strictly Positive]]
+}}
+{{eqn | r = \map \exp {\Re z}
+      | c = {{Defof|Real Part}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Laplace Transform of Exponential/Real Argument}
+Tags: Laplace Transforms, Exponential Function, Laplace Transform of Exponential
+
+\begin{theorem}
+Let $\laptrans f$ denote the [[Definition:Laplace Transform|Laplace transform]] of a [[Definition:Function|function]] $f$.
+Let $e^x$ be the [[Definition:Real Exponential Function|real exponential]].
+Then:
+:$\map {\laptrans {e^{a t} } } s = \dfrac 1 {s - a}$
+where $a \in \R$ is [[Definition:Constant|constant]], and $\map \Re s > \map \Re a$.
+\end{theorem}<|endoftext|>
+\section{Laplace Transform of Cosine}
+Tags: Laplace Transforms, Cosine Function, Laplace Transform of Cosine
+
+\begin{theorem}
+Let $\cos$ be the [[Definition:Real Cosine Function|real cosine function]].
+Let $\laptrans f$ denote the [[Definition:Laplace Transform|Laplace transform]] of the [[Definition:Real Function|real function]] $f$.
+Then:
+:$\laptrans {\cos a t} = \dfrac s {s^2 + a^2}$
+where $a \in \R_{>0}$ is [[Definition:Constant|constant]], and $\map \Re s > a$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \laptrans {e^{i a t} }
+      | r = \frac 1 {s - i a}
+      | c = [[Laplace Transform of Exponential]]
+}}
+{{eqn | r = \frac {s + i a} {s^2 + a^2}
+      | c = multiply top and bottom by $s + i a$
+}} 
+{{end-eqn}}
+Also:
+{{begin-eqn}}
+{{eqn | l = \laptrans {e^{i a t} }
+      | r = \laptrans {\cos a t + i \sin a t}
+      | c = [[Euler's Formula]]
+}}
+{{eqn | r = \laptrans {\cos a t} + i \laptrans {\sin a t}
+      | c = [[Linear Combination of Laplace Transforms]]
+}}
+{{end-eqn}}
+So:
+{{begin-eqn}}
+{{eqn | l = \laptrans {\cos a t}
+      | r = \map \Re {\laptrans {e^{i a t} } }
+}}
+{{eqn | r = \map \Re {\frac {s + i a} {s^2 + a^2} }
+}}
+{{eqn | r = \frac s {s^2 + a^2}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+By definition of the [[Definition:Laplace Transform|Laplace Transform]]:
+:$\displaystyle \laptrans {\cos at} = \int_0^{\to +\infty} e^{-s t} \cos at \rd t$
+From [[Integration by Parts]]:
+:$\displaystyle \int f g' \rd t = f g - \int f'g \rd t$
+Here:
+{{begin-eqn}}
+{{eqn | l = f
+      | r = \cos at
+}}
+{{eqn | ll= \leadsto
+      | l = f'
+      | r = -a \sin a t
+      | c = [[Derivative of Cosine Function]]
+}}
+{{eqn | l = g'
+      | r = e^{-s t}
+}}
+{{eqn | ll= \leadsto
+      | l = g
+      | r = -\frac 1 s e^{-s t}
+      | c = [[Primitive of Exponential Function]]
+}}
+{{end-eqn}}
+So:
+{{begin-eqn}}
+{{eqn | n = 1
+      | l = \int e^{-s t} \cos a t \rd t
+      | r = -\frac 1 s e^{-s t} \cos a t - \frac a s \int e^{-s t} \sin a t \rd t
+}}
+{{end-eqn}}
+Consider:
+:$\displaystyle \int e^{-s t} \sin a t \rd t$
+Again, using [[Integration by Parts]]:
+:$\displaystyle \int h j \,' \rd t = h j - \int h'j \rd t$
+Here:
+{{begin-eqn}}
+{{eqn | l = h
+      | r = \sin at
+}}
+{{eqn | ll= \leadsto
+      | l = h'
+      | r = a \cos at
+      | c = [[Derivative of Sine Function]]
+}}
+{{eqn | l = j\,'
+      | r = e^{-s t}
+}}
+{{eqn | ll= \leadsto
+      | l = j
+      | r = -\frac 1 s e^{-s t}
+      | c = [[Primitive of Exponential Function]]
+}}
+{{end-eqn}}
+So:
+{{begin-eqn}}
+{{eqn | l = \int e^{-s t} \sin a t \rd t
+      | r = -\frac 1 s e^{-s t} \sin a t + \frac a s \int e^{-s t} \cos a t \rd t
+}}
+{{end-eqn}}
+Substituting this into $(1)$:
+{{begin-eqn}}
+{{eqn | l = \int e^{-s t} \cos a t \rd t
+      | r = -\frac 1 s e^{-s t} \cos a t - \frac a s \paren {-\frac 1 s e^{-s t} \sin a t + \frac a s \int e^{-s t} \cos a t \rd t}
+}}
+{{eqn | r = -\frac 1 s e^{-s t} \cos a t + \frac a {s^2} e^{-s t} \sin a - \frac {a^2} {s^2} \int e^{-s t} \cos a t \rd t
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {1 + \frac {a^2} {s^2} } \int e^{-s t} \cos a t \rd t
+      | r = -\frac 1 s e^{-s t} \cos a t + \frac a {s^2} e^{-s t} \sin a t
+}}
+{{end-eqn}}
+Evaluating at $t = 0$ and $t \to +\infty$:
+{{begin-eqn}}
+{{eqn | l = \paren {1 + \frac {a^2} {s^2} } \laptrans {\cos at}
+      | r = \intlimits {-e^{-s t} \paren {\frac 1 s \cos a t - \frac a {s^2} \sin a t} } {t \mathop = 0} {t \mathop \to +\infty}
+}}
+{{eqn | r = 0 - \paren {-1 \paren {\frac 1 s \times 1 + \frac a {s^2} \times 0} }
+      | c = [[Boundedness of Real Sine and Cosine]], [[Complex Exponential Tends to Zero]]
+}}
+{{eqn | r = \frac 1 s
+}}
+{{eqn | ll= \leadsto
+      | l = \laptrans {\cos at}
+      | r = \frac 1 s \paren {1 + \frac {a^2} {s^2} }^{-1}
+}}
+{{eqn | r = \frac 1 s \paren {\frac {s^2} {a^2 + s^2} }
+}}
+{{eqn | r = \frac s {s^2 + a^2}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \laptrans {\cos a t}
+      | r = \laptrans {\frac {e^{i a t} + e^{-i a t} } 2}
+      | c = [[Cosine Exponential Formulation]]
+}}
+{{eqn | r = \frac 1 2 \paren {\laptrans {e^{i a t} } + \laptrans {e^{-i a t} } }
+      | c = [[Linear Combination of Laplace Transforms]]
+}}
+{{eqn | r = \frac 1 2 \paren {\frac 1 {s - i a} + \frac 1 {s + i a} }
+      | c = [[Laplace Transform of Exponential]]
+}}
+{{eqn | r = \frac 1 2 \paren {\frac {s + i a + s - i a} {s^2 + a^2} }
+      | c = simplifying
+}}
+{{eqn | r = \frac s {s^2 + a^2}
+      | c = simplifying
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map {\laptrans {\cos {a t} } } s
+      | r = \int_0^{\to +\infty} e^{-s t} \cos {a t} \rd t
+      | c = {{Defof|Laplace Transform}}
+}}
+{{eqn | r = \lim_{L \mathop \to \infty} \int_0^L e^{-s t} \cos {a t} \rd t
+      | c = {{Defof|Improper Integral}}
+}}
+{{eqn | r = \lim_{L \mathop \to \infty} \intlimits {\frac {e^{-s t} \paren {-s \cos a t + a \sin a t} } {\paren {-s}^2 + a^2} } 0 L
+      | c = [[Primitive of Exponential of a x by Cosine of b x|Primitive of $e^{a x} \cos b x$]]
+}}
+{{eqn | r = \lim_{L \mathop \to \infty} \paren {\frac {e^{-s L} \paren {-s \cos a L + a \sin a L} } {s^2 + a^2} - \frac {e^{-s \times 0} \paren {-s \, \map \cos {0 \times a} + a \, \map \sin {0 \times a} } } {s^2 + a^2} }
+      | c = 
+}}
+{{eqn | r = \lim_{L \mathop \to \infty} \paren {\frac {s \, \map \cos {0 \times a} - a \, \map \sin {0 \times a} } {s^2 + a^2} - \frac {e^{-s L} \paren {-s \cos a L + a \sin a L} } {s^2 + a^2} }
+      | c = 
+}}
+{{eqn | r = \frac {s \, \map \cos {0 \times a} - a \, \map \sin {0 \times a} } {s^2 + a^2} - 0
+      | c = [[Exponential Tends to Zero and Infinity|Exponential Tends to Zero]]
+}}
+{{eqn | r = \frac {s \cos 0 - a \sin 0} {s^2 + a^2}
+      | c = simplifying
+}}
+{{eqn | r = \frac s {s^2 + a^2}
+      | c = [[Sine of Zero is Zero]], [[Cosine of Zero is One]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $\map f t := \map \Ci t = \displaystyle \int_t^\infty \dfrac {\cos u} u \rd u$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \map {f'} t
+      | r = -\dfrac {\cos t} t
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = t \map {f'} t
+      | r = -\cos t
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \laptrans {t \map {f'} t}
+      | r = -\laptrans {\cos t}
+      | c = 
+}}
+{{eqn | r = -\dfrac s {s^2 + 1}
+      | c = [[Laplace Transform of Cosine]]
+}}
+{{eqn | ll= \leadsto
+      | l = -\dfrac \d {\d s} \laptrans {\map {f'} t}
+      | r = -\dfrac s {s^2 + 1}
+      | c = [[Derivative of Laplace Transform]]
+}}
+{{eqn | ll= \leadsto
+      | l = \map {\dfrac \d {\d s} } {s \laptrans {\map f t} - \map f 0}
+      | r = \dfrac s {s^2 + 1}
+      | c = [[Laplace Transform of Derivative]]
+}}
+{{eqn | ll= \leadsto
+      | l = s \laptrans {\map f t}
+      | r = \int \dfrac s {s^2 + 1} \rd s
+      | c = $\map f 0 = 0$, and integrating both sides {{WRT|Integration}} $s$
+}}
+{{eqn | ll= \leadsto
+      | l = s \laptrans {\map f t}
+      | r = \dfrac 1 2 \map \ln {s^2 + 1} + C
+      | c = [[Primitive of x over x squared plus a squared|Primitive of $\dfrac x {x^2 + a^2}$]]
+}}
+{{end-eqn}}
+By the [[Initial Value Theorem of Laplace Transform]]:
+:$\displaystyle \lim_{s \mathop \to \infty} s \laptrans {\map f t} = \lim_{t \mathop \to 0} \map f t = \map f 0 = 0$
+which leads to:
+:$c = 0$
+Thus:
+{{begin-eqn}}
+{{eqn | l = s \laptrans {\map f t}
+      | r = \dfrac 1 2 \map \ln {s^2 + 1}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \laptrans {\map f t}
+      | r = \dfrac {\map \ln {s^2 + 1} } {2 s}
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $\map f t = \sin \sqrt t$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \map {f'} t
+      | r = \dfrac {\cos \sqrt t} {2 \sqrt t}
+      | c = 
+}}
+{{eqn | l = \map f 0
+      | r = 0
+      | c = 
+}}
+{{end-eqn}}
+So:
+{{begin-eqn}}
+{{eqn | l = \laptrans {\map {f'} t}
+      | r = \dfrac 1 2 \laptrans {\dfrac {\cos \sqrt t} {\sqrt t} }
+      | c = 
+}}
+{{eqn | r = s \, \map F s - \map f 0
+      | c = [[Laplace Transform of Derivative]]
+}}
+{{eqn | r = \dfrac {\sqrt \pi} {2 s^{1/2} } \map \exp {-\dfrac 1 {4 s} }
+      | c = [[Laplace Transform of Sine of Root]]
+}}
+{{eqn | ll= \leadsto
+      | l = \laptrans {\dfrac {\cos \sqrt t} {\sqrt t} }
+      | r = \sqrt {\dfrac \pi s} \, \map \exp {-\dfrac 1 {4 s} }
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Laplace Transform of Sine}
+Tags: Laplace Transforms, Sine Function, Laplace Transform of Sine
+
+\begin{theorem}
+Let $\sin$ denote the [[Definition:Real Sine Function|real sine function]].
+Let $\laptrans f$ denote the [[Definition:Laplace Transform|Laplace transform]] of a [[Definition:Real Function|real function]] $f$.
+Then:
+:$\laptrans {\sin at} = \dfrac a {s^2 + a^2}$
+where $a \in \R_{>0}$ is [[Definition:Constant|constant]], and $\map \Re s > 0$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \laptrans {e^{i a t} }
+      | r = \frac 1 {s - i a}
+      | c = [[Laplace Transform of Exponential]]
+}}
+{{eqn | r = \frac {s + i a} {s^2 + a^2}
+      | c = multiplying [[Definition:Numerator|top]] and [[Definition:Denominator|bottom]] by $s + i a$
+}} 
+{{end-eqn}}
+Also:
+{{begin-eqn}}
+{{eqn | l = \laptrans {e^{i a t} }
+      | r = \laptrans {\cos a t + i \sin a t}
+      | c = [[Euler's Formula]]
+}}
+{{eqn | r = \laptrans {\cos a t} + i \laptrans {\sin a t}
+      | c = [[Linear Combination of Laplace Transforms]]
+}}
+{{end-eqn}}
+So:
+{{begin-eqn}}
+{{eqn | l = \laptrans {\sin a t}
+      | r = \map \Im {\laptrans {e^{i a t} } }
+}}
+{{eqn | r = \map \Im {\frac {s + i a} {s^2 + a^2} }
+}}
+{{eqn | r = \frac a {s^2 + a^2}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+By definition of the [[Definition:Laplace Transform|Laplace transform]]:
+:$\displaystyle \laptrans {\sin a t} = \int_0^{\to +\infty} e^{-s t} \sin a t \rd t$
+From [[Integration by Parts]]:
+:$\displaystyle \int f g' \rd t = f g - \int f' g \rd t$
+Here:
+{{begin-eqn}}
+{{eqn | l = f
+      | r = \sin a t
+}}
+{{eqn | ll= \leadsto
+      | l = f'
+      | r = a \cos a t
+      | c = [[Derivative of Sine of a x|Derivative of $\sin a x$]]
+}}
+{{eqn | l = g'
+      | r = e^{-s t}
+}}
+{{eqn | ll= \leadsto
+      | l = g
+      | r = -\frac 1 s e^{-s t}
+      | c = [[Primitive of Exponential of a x|Primitive of $e^{a x}$]]
+}}
+{{end-eqn}}
+So:
+{{begin-eqn}}
+{{eqn | n = 1
+      | l = \int e^{-s t} \sin a t \rd t
+      | r = -\frac 1 s e^{-s t} \sin a t + \frac a s \int e^{-s t} \cos a t \rd t
+}}
+{{end-eqn}}
+Consider:
+:$\displaystyle \int e^{-s t} \cos a t \rd t$
+Again, using [[Integration by Parts]]:
+:$\displaystyle \int h j\,' \rd t = h j - \int h' j \rd t$
+Here:
+{{begin-eqn}}
+{{eqn | l = h
+      | r = \cos a t
+}}
+{{eqn | ll= \leadsto
+      | l = h'
+      | r = -a \sin a t
+      | c = [[Derivative of Cosine Function]]
+}}
+{{eqn | l = j\,'
+      | r = e^{-s t}
+}}
+{{eqn | ll= \leadsto
+      | l = j
+      | r = -\frac 1 s e^{-s t}
+      | c = [[Primitive of Exponential Function]]
+}}
+{{end-eqn}}
+So:
+{{begin-eqn}}
+{{eqn | l = \int e^{-s t} \cos a t \rd t
+      | r = -\frac 1 s e^{-st} \cos a t - \frac a s \int e^{-s t} \sin a t \rd t
+}}
+{{end-eqn}}
+Substituting this into $(1)$:
+{{begin-eqn}}
+{{eqn | l = \int e^{-s t} \sin a t \rd t
+      | r = -\frac 1 s e^{-s t} \sin a t + \frac a s \paren {-\frac 1 s e^{-s t} \cos a t - \frac a s \int e^{-s t} \sin a t \rd t}
+}}
+{{eqn | r = -\frac 1 s e^{-s t} \sin a t - \frac a {s^2} e^{-s t} \cos a t - \frac {a^2} {s^2} \int e^{-s t} \sin a t \rd t
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {1 + \frac {a^2} {s^2} } \int e^{-s t} \sin a t \rd t
+      | r = -e^{-s t} \paren {\frac 1 s \sin a t + \frac a {s^2} \cos a t}
+}}
+{{end-eqn}}
+Evaluating at $t = 0$ and $t \to +\infty$:
+{{begin-eqn}}
+{{eqn | l = \paren {1 + \frac {a^2} {s^2} } \laptrans {\sin a t} 
+      | r = \intlimits {-e^{-s t} \paren {\frac 1 s \sin a t + \frac a {s^2} \cos a t} } {t \mathop = 0} {t \mathop \to +\infty}
+}}
+{{eqn | r = 0 - \paren {-1 \paren {\frac 1 s \times 0 + \frac a {s^2} \times 1} }
+      | c = [[Boundedness of Real Sine and Cosine]], [[Complex Exponential Tends to Zero]]
+}}
+{{eqn | r = \frac a {s^2}
+}}
+{{eqn | ll= \leadsto
+      | l = \laptrans {\sin a t}
+      | r = \frac a {s^2} \paren {1 + \frac {a^2} {s^2} }^{-1}
+}}
+{{eqn | r = \frac a {s^2} \paren {\frac {s^2} {a^2 + s^2} }
+}}
+{{eqn | r = \frac a {s^2 + a^2}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \laptrans {\sin a t}
+      | r = \laptrans {\frac {e^{i a t} - e^{-i a t} } {2 i} }
+      | c = [[Sine Exponential Formulation]]
+}}
+{{eqn | r = \frac 1 {2 i} \paren {\laptrans {e^{i a t} } - \laptrans {e^{-i a t} } }
+      | c = [[Linear Combination of Laplace Transforms]]
+}}
+{{eqn | r = \frac 1 {2 i} \paren {\frac 1 {s - i a} - \frac 1 {s + i a} }
+      | c = [[Laplace Transform of Exponential]]
+}}
+{{eqn | r = \frac 1 {2 i} \paren {\frac {s + i a - s + i a} {s^2 + a^2} }
+      | c = simplifying
+}}
+{{eqn | r = \frac 1 {2 i} \paren {\frac {2 i a} {s^2 + a^2} }
+      | c = simplifying
+}}
+{{eqn | r = \frac a {s^2 + a^2}
+      | c = simplifying
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map {\laptrans {\sin {a t} } } s
+      | r = \int_0^{\to +\infty} e^{-s t} \sin {a t} \rd t
+      | c = {{Defof|Laplace Transform}}
+}}
+{{eqn | r = \lim_{L \mathop \to \infty} \int_0^L e^{-s t} \sin {a t} \rd t
+      | c = {{Defof|Improper Integral}}
+}}
+{{eqn | r = \lim_{L \mathop \to \infty} \intlimits {\frac {e^{-s t} \paren {-s \sin a t - a \cos a t} } {\paren {-s}^2 + a^2} } 0 L
+      | c = [[Primitive of Exponential of a x by Sine of b x|Primitive of $e^{a x} \sin b x$]]
+}}
+{{eqn | r = \lim_{L \mathop \to \infty} \paren {\dfrac {e^{-s L} \paren {-s \sin a L - a \cos a L} } {s^2 + a^2} - \dfrac {e^{-s \times 0} \paren {-s \, \map \sin {0 \times a} - a \, \map \cos {0 \times a} } } {s^2 + a^2} }
+      | c = 
+}}
+{{eqn | r = \lim_{L \mathop \to \infty} \paren {\dfrac {s \sin 0 + a \cos 0} {s^2 + a^2} - \dfrac {e^{-s L} \paren {s \sin a L + a \cos a L} } {s^2 + a^2} }
+      | c = [[Exponential of Zero]] and rearranging
+}}
+{{eqn | r = \dfrac {s \sin 0 + a \cos 0} {s^2 + a^2} - 0
+      | c = [[Exponential Tends to Zero and Infinity|Exponential Tends to Zero]]
+}}
+{{eqn | r = \frac a {s^2 + a^2}
+      | c = [[Sine of Zero is Zero]], [[Cosine of Zero is One]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $\map f t := \map \Si t = \displaystyle \int_0^t \dfrac {\sin u} u \rd u$.
+Then:
+:$\map f 0 = 0$
+and:
+{{begin-eqn}}
+{{eqn | l = \map {f'} t
+      | r = \dfrac {\sin t} t
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = t \map {f'} t
+      | r = \sin t
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \laptrans {t \map {f'} t}
+      | r = \laptrans {\sin t}
+      | c = 
+}}
+{{eqn | r = \dfrac 1 {s^2 + 1}
+      | c = [[Laplace Transform of Sine]]
+}}
+{{eqn | ll= \leadsto
+      | l = -\dfrac \d {\d s} \laptrans {\map {f'} t}
+      | r = \dfrac 1 {s^2 + 1}
+      | c = [[Derivative of Laplace Transform]]
+}}
+{{eqn | ll= \leadsto
+      | l = \map {\dfrac \d {\d s} } {s \laptrans {\map f t} - \map f 0}
+      | r = -\dfrac 1 {s^2 + 1}
+      | c = [[Laplace Transform of Derivative]]
+}}
+{{eqn | ll= \leadsto
+      | l = s \laptrans {\map f t}
+      | r = -\int \dfrac 1 {s^2 + 1} \rd s
+      | c = $\map f 0 = 0$, and integrating both sides {{WRT|Integration}} $s$
+}}
+{{eqn | ll= \leadsto
+      | l = s \laptrans {\map f t}
+      | r = -\arctan s + C
+      | c = [[Primitive of Reciprocal of x squared plus a squared/Arctangent Form|Primitive of $\dfrac 1 {x^2 + a^2}$]]
+}}
+{{end-eqn}}
+By the [[Initial Value Theorem of Laplace Transform]]:
+:$\displaystyle \lim_{s \mathop \to \infty} s \laptrans {\map f t} = \lim_{t \mathop \to 0} \map f t = \map f 0 = 0$
+which leads to:
+:$c = \dfrac \pi 2$
+Thus:
+{{begin-eqn}}
+{{eqn | l = s \laptrans {\map f t}
+      | r = \dfrac \pi 2 - \arctan s
+      | c = 
+}}
+{{eqn | r = \arccot s
+      | c = [[Sum of Arctangent and Arccotangent]]
+}}
+{{eqn | r = \arctan \dfrac 1 s
+      | c = [[Arctangent of Reciprocal equals Arccotangent]]
+}}
+{{eqn | ll= \leadsto
+      | l = \laptrans {\map f t}
+      | r = \dfrac 1 s \arctan \dfrac 1 s
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $\map f t := \map \Si t = \displaystyle \int_0^t \dfrac {\sin u} u \rd u$.
+Then:
+:$\map f 0 = 0$
+and:
+{{begin-eqn}}
+{{eqn | l = \map \Si t
+      | r = \int_0^t \dfrac {\sin u} u \rd u
+      | c = {{Defof|Sine Integral Function}}
+}}
+{{eqn | r = \int_0^t \dfrac 1 u \paren {u - \dfrac {u^3} {3!} + \dfrac {u^5} {5!} - \dfrac {u^7} {7!} + \dotsb} \rd u
+      | c = {{Defof|Real Sine Function}}
+}}
+{{eqn | r = t - \dfrac {t^3} {3 \times 3!} + \dfrac {t^5} {5 \times 5!} - \dfrac {t^7} {7 \times 7!} + \dotsb
+      | c = [[Primitive of Power]]
+}}
+{{eqn | ll= \leadsto
+      | l = \laptrans {\map \Si t}
+      | r = \laptrans {t - \dfrac {t^3} {3 \times 3!} + \dfrac {t^5} {5 \times 5!} - \dfrac {t^7} {7 \times 7!} + \dotsb}
+      | c = 
+}}
+{{eqn | r = \dfrac 1 {s^2} - \dfrac 1 {3 \times 3!} \dfrac {3!} {s^4} + \dfrac 1 {5 \times 5!} \dfrac {5!} {s^6} - \dfrac 1 {7 \times 7!} \dfrac {7!} {s^8} + \dotsb
+      | c = [[Laplace Transform of Positive Integer Power]]
+}}
+{{eqn | r = \dfrac 1 {s^2} - \dfrac 1 {3 s^4} + \dfrac 1 {5 s^6} - \dfrac 1 {7 s^8} + \dotsb
+      | c = simplifying
+}}
+{{eqn | r = \dfrac 1 s \paren {\dfrac {\paren {1 / s} } 1 - \dfrac {\paren {1 / s}^3} 3 + \dfrac {\paren {1 / s}^5} 5 - \dfrac {\paren {1 / s}^7} 7 + \dotsb}
+      | c = rearranging
+}}
+{{eqn | r = \dfrac 1 s \arctan \dfrac 1 s
+      | c = [[Power Series Expansion for Real Arctangent Function]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $\map f t := \map \Si t = \displaystyle \int_0^t \dfrac {\sin u} u \rd u$.
+Then:
+:$\map f 0 = 0$
+and:
+{{begin-eqn}}
+{{eqn | l = \map \Si t
+      | r = \int_0^t \dfrac {\sin u} u \rd u
+      | c = {{Defof|Sine Integral Function}}
+}}
+{{eqn | r = \int_0^1 \dfrac {\sin t v} v \rd v
+      | c = [[Integration by Substitution]] $u = t v$
+}}
+{{eqn | ll= \leadsto
+      | l = \laptrans {\map \Si t}
+      | r = \laptrans {\int_0^1 \dfrac {\sin t v} v \rd v}
+      | c = 
+}}
+{{eqn | r = \int_0^\infty e^{-s t} \paren {\int_0^1 \dfrac {\sin t v} v \rd v} \rd t
+      | c = {{Defof|Laplace Transform}}
+}}
+{{eqn | r = \int_0^1 \dfrac 1 v \paren {\int_0^\infty e^{-s t} \sin t v \rd t} \rd v
+      | c = exchanging order of integration
+}}
+{{eqn | r = \int_0^1 \dfrac {\laptrans {\sin t v} } v \rd v
+      | c = {{Defof|Laplace Transform}}
+}}
+{{eqn | r = \int_0^1 \dfrac {\d v} {s^2 + v^2}
+      | c = [[Laplace Transform of Sine]]
+}}
+{{eqn | r = \intlimits {\dfrac 1 s \arctan \dfrac v s} 0 1
+      | c = [[Primitive of Reciprocal of x squared plus a squared/Arctangent Form|Primitive of $\dfrac 1 {x^2 + a^2}$]]
+}}
+{{eqn | r = \dfrac 1 s \arctan \dfrac 1 s
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \sin \sqrt t
+      | r = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {\paren {\sqrt t}^{2 n + 1} } {\paren {2 n + 1}!}
+      | c = {{Defof|Real Sine Function}}
+}}
+{{eqn | r = \sum_{n \mathop = 0}^\infty \frac {\paren {-1}^n} {\paren {2 n + 1}!} t^{n + \frac 1 2}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \laptrans {\sin \sqrt t}
+      | r = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {\map \Gamma {n + \frac 3 2} } {\paren {2 n + 1}! s^{n + \frac 3 2} }
+      | c = [[Laplace Transform of Power]], [[Linear Combination of Laplace Transforms]]
+}}
+{{eqn | r = \frac 1 {s^{3/2} } \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {\paren {n + \frac 1 2} \map \Gamma {n + \frac 1 2} } {\paren {2 n + 1}! s^n}
+      | c = [[Gamma Difference Equation]]
+}}
+{{eqn | r = \frac {\sqrt \pi} {2 s^{3/2} } \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {\paren {2 n + 1} \paren {2 n}!} {2^{2 n} n! \paren {2 n + 1}! s^n}
+      | c = [[Gamma Function of Positive Half-Integer]]
+}}
+{{eqn | r = \frac {\sqrt \pi} {2 s^{3/2} } \sum_{n \mathop = 0}^\infty \frac {\paren {-1}^n} {2^{2 n} n! s^n}
+      | c = 
+}}
+{{eqn | r = \frac {\sqrt \pi} {2 s^{3/2} } \sum_{n \mathop = 0}^\infty \frac 1 {n!} \paren {-\frac 1 {2^2 s} }^n
+      | c = 
+}}
+{{eqn | r = \dfrac {\sqrt \pi} {2 s^{3/2} } e^{-1/\paren {2^2 s} }
+      | c = {{Defof|Exponential Function/Real|subdef = Sum of Series|Exponential Function}}
+}}
+{{eqn | r = \dfrac {\sqrt \pi} {2 s^{3/2} } \map \exp {-\dfrac 1 {4 s} }
+      | c = simplifying
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sum Rule for Derivatives/General Result}
+Tags: Differential Calculus, Sum Rule for Derivatives
+
+\begin{theorem}
+Let $f_1 \left({x}\right), f_2 \left({x}\right), \ldots, f_n \left({x}\right)$ be [[Definition:Real Function|real functions]] all [[Definition:Differentiable on Interval|differentiable]].
+Then for all $n \in \N_{>0}$:
+: $\displaystyle D_x \left({\sum_{i \mathop = 1}^n f_i \left({x}\right)}\right) = \sum_{i \mathop = 1}^n D_x \left({f_i \left({x}\right)}\right)$
+\end{theorem}
+
+\begin{proof}
+The proof proceeds by [[Principle of Mathematical Induction|induction]].
+For all $n \in \N_{> 0}$, let $P \left({n}\right)$ be the [[Definition:Proposition|proposition]]:
+:$\displaystyle D_x \left({\sum_{i \mathop = 1}^n f_i \left({x}\right)}\right) = \sum_{i \mathop = 1}^n D_x \left({f_i \left({x}\right)}\right)$
+$P \left({1}\right)$ is true, as this just says:
+:$D_x \left({f_1 \left({x}\right)}\right) = D_x \left({f_1 \left({x}\right)}\right)$
+which is trivially true.
+=== Basis for the Induction ===
+$P \left({2}\right)$ is the case:
+:$\displaystyle D_x \left({f_1 \left({x}\right) + f_2 \left({x}\right)}\right) = D_x \left({f_1 \left({x}\right)}\right) + D_x  \left({f_2 \left({x}\right)}\right)$
+which has been proved in [[Sum Rule for Derivatives]].
+This is the [[Principle of Mathematical Induction#Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Now it needs to be shown that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k + 1}\right)$ is true.
+So this is the [[Principle of Mathematical Induction#Induction Hypothesis|induction hypothesis]]:
+:$\displaystyle D_x \left({\sum_{i \mathop = 1}^k f_i \left({x}\right)}\right) = \sum_{i \mathop = 1}^k D_x \left({f_i \left({x}\right)}\right)$
+from which it is to be shown that:
+:$\displaystyle D_x \left({\sum_{i \mathop = 1}^{k + 1} f_i \left({x}\right)}\right) = \sum_{i \mathop = 1}^{k + 1} D_x \left({f_i \left({x}\right)}\right)$
+=== Induction Step ===
+This is the [[Principle of Mathematical Induction#Induction Step|induction step]]:
+{{begin-eqn}}
+{{eqn | l = D_x \left({\sum_{i \mathop = 1}^{k + 1} f_i \left({x}\right)}\right)
+      | r = D_x \left({\sum_{i \mathop = 1}^k f_i \left({x}\right) + f_{k+1} \left({x}\right)}\right)
+      | c = Definition of [[Definition:Summation|Summation]]
+}}
+{{eqn | r = D_x \left({\sum_{i \mathop = 1}^k f_i \left({x}\right)}\right) + D_x \left({f_{k+1} \left({x}\right)}\right)
+      | c = [[Sum Rule for Derivatives/General Result#Basis of the Induction|Basis of the Induction]]
+}}
+{{eqn | r = \sum_{i \mathop = 1}^k D_x \left({f_i \left({x}\right)}\right) + D_x \left({f_{k+1} \left({x}\right)}\right)
+      | c = [[Sum Rule for Derivatives/General Result#Basis of the Induction|Induction Hypothesis]]
+}}
+{{eqn | r = \sum_{i \mathop = 1}^{k + 1} D_x \left({f_i \left({x}\right)}\right)
+      | c = Definition of [[Definition:Summation|Summation]]
+}}
+{{end-eqn}}
+So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\displaystyle \forall n \in \N_{>0}: D_x \left({\sum_{i \mathop = 1}^n f_i \left({x}\right)}\right) = \sum_{i \mathop = 1}^n D_x \left({f_i \left({x}\right)}\right)$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Product Rule for Derivatives/General Result}
+Tags: Differential Calculus
+
+\begin{theorem}
+Let $f_1 \left({x}\right), f_2 \left({x}\right), \ldots, f_n \left({x}\right)$ be [[Definition:Real Function|real functions]] [[Definition:Differentiable on Interval|differentiable]] on the [[Definition:Open Real Interval|open interval]] $I$.
+then:
+:$\forall x \in I: \displaystyle D_x \left({\prod_{i \mathop = 1}^n f_i \left({x}\right)}\right) = \sum_{i \mathop = 1}^n \left({D_x \left({f_i \left({x}\right)}\right) \prod_{j \mathop \ne i} f_j \left({x}\right)}\right)$
+\end{theorem}
+
+\begin{proof}
+Proof by [[Principle of Mathematical Induction|induction]]:
+For all $n \in \N_{\ge 1}$, let $P \left({n}\right)$ be the [[Definition:Proposition|proposition]]:
+: $\displaystyle D_x \left({\prod_{i \mathop = 1}^n f_i \left({x}\right)}\right) = \sum_{i \mathop = 1}^n \left({D_x \left({f_i \left({x}\right)}\right) \prod_{j \mathop \ne i} f_j \left({x}\right)}\right)$
+$P(1)$ is true, as this just says:
+: $D_x \left({f_1 \left({x}\right)}\right) = D_x \left({f_1 \left({x}\right)}\right)$
+=== Basis for the Induction ===
+$P(2)$ is the case:
+: $D_x \left({f_1 \left({x}\right) f_2 \left({x}\right)}\right) = D_x \left({f_1 \left({x}\right)}\right) f_2 \left({x}\right) + f_1 \left({x}\right) D_x \left({f_2 \left({x}\right)}\right)$
+which has been proved in [[Product Rule for Derivatives]].
+This is our [[Principle of Mathematical Induction#Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k+1}\right)$ is true.
+So this is our [[Principle of Mathematical Induction#Induction Hypothesis|induction hypothesis]]:
+:$\displaystyle D_x \left({\prod_{i \mathop = 1}^k f_i \left({x}\right)}\right) = \sum_{i \mathop = 1}^k \left({D_x \left({f_i \left({x}\right)}\right) \prod_{j \mathop \ne i} f_j \left({x}\right)}\right)$
+Then we need to show:
+:$\displaystyle D_x \left({\prod_{i \mathop = 1}^{k+1} f_i \left({x}\right)}\right) = \sum_{i \mathop = 1}^{k+1} \left({D_x \left({f_i \left({x}\right)}\right) \prod_{j \mathop \ne i} f_j \left({x}\right)}\right)$
+=== Induction Step ===
+This is our [[Principle of Mathematical Induction#Induction Step|induction step]]:
+{{begin-eqn}}
+{{eqn | l = D_x \left({\prod_{i \mathop = 1}^{k+1} f_i \left({x}\right)}\right)
+      | r = D_x \left({\left({\prod_{i \mathop = 1}^k f_i \left({x}\right)}\right) f_{k+1} \left({x}\right)}\right)
+      | c = 
+}}
+{{eqn | r = D_x \left({f_{k+1} \left({x}\right)}\right) \left({\prod_{i \mathop = 1}^k f_i \left({x}\right)}\right) + D_x \left({\prod_{i \mathop = 1}^k f_i \left({x}\right)}\right) f_{k+1} \left({x}\right)
+      | c = [[Product Rule for Derivatives/General Result#Basis for the Induction|Basis for the Induction]]
+}}
+{{eqn | r = D_x \left({f_{k+1} \left({x}\right)}\right) \left({\prod_{i \mathop = 1}^k f_i \left({x}\right)}\right) + \left({\sum_{i \mathop = 1}^k \left({D_x \left({f_i \left({x}\right)}\right) \prod_{j \mathop \ne i} f_j \left({x}\right)}\right)}\right) f_{k+1} \left({x}\right)
+      | c = [[Product Rule for Derivatives/General Result#Induction Hypothesis|Induction Hypothesis]]
+}}
+{{eqn | r = \sum_{i \mathop = 1}^{k+1} \left({D_x \left({f_i \left({x}\right)}\right) \prod_{j \mathop \ne i} f_j \left({x}\right)}\right)
+      | c = 
+}}
+{{end-eqn}}
+So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\displaystyle D_x \left({\prod_{i \mathop = 1}^n f_i \left({x}\right)}\right) = \sum_{i \mathop = 1}^n \left({D_x \left({f_i \left({x}\right)}\right) \prod_{j \mathop \ne i} f_j \left({x}\right)}\right)$ for all $n \in \N$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Derivative of Arccotangent Function}
+Tags: Derivatives of Inverse Trigonometric Functions, Arccotangent Function, Derivative of Arccotangent Function
+
+\begin{theorem}
+:$\dfrac {\map \d {\arccot x} } {\d x} = \dfrac {-1} {1 + x^2}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = y
+      | r = \arccot x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x
+      | r = \cot y
+      | c = {{Defof|Real Arccotangent}}
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d x} {\d y}
+      | r = -\csc^2 y
+      | c = [[Derivative of Cotangent Function]]
+}}
+{{eqn | r = -\paren {1 + \cot^2 y}
+      | c = [[Difference of Squares of Cosecant and Cotangent]]
+}}
+{{eqn | r = -\paren {1 + x^2}
+      | c = Definition of $x$
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d y} {\d x}
+      | r = \frac {-1} {1 + x^2}
+      | c = [[Derivative of Inverse Function]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \frac {\map \d {\arccot x} } {\d x}
+      | r = \map {\frac \d {\d x} } {\frac \pi 2 - \arctan x}
+      | c = [[Tangent of Complement equals Cotangent]]
+}}
+{{eqn | r = -\frac 1 {1 + x^2}
+      | c = [[Derivative of Arctangent Function]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Derivative of Arcsecant Function/Corollary 2}
+Tags: Derivative of Arcsecant Function
+
+\begin{theorem}
+:$\dfrac {\mathrm d \left({\operatorname{arcsec} x }\right)} {\mathrm d x} = \dfrac 1 {x^2 \sqrt {1 - \frac 1 {x^2}}}$
+\end{theorem}
+
+\begin{proof}
+From [[Derivative of Arcsecant Function]]:
+:$\dfrac {\mathrm d \left({\operatorname{arcsec} x}\right)} {\mathrm d x} = \dfrac 1 {\left|{x}\right| \sqrt {x^2 - 1} }$
+Since for all $x \in \R$, we have $\left|{x}\right| = \sqrt{x^2}$, we can write:
+:$\dfrac {\mathrm d \left({\operatorname{arcsec} x}\right)} {\mathrm d x} = \dfrac 1 {\sqrt {x^2} \sqrt {x^2 - 1}}$
+Multiplying the denominator by $1 = \dfrac {\sqrt{x^2}} {\sqrt{x^2}}$ yields:
+:$\dfrac {\mathrm d \left({\operatorname{arcsec} x}\right)} {\mathrm d x} = \dfrac 1 {x^2 \sqrt {1 - \frac 1 {x^2}}}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Asymptotic Formula for Bernoulli Numbers}
+Tags: Bernoulli Numbers, Asymptotics
+
+\begin{theorem}
+The [[Definition:Bernoulli Numbers|Bernoulli numbers]] with [[Definition:Even Integer|even]] index can be approximated by the asymptotic formula:
+:$B_{2 n} \sim \paren {-1}^{n + 1} 4 \sqrt {\pi n} \paren {\dfrac n {\pi e} }^{2 n}$
+where:
+:$B_n$ denotes the $n$th [[Definition:Bernoulli Numbers|Bernoulli number]]
+:$\sim$ denotes [[Definition:Asymptotically Equal|asymptotically equal]].
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \lim_{n \mathop \to \infty} \frac {B_{2 n} } {\paren {-1}^{n + 1} 4 \sqrt {\pi n} \paren {\frac n {\pi e} }^{2 n} } 
+      | r = \lim_{n \mathop \to \infty} \frac {\paren {-1}^{n + 1} \paren {2 n}! \map \zeta {2 n} } {2^{2 n - 1} \pi^{2 n} \paren {-1}^{n + 1} 4 \sqrt {\pi n} \paren {\frac n {\pi e} }^{2 n} }
+      | c = [[Riemann Zeta Function at Even Integers]]
+}}
+{{eqn | r = \lim_{n \mathop \to \infty} \frac {\paren {2 n}! \map \zeta {2 n} } {2^{2 n + 1} \sqrt {\pi n} \paren {\frac n e}^{2 n} }
+}}
+{{eqn | r = \lim_{n \mathop \to \infty} \frac {2 \sqrt {\pi n} \paren {\frac {2 n} e}^{2 n} \map \zeta {2 n} } {2^{2 n + 1} \sqrt {\pi n} \paren {\frac n e}^{2 n} }
+      | c = [[Stirling's Formula]]
+}}
+{{eqn | r = \lim_{n \mathop \to \infty} \map \zeta {2 n}
+      | c = 
+}}
+{{eqn | r = 1
+      | c =
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cotangent Minus Tangent}
+Tags: Trigonometric Identities
+
+\begin{theorem}
+:$\cot x - \tan x = 2 \cot 2 x$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \cot x - \tan x
+      | r = \frac {\cos x} {\sin x} - \frac {\sin x} {\cos x}
+      | c = {{Defof|Tangent Function|Tangent}} and [[Definition:Cotangent|Cotangent]]
+}}
+{{eqn | r = \frac {\cos^2 x - \sin^2 x} {\sin x \cos x}
+      | c = 
+}}
+{{eqn | r = 2 \frac {\cos^2 x - \sin^2 x} {2 \sin x \cos x}
+      | c = 
+}}
+{{eqn | r = 2 \frac {\cos 2 x} {\sin 2 x}
+      | c = [[Double Angle Formula for Sine]] and [[Double Angle Formula for Cosine]]
+}}
+{{eqn | r = 2 \cot 2 x
+      | c = {{Defof|Cotangent}}
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Trigonometric Identities]]
+ahdgo2n5ikx91lw9qfq4541fnvulqgr
+\end{proof}<|endoftext|>
+\section{Derivative of Hyperbolic Cotangent Function}
+Tags: Derivatives of Hyperbolic Functions, Hyperbolic Cotangent Function
+
+\begin{theorem}
+:$\map {\dfrac \d {\d x} } {\coth u} = -\csch^2 u \dfrac {\d u} {\d x}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map {\frac \d {\d x} } {\coth u}
+      | r = \map {\frac \d {\d u} } {\coth u} \frac {\d u} {\d x}
+      | c = [[Chain Rule for Derivatives]]
+}}
+{{eqn | r = -\csch^2 u \frac {\d u} {\d x}
+      | c = [[Derivative of Hyperbolic Cotangent]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Derivative of Hyperbolic Cosecant Function}
+Tags: Derivatives of Hyperbolic Functions, Hyperbolic Cosecant Function
+
+\begin{theorem}
+:$\map {\dfrac \d {\d x} } {\csch u} = -\csch u \coth u \dfrac {\d u} {\d x}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map {\frac \d {\d x} } {\csch u}
+      | r = \map {\frac \d {\d u} } {\csch u} \frac {\d u} {\d x}
+      | c = [[Chain Rule for Derivatives]]
+}}
+{{eqn | r = -\csch u \coth u \frac {\d u} {\d x}
+      | c = [[Derivative of Hyperbolic Cotangent]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Derivative of Inverse Hyperbolic Sine}
+Tags: Derivatives of Inverse Hyperbolic Functions, Inverse Hyperbolic Sine, Derivative of Inverse Hyperbolic Sine
+
+\begin{theorem}
+:$\map {\dfrac \d {\d x} } {\sinh^{-1} x} = \dfrac 1 {\sqrt {x^2 + 1} }$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = y
+      | r = \sinh^{-1} x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x
+      | r = \sinh y
+      | c = {{Defof|Real Inverse Hyperbolic Sine}}
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d x} {\d y}
+      | r = \cosh y
+      | c = [[Derivative of Hyperbolic Sine]]
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d y} {\d x}
+      | r = \frac 1 {\cosh y}
+      | c = [[Derivative of Inverse Function]]
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d y} {\d x}
+      | r = \pm \frac 1 {\sqrt {\sinh^2 y + 1} }
+      | c = [[Difference of Squares of Hyperbolic Cosine and Sine]]
+}}
+{{end-eqn}}
+For all $x \in \R$ we have that $\cosh y \ge 1$.
+Thus it follows that it is necessary to take the positive root of $\sqrt {\sinh^2 y + 1}$.
+So:
+{{begin-eqn}}
+{{eqn | l = \frac {\d y} {\d x}
+      | r = \frac 1 {\sqrt {\sinh^2 y + 1} }
+      | c =
+}}
+{{eqn | ll= \leadsto
+      | l = \map {\frac \d {\d x} } {\sinh^{-1} x}
+      | r = \frac 1 {\sqrt {x^2 + 1} }
+      | c = Definition of $x$ and $y$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Derivative of Inverse Hyperbolic Cosine}
+Tags: Derivatives of Inverse Hyperbolic Functions, Inverse Hyperbolic Cosine, Derivative of Inverse Hyperbolic Cosine
+
+\begin{theorem}
+:$\map {\dfrac \d {\d x} } {\cosh^{-1} x} = \dfrac 1 {\sqrt {x^2 - 1} }$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = y
+      | r = \cosh^{-1} x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x
+      | r = \cosh y
+      | c = {{Defof|Real Inverse Hyperbolic Cosine}}
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d x} {\d y}
+      | r = \sinh y
+      | c = [[Derivative of Hyperbolic Cosine]]
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d y} {\d x}
+      | r = \frac 1 {\sinh y}
+      | c = [[Derivative of Inverse Function]]
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d y} {\d x}
+      | r = \pm \frac 1 {\sqrt {\cosh^2 y - 1} }
+      | c = [[Difference of Squares of Hyperbolic Cosine and Sine]]
+}}
+{{end-eqn}}
+Note that when $y = 0$, $\cosh y$ is defined and equals $1$.
+But from [[Derivative of Hyperbolic Cosine]]:
+:$\valueat {\dfrac \d {\d y} \cosh y} {y \mathop = 0} = \sinh 0 = 0$
+Thus $\dfrac {\d y} {\d x} = \dfrac 1 {\sinh y}$ is not defined at $y = 0$.
+Hence the limitation of the [[Definition:Domain of Mapping|domain]] of $\map {\dfrac \d {\d x} } {\cosh^{-1} x}$ to exclude $x = 1$.
+Now it is necessary to determine the [[Definition:Sign of Number|sign]] of $\dfrac {\d y} {\d x}$.
+From:
+:[[Real Inverse Hyperbolic Cosine is Strictly Increasing]]
+:[[Derivative of Strictly Increasing Real Function is Strictly Positive]]
+it follows that $\map {\dfrac \d {\d x} } {\cosh^{-1} x} > 0$ on $\R_{>1}$.
+Thus:
+:$\dfrac {\d y} {\d x} = \dfrac 1 {\sqrt {\cosh^2 y - 1} }$
+where $\sqrt {\cosh^2 y - 1}$ denotes the [[Definition:Positive Square Root|positive square root]] of $\cosh^2 y - 1$.
+Hence by definition of $x$ and $y$ above:
+:$\map {\dfrac \d {\d x} } {\cosh^{-1} x} = \dfrac 1 {\sqrt {x^2 - 1} }$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Derivative of Inverse Hyperbolic Tangent}
+Tags: Derivatives of Inverse Hyperbolic Functions, Inverse Hyperbolic Tangent, Derivative of Inverse Hyperbolic Tangent
+
+\begin{theorem}
+:$\map {\dfrac \d {\d x} } {\tanh^{-1} x} = \dfrac 1 {1 - x^2}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = y
+      | r = \tanh^{-1} x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x
+      | r = \tanh y
+      | c = {{Defof|Real Inverse Hyperbolic Tangent}}
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d x} {\d y}
+      | r = \sech^2 y
+      | c = [[Derivative of Hyperbolic Tangent]]
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d y} {\d x}
+      | r = \frac 1 {\sech^2 y}
+      | c = [[Derivative of Inverse Function]]
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d y} {\d x}
+      | r = \frac 1 {1 - \tanh^2 y}
+      | c = [[Sum of Squares of Hyperbolic Secant and Tangent]]
+}}
+{{eqn | ll= \leadsto
+      | l = \map {\frac \d {\d x} } {\tanh^{-1} x}
+      | r = \frac 1 {1 - x^2}
+      | c = Definition of $x$ and $y$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Derivative of Inverse Hyperbolic Cotangent}
+Tags: Derivatives of Inverse Hyperbolic Functions, Inverse Hyperbolic Cotangent, Derivative of Inverse Hyperbolic Cotangent
+
+\begin{theorem}
+:$\map {\dfrac \d {\d x} } {\coth^{-1} x} = \dfrac {-1} {x^2 - 1}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = y
+      | r = \coth^{-1} x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x
+      | r = \coth y
+      | c = {{Defof|Real Inverse Hyperbolic Cotangent}}
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d x} {\d y}
+      | r = -\csch^2 y
+      | c = [[Derivative of Hyperbolic Cotangent]]
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d y} {\d x}
+      | r = \frac {-1} {\csch^2 y}
+      | c = [[Derivative of Inverse Function]]
+}}
+{{eqn | r = \frac {-1} {\coth^2 y - 1}
+      | c = [[Difference of Squares of Hyperbolic Cotangent and Cosecant]]
+}}
+{{eqn | ll= \leadsto
+      | l = \map {\frac \d {\d x} } {\coth^{-1} x}
+      | r = \frac {-1} {x^2 - 1}
+      | c = Definition of $x$ and $y$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Derivative of Inverse Hyperbolic Secant}
+Tags: Derivatives of Inverse Hyperbolic Functions, Inverse Hyperbolic Secant
+
+\begin{theorem}
+:$\map {\dfrac \d {\d x} } {\sech^{-1} x} = \dfrac {-1} {x \sqrt{1 - x^2} }$
+\end{theorem}
+
+\begin{proof}
+[[File:Arcsech.png|300px|right]]
+$\sech^{-1} x$ is defined only on the [[Definition:Half-Open Real Interval|half-open real interval]] $\hointl 0 1$.
+Thus on $\hointl 0 1$:
+{{begin-eqn}}
+{{eqn | l = y
+      | r = \sech^{-1} x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x
+      | r = \sech y
+      | c = where $y \in \R_{>0}$
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d x} {\d y}
+      | r = -\sech y \tanh y
+      | c = [[Derivative of Hyperbolic Secant]]
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d y} {\d x}
+      | r = \dfrac {-1} {\sech y \ \tanh y}
+      | c = [[Derivative of Inverse Function]]
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d y} {\d x}
+      | r = \frac {-1} {\sech y \sqrt {1 - \sech^2 y} }
+      | c = [[Sum of Squares of Hyperbolic Secant and Tangent]]
+}}
+{{eqn | ll= \leadsto
+      | l = \map {\frac \d {\d x} } {\sech^{-1} x}
+      | r = \frac {-1} {x \sqrt {1 - x^2} }
+      | c = Definition of $x$ and $y$
+}}
+{{end-eqn}}
+When $x = 1$, however, $\sqrt{1 - x^2} = 0$ and so $\dfrac {-1} {x \sqrt {1 - x^2} }$ is undefined.
+Hence $\sech^{-1} x$ can be defined only on $\openint 0 1$.
+{{explain|Explain why positive square root is taken at [[Sum of Squares of Hyperbolic Secant and Tangent]]}}
+{{qed}}
+[[Category:Derivatives of Inverse Hyperbolic Functions]]
+[[Category:Inverse Hyperbolic Secant]]
+0i0nkjd488kcr56ejx2a18l91hr94ww
+\end{proof}<|endoftext|>
+\section{Derivative of Inverse Hyperbolic Cosecant}
+Tags: Derivatives of Inverse Hyperbolic Functions, Inverse Hyperbolic Cosecant
+
+\begin{theorem}
+:$\map {\dfrac \d {\d x} } {\csch^{-1} x} = \dfrac {-1} {\size x \sqrt {1 + x^2} }$
+\end{theorem}
+
+\begin{proof}
+[[File:Arccosech.png|600px|right]]
+Let $x > 1$.
+Then we have:
+{{begin-eqn}}
+{{eqn | l = y
+      | r = \csch^{-1} x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x
+      | r = \csch y
+      | c = where $y \ne 0$
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d x} {\d y}
+      | r = -\csch y \coth y
+      | c = [[Derivative of Hyperbolic Cosecant]]
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d y} {\d x}
+      | r = \dfrac {-1} {\csch y \coth y}
+      | c = [[Derivative of Inverse Function]]
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d y} {\d x}
+      | r = \frac {-1} {\csch y \sqrt {1 + \csch^2 y} }
+      | c = [[Difference of Squares of Hyperbolic Cotangent and Cosecant]]
+}}
+{{eqn | ll= \leadsto
+      | l = \map {\frac \d {\d x} } {\csch^{-1} x}
+      | r = \frac {-1} {x \sqrt {1 + x^2} }
+      | c = Definition of $x$ and $y$
+}}
+{{eqn | r = \frac {-1} {\size x \sqrt {1 + x^2} }
+      | c = as $x > 0$
+}}
+{{end-eqn}}
+We have that [[Inverse Hyperbolic Cosecant is Odd Function]].
+Hence from [[Derivative of Odd Function is Even]], $\map {\dfrac \d {\d x} } {\csch^{-1} x}$ is [[Definition:Even Function|even]].
+Hence for $x < -1$ we have that:
+:$\map {\dfrac \d {\d x} } {\csch^{-1} x} = \dfrac {-1} {\paren {-x} \sqrt {1 + x^2} }$
+and so for $x < -1$:
+:$\map {\dfrac \d {\d x} } {\csch^{-1} x} = \dfrac {-1} {\size x \sqrt {1 + x^2} }$
+and the result follows.
+{{qed}}
+[[Category:Derivatives of Inverse Hyperbolic Functions]]
+[[Category:Inverse Hyperbolic Cosecant]]
+q0u3omwylgs3hn10jhsv7daovgjgqwd
+\end{proof}<|endoftext|>
+\section{Sum of Cosecant and Cotangent}
+Tags: Trigonometric Identities
+
+\begin{theorem}
+:$\displaystyle \csc x + \cot x = \cot {\frac x 2}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn |l = \csc x + \cot x
+      |r = \frac 1 {\sin x} + \frac {\cos x} {\sin x}
+      |c = by definition of [[Definition:Cosecant|Cosecant]] and [[Definition:Cotangent|Cotangent]]
+}}
+{{eqn |r = \frac {1 + \cos x} {\sin x}
+      |c =
+}}
+{{eqn |r = \frac {2 \cos^2 {\frac x 2} } {2 \sin {\frac x 2} \cos {\frac x 2} }
+      |c = [[Double Angle Formula for Sine]] and [[Double Angle Formulas/Cosine/Corollary 1|Double Angle Formula for Cosine]]
+}}
+{{eqn |r = \frac {\cos {\frac x 2} } {\sin {\frac x 2} }
+      |c =
+}}
+{{eqn |r = \cot {\frac x 2}
+      |c = by definition of [[Definition:Cotangent|Cotangent]]
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Trigonometric Identities]]
+cxkxef0xrf4mz2sgewslcm4tji8i49q
+\end{proof}<|endoftext|>
+\section{Leibniz's Rule/One Variable/Second Derivative}
+Tags: Leibniz's Rule
+
+\begin{theorem}
+Let $f$ and $g$ be [[Definition:Real Function|real functions]] defined on the [[Definition:Open Real Interval|open interval]] $I$.
+Let $x \in I$ be a point in $I$ at which both $f$ and $g$ are [[Definition:Second Derivative|twice differentiable]].
+Then:
+:$\paren {\map f x \, \map g x}'' = \map f x \, \map {g''} x + 2 \map {f'} x \, \map {g'} x + \map {f''} x \, \map g x$
+\end{theorem}
+
+\begin{proof}
+From [[Leibniz's Rule/One Variable|Leibniz's Rule]]:
+:$\displaystyle \paren {\map f x \, \map g x}^{\paren n} = \sum_{k \mathop = 0}^n \binom n k \map {f^{\paren k} } x \, \map {g^{\paren {n - k} } } x$
+where $\paren n$ denotes the [[Definition:Order of Derivative|order of the derivative]].
+Setting $n = 2$:
+{{begin-eqn}}
+{{eqn | l = \paren {\map f x \, \map g x}''
+      | r = \paren {\map f x \, \map g x}^{\paren 2}
+      | c = {{Defof|Nth Derivative}}
+}}
+{{eqn | r = \sum_{k \mathop = 0}^2 \binom n k \map {f^{\paren k} } x \, \map {g^{\paren {n - k} } } x
+      | c = [[Leibniz's Rule/One Variable|Leibniz's Rule]]
+}}
+{{eqn | r = \binom 2 0 \map {f^{\paren 0} } x \, \map {g^{\paren 2} } x + \binom 2 1 \map {f^{\paren 1} } x \, \map {g^{\paren 1} } x + \binom 2 2 {f^{\paren 2} } x \, \map {g^{\paren 0} } x
+      | c = 
+}}
+{{eqn | r = \map {f^{\paren 0} } x \, \map {g^{\paren 2 } } x + 2 \map {f^{\paren 1} } x \, \map {g^{\paren 1} } x + \map {f^{\paren 2} } x \, \map {g^{\paren 0} } x
+      | c = {{Defof|Binomial Coefficient}}
+}}
+{{eqn | r = \map f x \, \map {g''} x + 2 \map {f'} x \, \map {g'} x + \map {f''} x \, \map g x
+      | c = {{Defof|Nth Derivative}}
+}}
+{{end-eqn}}
+{{qed}}
+{{Namedfor|Gottfried Wilhelm von Leibniz}}
+\end{proof}<|endoftext|>
+\section{Leibniz's Rule/One Variable/Third Derivative}
+Tags: Leibniz's Rule
+
+\begin{theorem}
+Let $f$ and $g$ be [[Definition:Real Function|real functions]] defined on the [[Definition:Open Real Interval|open interval]] $I$.
+Let $x \in I$ be a point in $I$ at which both $f$ and $g$ are [[Definition:Third Derivative|thrice differentiable]].
+Then:
+:$\paren {\map f x \, \map g x}''' = \map f x \, \map {g'''} x + 3 \map {f'} x \, \map {g''} x + 3 \map {f''} x \, \map {g'} x + \map {f'''} x \, \map g x$
+\end{theorem}
+
+\begin{proof}
+From [[Leibniz's Rule/One Variable|Leibniz's Rule]]:
+:$\displaystyle \paren {\map f x \, \map g x}^{\paren n} = \sum_{k \mathop = 0}^n \binom n k \map {f^{\paren k} } x \, \map {g^{\paren {n - k} } } x$
+where $\paren n$ denotes the [[Definition:Order of Derivative|order of the derivative]].
+Setting $n = 3$:
+{{begin-eqn}}
+{{eqn | l = \paren {\map f x \, \map g x}'''
+      | r = \paren {\map f x \, \map g x}^{\paren 3}
+      | c = {{Defof|Nth Derivative}}
+}}
+{{eqn | r = \sum_{k \mathop = 0}^3 \binom n k \map {f^{\paren k} } x \, \map {g^{\paren {n - k} } } x
+      | c = [[Leibniz's Rule/One Variable|Leibniz's Rule]]
+}}
+{{eqn | r = \binom 3 0 \map {f^{\paren 0} } x \, \map {g^{\paren 3} } x + \binom 3 1 \map {f^{\paren 1} } x \, \map {g^{\paren 2} } x + \binom 3 2 \map {f^{\paren 2} } x \map {g^{\paren 1} } x + \binom 3 3 \map {f^{\paren 3} } x \, \map {g^{\paren 0} } x
+      | c = 
+}}
+{{eqn | r = \map {f^{\paren 0} } x \, \map {g^{\paren 3} } x + 3 \map {f^{\paren 1} } x \, \map {g^{\paren 2} } x + 3 \map {f^{\paren 2} } x \, \map {g^{\paren 1} } x + \map {f^{\paren 3} } x \, \map {g^{\paren 0} } x
+      | c = {{Defof|Binomial Coefficient}}
+}}
+{{eqn | r = \map f x \, \map {g'''} x + 3 \map {f'} x \, \map {g''} x + 3 \map {f''} x \, \map {g'} x + \map {f'''} x \, \map g x
+      | c = {{Defof|Nth Derivative}}
+}}
+{{end-eqn}}
+{{qed}}
+{{Namedfor|Gottfried Wilhelm von Leibniz}}
+\end{proof}<|endoftext|>
+\section{Power Series Expansion for Tangent Function}
+Tags: Examples of Power Series, Bernoulli Numbers, Tangent Function, Power Series Expansion for Tangent Function
+
+\begin{theorem}
+The [[Definition:Tangent Function|tangent function]] has a [[Definition:Taylor Series|Taylor series expansion]]:
+{{begin-eqn}}
+{{eqn | l = \tan x
+      | r = \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n - 1} 2^{2 n} \paren {2^{2 n} - 1} B_{2 n} \, x^{2 n - 1} } {\paren {2 n}!}
+      | c = 
+}}
+{{eqn | r = x + \frac {x^3} 3 + \frac {2 x^5} {15} + \frac {17 x^7} {315} + \cdots
+      | c = 
+}}
+{{end-eqn}}
+where $B_{2 n}$ denotes the [[Definition:Bernoulli Numbers|Bernoulli numbers]].
+This [[Definition:Convergent Series|converges]] for $\size x < \dfrac \pi 2$.
+\end{theorem}
+
+\begin{proof}
+By [[Combination Theorem for Limits of Functions]] we can deduce the following.
+{{begin-eqn}}
+{{eqn | o = 
+      | r = \lim_{n \mathop \to \infty} \size {\frac {\frac {\paren {-1}^n 2^{2 n + 2} \paren {2^{2 n + 2} - 1} B_{2 n + 2} } {\paren {2 n + 2}!} x^{2 n + 1} } {\frac {\paren {-1}^{n - 1} 2^{2 n} \paren {2^{2 n} - 1} B_{2 n} } {\paren {2 n}!} x^{2 n - 1} } }
+      | c =
+}}
+{{eqn | r = \lim_{n \mathop \to \infty} \size {\frac {\paren {2^{2 n + 2} - 1} } {\paren {2^{2 n} - 1} } \frac 1 {\paren {2 n + 1} \paren {2 n + 2} } \frac {B_{2 n + 2} } {B_{2 n} } } 4 x^2
+      | c =
+}}
+{{eqn | r = \lim_{n \mathop \to \infty} \size {\frac {2^{2 n + 2} - 1} {2^{2 n} - 1} } \size {\frac 1 {\paren {2 n + 1} \paren {n + 1} } \frac {B_{2 n + 2} } {B_{2 n} } } 2 x^2
+      | c =
+}}
+{{eqn | r = \lim_{n \mathop \to \infty} \size {4 \frac {2^{2 n} } {2^{2 n} - 1} - \frac 1 {2^{2 n} - 1} } \size {\frac 1 {\paren {2 n + 1} \paren {n + 1} } \frac {B_{2 n + 2} } {B_{2 n} } } 2 x^2
+      | c =
+}}
+{{eqn | r = \lim_{n \mathop \to \infty} \size {4 + \frac 4 {2^{2 n} - 1} - \frac 1 {2^{2 n} - 1} } \size {\frac 1 {\paren {2 n + 1} \paren {n + 1} } \frac {B_{2 n + 2} } {B_{2 n} } } 2 x^2
+      | c =
+}}
+{{eqn | r = \lim_{n \mathop \to \infty} \size {\frac 1 {\paren {2 n + 1} \paren {n + 1} } \frac {B_{2 n + 2} } { B_{2 n} } } 8 x^2
+      | c =
+}}
+{{eqn | r = \lim_{n \mathop \to \infty} \size {\frac 1 {\paren {2 n + 1} \paren {n + 1} } \frac {\paren {-1}^{n + 2} 4 \sqrt {\pi \paren {n + 1} } \paren {\frac {n + 1} {\pi e} }^{2 n + 2} } {\paren {-1}^{n + 1} 4 \sqrt {\pi n} \paren {\frac n {\pi e} }^{2 n} } } 8 x^2
+      | c = [[Asymptotic Formula for Bernoulli Numbers]]
+}}
+{{eqn | r = \lim_{n \mathop \to \infty} \size {\frac {\paren {n + 1}^2} {\paren {2 n + 1} \paren {n + 1} } \sqrt {\frac {n + 1} n } \paren {\frac {n + 1} n}^{2 n} } \frac 8 {\pi^2 e^2} x^2
+      | c =
+}}
+{{eqn | r = \lim_{n \mathop \to \infty} \size {\paren {\frac {n + 1} n}^{2 n} } \frac 4 {\pi^2 e^2} x^2
+      | c =
+}}
+{{eqn | r = \lim_{n \mathop \to \infty} \size {\paren {\paren {1 + \frac 1 n}^n}^2} \frac 4 {\pi^2 e^2} x^2
+      | c =
+}}
+{{eqn | r = \frac {4 e^2} {\pi^2 e^2} x^2
+      | c = {{Defof|Euler's Number/Limit of Sequence|Euler's Number}}
+}}
+{{eqn | r = \frac 4 {\pi^2} x^2
+      | c =
+}}
+{{end-eqn}}
+This is less than $1$ {{iff}}:
+:$\size x < \dfrac \pi 2$
+Hence by the [[Ratio Test]], the series [[Definition:Convergent Series|converges]] for $\size x < \dfrac \pi 2$.
+\end{proof}
+
+\begin{proof}
+From [[Power Series Expansion for Cotangent Function]]:
+:$(1): \quad \cot x = \displaystyle \sum_{n \mathop = 0}^\infty \frac {\left({- 1}\right)^n 2^{2 n} B_{2 n} \, x^{2 n - 1} } {\left({2 n}\right)!}$
+Then:
+{{begin-eqn}}
+{{eqn | l = \tan x
+      | r = \cot x - 2 \cot 2 x
+      | c = [[Cotangent Minus Tangent]]
+}}
+{{eqn | r = \sum_{n \mathop = 0}^\infty \frac {\left({- 1}\right)^n 2^{2 n} B_{2 n} \, x^{2 n - 1} } {\left({2 n}\right)!} - 2 \sum_{n \mathop = 0}^\infty \frac {\left({-1}\right)^n 2^{2 n} B_{2 n} \, \left({2 x}\right)^{2 n - 1} } {\left({2 n}\right)!} 
+      | c = by $(1)$
+}}
+{{eqn | r =  \sum_{n \mathop = 0}^\infty \frac {\left({- 1}\right)^n 2^{2 n} \left({1 - 2^{2 n} }\right) B_{2 n} \, x^{2 n - 1} } {\left({2 n}\right)!}
+      | c =
+}}
+{{eqn | r =  \sum_{n \mathop = 1}^\infty \frac {\left({- 1}\right)^{n - 1} 2^{2 n} \left({2^{2 n} - 1}\right) B_{2 n} \, x^{2 n - 1} } {\left({2 n}\right)!}
+      | c =
+}}
+{{end-eqn}}
+{{qed|lemma}}
+\end{proof}
+
+\begin{proof}
+We have:
+{{begin-eqn}}
+{{eqn | l = \frac x {e^x - 1}
+      | r = \frac x 2 \left({\frac 2 {e^x -  1} }\right)
+      | c = 
+}}
+{{eqn | r = \frac x 2 \left({\frac {e^x - e^x + 2} {e^x -  1} }\right)
+      | c = 
+}}
+{{eqn | r = \frac x 2 \left({\frac {\left({e^x + 1}\right) - \left({e^x - 1}\right)} {e^x -  1} }\right)
+      | c = 
+}}
+{{eqn | r = \frac x 2 \left({\frac {e^x + 1} {e^x - 1} - 1}\right)
+      | c = 
+}}
+{{eqn | r = -\frac x 2 + \frac x 2 \left({\frac {e^x + 1} {e^x - 1} }\right)
+      | c = 
+}}
+{{eqn | r = -\frac x 2 + \frac x 2 \left({\frac {e^{x/2} + e^{-x/2} } {e^{x/2} - e^{-x/2} } }\right)
+      | c = multiplying [[Definition:Numerator|top]] and [[Definition:Denominator|bottom]] by $e^{-x/2}$
+}}
+{{end-eqn}}
+Thus:
+{{begin-eqn}}
+{{eqn | n = 1
+      | l = \frac x 2 \left({\frac {e^{x/2} + e^{-x/2} } {e^{x/2} - e^{-x/2} } }\right)
+      | r = \sum_{n \mathop = 0}^\infty \frac {B_{2 n} } {\left({2 n!}\right)} x^{2 n}
+      | c = {{Defof|Bernoulli Numbers/Generating Function|Bernoulli numbers}}
+}}
+{{end-eqn}}
+Replacing $x$ with $2 i x$ in the {{LHS}} $(1)$:
+{{begin-eqn}}
+{{eqn | o = 
+      | r = \frac {2 i x} 2 \left({\frac {e^{2 i x / 2} + e^{-2 i x / 2} } {e^{2 i x / 2} - e^{-2 i x / 2} } }\right)
+      | c = 
+}}
+{{eqn | r = i x \left({\frac {e^{i x} + e^{-i x} } {e^{i x} - e^{-i x} } }\right)
+      | c = 
+}}
+{{eqn | r = x \cot x
+      | c = [[Cotangent Exponential Formulation]]
+}}
+{{end-eqn}}
+Replacing $x$ with $2 i x$ in the {{RHS}} $(1)$:
+{{begin-eqn}}
+{{eqn | o = 
+      | r = \sum_{n \mathop = 0}^\infty \frac {B_{2 n} } {\left({2 n!}\right)} \left({2 i x}\right)^{2 n}
+      | c = 
+}}
+{{eqn | r = \sum_{n \mathop = 0}^\infty \left({-1}\right)^n \frac {2^{2 n} B_{2 n} } {\left({2 n!}\right)} x^{2 n}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x \cot x
+      | r = \sum_{n \mathop = 0}^\infty \left({-1}\right)^n \frac {2^{2 n} B_{2 n} } {\left({2 n!}\right)} x^{2 n}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \cot x
+      | r = \sum_{n \mathop = 0}^\infty \left({-1}\right)^n \frac {2^{2 n} B_{2 n} } {\left({2 n!}\right)} x^{2 n - 1}
+      | c = 
+}}
+{{end-eqn}}
+Then from [[Cotangent Minus Tangent]]:
+:$\tan x = \cot x - 2 \cot 2 x$
+from which:
+{{begin-eqn}}
+{{eqn | l = \tan x
+      | r = \sum_{n \mathop = 0}^\infty \frac {\left({- 1}\right)^n 2^{2 n} B_{2 n} } {\left({2 n}\right)!} x^{2 n - 1} - 2 \sum_{n \mathop = 0}^\infty \frac {\left({-1}\right)^n 2^{2 n} B_{2 n} } {\left({2 n}\right)!} \left({2 x}\right)^{2 n - 1}
+      | c = by $(1)$
+}}
+{{eqn | r = \sum_{n \mathop = 0}^\infty \frac {\left({- 1}\right)^n 2^{2 n} (1 - 2^{2 n}) B_{2 n} } {\left({2 n}\right)!} x^{2 n - 1}
+      | c =
+}}
+{{eqn | r = \sum_{n \mathop = 1}^\infty \frac {\left({- 1}\right)^{n - 1} 2^{2 n} (2^{2 n} - 1) B_{2 n} } {\left({2 n}\right)!} x^{2 n - 1}
+      | c =
+}}
+{{end-eqn}}
+{{qed|lemma}}
+\end{proof}<|endoftext|>
+\section{Partial Differentiation Operator is Commutative for Continuous Functions}
+Tags: Partial Differentiation
+
+\begin{theorem}
+Let $\map f {x, y}$ be a [[Definition:Real-Valued Function|function]] of the two [[Definition:Real Independent Variable|independent variables]] $x$ and $y$.
+Let $\map f {x, y}$ be [[Definition:Continuous Real-Valued Function|continuous]].
+Let the [[Definition:Partial Derivative|partial deriviatives]] of $f$ also be [[Definition:Continuous Real-Valued Function|continuous]].
+Then:
+:$\dfrac {\partial^2 f} {\partial x \partial y} = \dfrac {\partial^2 f} {\partial y \partial x}$
+\end{theorem}<|endoftext|>
+\section{Primitive of Constant Multiple of Function}
+Tags: Primitives, Primitive of Constant Multiple of Function
+
+\begin{theorem}
+Let $f$ be a [[Definition:Real Function|real function]] which is [[Definition:Integrable Function|integrable]].
+Let $c$ be a [[Definition:Constant|constant]].
+Then:
+:$\displaystyle \int c \map f x \rd x = c \int \map f x \rd x$
+\end{theorem}
+
+\begin{proof}
+From [[Derivative of Constant Multiple]]:
+:$\map {\dfrac \d {\d x} } {c \map f x} = c \map {\dfrac \d {\d x} } {\map f x}$
+The result follows from the definition of [[Definition:Primitive (Calculus)|primitive]].
+\end{proof}<|endoftext|>
+\section{Primitive of Pointwise Sum of Functions}
+Tags: Primitives
+
+\begin{theorem}
+Let $f_1, f_2, \ldots, f_n$ be [[Definition:Real Function|real functions]] which are [[Definition:Integrable Function|integrable]].
+Then:
+:$\ds \int \map {\paren {f_1 \pm f_2 \pm \, \cdots \pm f_n} } x \rd x = \int \map {f_1} x \rd x \pm \int \map {f_2} x \rd x \pm \, \cdots \pm \int \map {f_n} x \rd x$
+\end{theorem}
+
+\begin{proof}
+Proof by [[Principle of Mathematical Induction|induction]]:
+For all $n \in \N_{> 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
+:$\ds \int \map {\paren {f_1 \pm f_2 \pm \, \cdots \pm f_n} } x \rd x = \int \map {f_1} x \rd x \pm \int \map {f_2} x \rd x \pm \, \cdots \pm \int \map {f_n} x \rd x$
+$\map P 1$ is true, as this just says:
+:$\ds \int \map {f_1} x \rd x = \int \map {f_1} x \rd x$
+=== Basis for the Induction ===
+$\map P 2$ is the case:
+{{begin-eqn}}
+{{eqn | l = \int \map {\paren {f_1 \pm f_2} } x \rd x
+      | r = \int \paren {\map {f_1} x \pm \map {f_2} x} \rd x
+      | c = {{Defof|Pointwise Addition of Real-Valued Functions}}
+}}
+{{eqn | r = \int \map {f_1} x \rd x \pm \map {f_2} x \rd x
+      | c = [[Linear Combination of Primitives]]
+}}
+{{end-eqn}}
+This is our [[Definition:Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.
+So this is our [[Definition:Induction Hypothesis|induction hypothesis]]:
+:$\ds \int \map {\paren {f_1 \pm f_2 \pm \, \cdots \pm f_k} } x \rd x = \int \map {f_1} x \rd x \pm \int \map {f_2} x \rd x \pm \, \cdots \pm \int \map {f_k} x \rd x$
+Then we need to show:
+:$\ds \int \map {\paren {f_1 \pm f_2 \pm \, \cdots \pm f_{k + 1} } } x \rd x = \int \map {f_1} x \rd x \pm \int \map {f_2} x \rd x \pm \, \cdots \pm \int \map {f_{k + 1} } x \rd x$
+=== Induction Step ===
+This is our [[Definition:Induction Step|induction step]]:
+{{begin-eqn}}
+{{eqn | o = 
+      | r = \int \map {\paren {f_1 \pm f_2 \pm \, \cdots \pm f_{k + 1} } } x \rd x
+      | c = 
+}}
+{{eqn | r = \int \paren {\map {\paren {f_1 \pm f_2 \pm \, \cdots \pm f_k} } x + \map {f_{k + 1} } x} \rd x
+      | c = {{Defof|Pointwise Addition of Real-Valued Functions}}
+}}
+{{eqn | r = \int \map {\paren {f_1 \pm f_2 \pm \, \cdots \pm f_k} } x \rd x + \int \map {f_{k + 1} } x \rd x
+      | c = [[Primitive of Pointwise Sum of Functions#Basis for the Induction|Basis for the Induction]]
+}}
+{{eqn | r = \int \map {f_1} x \rd x \pm \int \map {f_2} x \rd x \pm \, \cdots \pm \int \map {f_k} x \rd x + \int \map {f_{k + 1} } x \rd x
+      | c = [[Primitive of Pointwise Sum of Functions#Induction Hypothesis|Induction Hypothesis]]
+}}
+{{end-eqn}}
+So $\map P k \implies \map P {k + 1}$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\forall n \in \N: \ds \int \map {\paren {f_1 \pm f_2 \pm \, \cdots \pm f_n} } x \rd x = \int \map {f_1} x \rd x \pm \int \map {f_2} x \rd x \pm \, \cdots \pm \int \map {f_n} x \rd x$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Hurwitz's Theorem (Number Theory)}
+Tags: Number Theory, Hurwitz's Theorem (Number Theory)
+
+\begin{theorem}
+Let $\xi$ be an [[Definition:Irrational Number|irrational number]].
+Then there are [[Definition:Infinite Set|infinitely]] many [[Definition:Coprime Integers|relatively prime integers]] $m, n \in \Z$ such that:
+:$\left|{\xi - \dfrac p q}\right| < \dfrac 1 {\sqrt 5 \, q^2}$
+\end{theorem}
+
+\begin{proof}
+=== [[Hurwitz's Theorem (Number Theory)/Lemma 1|Lemma 1]] ===
+{{:Hurwitz's Theorem (Number Theory)/Lemma 1}}{{qed|lemma}}
+=== [[Hurwitz's Theorem (Number Theory)/Lemma 2|Lemma 2]] ===
+{{:Hurwitz's Theorem (Number Theory)/Lemma 2}}{{qed|lemma}}
+There are an [[Definition:Infinite Set|infinite number]] of [[Definition:Convergent of Continued Fraction|convergents]] to $\xi$.
+Taking these in sets of $3$ at a time, it can be seen from [[Hurwitz's Theorem (Number Theory)/Lemma 2|Lemma 2]] that there are an [[Definition:Infinite Set|infinite number]] of approximations that satisfy the given inequality.
+From [[Hurwitz's Theorem (Number Theory)/Lemma 1|Lemma 1]] it is seen that this inequality is the best possible.
+{{qed}}
+{{Namedfor|Adolf Hurwitz|cat = Hurwitz}}
+\end{proof}<|endoftext|>
+\section{Primitive of Function of Constant Multiple}
+Tags: Primitives
+
+\begin{theorem}
+Let $f$ be a [[Definition:Real Function|real function]] which is [[Definition:Integrable Function|integrable]].
+Let $c$ be a [[Definition:Constant|constant]].
+Then:
+:$\displaystyle \int \map f {c x} \rd x = \frac 1 c \int \map f u \d u$
+where $u = c x$.
+\end{theorem}
+
+\begin{proof}
+Let $u = c x$.
+By [[Derivative of Identity Function/Corollary|Derivative of Identity Function: Corollary]]:
+:$\dfrac {\d u} {\d x} = c$
+Thus:
+{{begin-eqn}}
+{{eqn | l = \int \map f {c x} \rd x
+      | r = \int \frac {\map f u} c \rd u
+      | c = [[Primitive of Composite Function]]
+}}
+{{eqn | r = \frac 1 c \int \map f u \rd u
+      | c = [[Primitive of Constant Multiple of Function]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Composite Function}
+Tags: Primitives
+
+\begin{theorem}
+Let $f$ and $g$ be a [[Definition:Real Function|real functions]] which are [[Definition:Integrable Function|integrable]].
+Let the [[Definition:Composition of Mappings|composite function]] $g \circ f$ also be [[Definition:Integrable Function|integrable]].
+Then:
+{{begin-eqn}}
+{{eqn | l = \int \map {\paren {g \circ f} } x \rd x
+      | r = \int \map g u \frac {\d x} {\d u} \rd u
+      | c = 
+}}
+{{eqn | r = \int \frac {\map g u} {\map {f'} x} \rd u
+      | c = 
+}}
+{{end-eqn}}
+where $u = \map f x$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map F x
+      | r = \int \map {\paren {g \circ f} } x \rd x
+}}
+{{eqn | r = \int \map g {\map f x} \rd x
+      | c = {{Defof|Composition of Mappings}}
+}}
+{{eqn | r = \int \map g u \rd x
+      | c = where $u = \map f x$
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d F} {\d x}
+      | r = \map g u
+      | c = {{Defof|Primitive (Calculus)}}
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d F} {\d u} \frac {\d u} {\d x}
+      | r = \map g u
+      | c = [[Chain Rule for Derivatives]]
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d F} {\d u} \frac \d {\d x} \map f x
+      | r = \map g u
+      | c = Definition of $u$
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d F} {\d u} \map {f'} x
+      | r = \map g u
+      | c = {{Defof|Derivative}}
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d F} {\d u}
+      | r = \frac {\map g u} {\map {f'} x}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = F
+      | r = \int \frac {\map g u} {\map {f'} x} \rd u
+      | c = {{Defof|Primitive (Calculus)}}
+}}
+{{eqn | r = \int \map g u \frac {\d x} {\d u} \rd u
+      | c = [[Derivative of Inverse Function]]: $\dfrac {\d x} {\d u} = \dfrac 1 {\d u / \d x}$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Weierstrass Product Inequality}
+Tags: Inequalities, Infinite Products
+
+\begin{theorem}
+For $n \ge 1$:
+:$\displaystyle \prod_{i \mathop = 1}^n \left({1 - a_i}\right) \ge 1 - \sum_{i \mathop = 1}^n a_i$
+where all of $a_i$ are in the [[Definition:Real Interval/Closed|closed interval]] $\left[{0 \,.\,.\,1 }\right]$.
+\end{theorem}
+
+\begin{proof}
+For $n = 1$ we have:
+:$1 - a_1 \ge 1 - a_1$
+which is clearly true. 
+Suppose the proposition is true for $n = k$, that is:
+:$\displaystyle \prod_{i \mathop = 1}^k \left({1 - a_i}\right) \ge 1 - \sum_{i \mathop = 1}^k a_i$
+Then:
+{{begin-eqn}}
+{{eqn | l = \prod_{i \mathop = 1}^{k + 1} \left({1 - a_i}\right) 
+      | r = \left({1 - a_{k + 1} }\right) \prod_{i \mathop = 1}^k \left({1 - a_i}\right)
+      | c = 
+}}
+{{eqn | o = \ge
+      | r = \left({1 - a_{k + 1} }\right) \left({1 - \sum_{i \mathop = 1}^k a_i}\right)
+      | c = as $1 - a_{k + 1} \ge 0$
+}}
+{{eqn | r = 1 - a_{k + 1} - \sum_{i \mathop = 1}^k a_i + a_{k + 1} \sum_{i \mathop = 1}^k a_i
+      | c = 
+}}
+{{eqn | o = \ge
+      | r = 1 - a_{k + 1} - \sum_{i \mathop = 1}^k a_i
+      | c = as $a_i \ge 0$
+}}
+{{eqn | r = 1 - \sum_{i \mathop = 1}^{k + 1} a_i
+      | c = 
+}}
+{{end-eqn}}
+Thus, by [[Principle of Mathematical Induction]], the proof is complete. 
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Power}
+Tags: Primitives
+
+\begin{theorem}
+Let $n \in \R: n \ne -1$.
+Then:
+:$\ds \int x^n \rd x = \frac {x^{n + 1} } {n + 1} + C$
+where $C$ is an [[Definition:Arbitrary Constant (Calculus)|arbitrary constant]].
+That is:
+:$\dfrac {x^{n + 1} } {n + 1}$ is a [[Definition:Primitive (Calculus)|primitive]] of $x^n$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map {\frac \d {\d x} } {\dfrac {x^{n + 1} } {n + 1} }
+      | r = \paren {n + 1} \paren {\dfrac {x^{\paren {n + 1} - 1} } {n + 1} }
+      | c = [[Power Rule for Derivatives]]
+}}
+{{eqn | r = x^n
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \int x^n \rd x
+      | r = \frac {x^{n + 1} } {n + 1} + C
+      | c = {{Defof|Primitive (Calculus)}}
+}}
+{{end-eqn}}
+When $n = -1$ we have $n + 1 = 0$, and $\dfrac {x^{n + 1} } {n + 1} = \dfrac {x^0} 0$ is undefined.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $u = a x + b$.
+Then:
+:$\dfrac {\d u} {\d x} = a$
+Then:
+{{begin-eqn}}
+{{eqn | l = \int \paren {a x + b}^n \rd x
+      | r = \int \dfrac {u^n} a \rd u
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \dfrac 1 a \dfrac {u^{n + 1} } {n + 1}
+      | c = [[Primitive of Power]]
+}}
+{{eqn | r = \frac {\paren {a x + b}^{n + 1} } {\paren {n + 1} a} + C
+      | c = substituting back for $u$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sine Inequality}
+Tags: Sine Function, Named Theorems
+
+\begin{theorem}
+:$\size {\sin x} \le \size x$
+for all $x \in \R$.
+\end{theorem}
+
+\begin{proof}
+Let $\map f x = x - \sin x$.
+By [[Derivative of Sine Function]]:
+:$\map {f'} x = 1 - \cos x$
+From [[Real Cosine Function is Bounded]] we know $\cos x \le 1$ for all $x$.
+Hence $\map f x \ge 0$ for all $x$.
+From [[Derivative of Monotone Function]], $\map f x$ is [[Definition:Increasing|increasing]]. 
+By [[Sine of Zero is Zero]], $\map f x = 0$.
+It follows that $\map f x \ge 0$ for all $x \ge 0$.
+{{qed|lemma}}
+Now let $\map g x = x^2 - \sin^2 x$.
+{{begin-eqn}}
+{{eqn | l = \map {g'} x
+      | r = 2 x - 2 \sin x \cos x
+      | c = [[Chain Rule for Derivatives]] and [[Derivative of Sine Function]]
+}}
+{{eqn | r = 2 x - \sin 2 x
+      | c = [[Double Angle Formula for Sine]]
+}}
+{{eqn | o = \ge
+      | r = 0
+      | c = for $x \ge 0$, as shown above
+}}
+{{end-eqn}}
+From [[Derivative of Monotone Function]], $\map g x$ is [[Definition:Increasing|increasing]] for $x \ge 0$. 
+By [[Sine of Zero is Zero]], $\map g x = 0$.
+It follows that $\map g x \ge 0$ for all $x \ge 0$.
+Observe that $\map g x$ is an [[Definition:Even Function|even function]].
+This implies $\map g x \ge 0$ for all $x \in \R$.
+Finally note that $\sin^2 x \le x^2 \iff \size {\sin x} \le \size x$.
+{{qed}}
+[[Category:Sine Function]]
+[[Category:Named Theorems]]
+lnyrzyp3giyxd41fhu5add5w5q0lmlt
+\end{proof}<|endoftext|>
+\section{Jordan's Inequality}
+Tags: Sine Function
+
+\begin{theorem}
+:$\dfrac 2 {\pi} x \le \sin x \le x$
+for all $x$ in the [[Definition:Real Interval/Closed|interval]] $\left[{0 \,.\,.\, \dfrac {\pi} 2}\right]$
+\end{theorem}
+
+\begin{proof}
+The right inequality is true by [[Sine Inequality]].
+The left inequality is true for $x = 0$ and $x = \dfrac {\pi} 2$, where we have equality.
+Now consider $x \in \left({0 \,.\,.\, \dfrac {\pi} 2}\right)$.
+From [[Shape of Sine Function]], $\sin x$ is concave on the [[Definition:Closed Real Interval|interval]] $\left[{0 \,.\,.\, \pi }\right]$.
+Letting $x_1 = 0$, $x_2 = x$ and $x_3 = \dfrac {\pi} 2$ in the definition of [[Definition:Concave Real Function/Definition 2|Concave Real Function]] we obtain:
+:$\dfrac {\sin x} x \ge \dfrac {1 - \sin x} {\dfrac {\pi} 2 - x}$
+Rearranging gives the desired inequality.
+{{qed}}
+{{namedfor|Marie Ennemond Camille Jordan}}
+\end{proof}<|endoftext|>
+\section{Cosine Inequality}
+Tags: Cosine Function
+
+\begin{theorem}
+:$1 - \dfrac {x^2} 2 \le \cos x$
+for all $x \in \R$.
+\end{theorem}
+
+\begin{proof}
+Let $\map f x = \cos x - \paren {1 - \dfrac {x^2} 2}$.
+By [[Derivative of Cosine Function]]:
+:$\map {f'} x = x - \sin x$
+From [[Sine Inequality]], we know $\sin x \le x$ for $x \ge 0$.
+Hence $\map {f'} x \ge 0$ for $x \ge 0$.
+From [[Derivative of Monotone Function]], $\map f x$ is [[Definition:Increasing|increasing]] for $x \ge 0$.
+By [[Cosine of Zero is One]], $\map f 0 = 0$.
+It follows that $\map f x \ge 0$ for $x \ge 0$.
+By [[Cosine Function is Even]], $\map f x$ is an [[Definition:Even Function|even function]].
+This implies $\map f x \ge 0$ for all $x \in \R$.
+{{qed}}
+[[Category:Cosine Function]]
+sxm3zaovh90d8feog19hsx3m5qogudc
+\end{proof}<|endoftext|>
+\section{Laplace Transform of Constant Mapping}
+Tags: Laplace Transforms, Constant Mappings, Laplace Transform of Constant Mapping
+
+\begin{theorem}
+Let $a \in \R$ be a [[Definition:Real Number|real]] [[Definition:Constant|constant]].
+Let $f_a: \R \to \R$ or $\C$ be the [[Definition:Constant Mapping|constant mapping]], defined as:
+:$\forall t \in \R: \map {f_a} t = a$
+Let $\laptrans {f_a}$ be the [[Definition:Laplace Transform|Laplace transform]] of $f_a$.
+Then:
+:$\laptrans {\map {f_a} t} = \dfrac a s$
+for $\map \Re s > a$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \laptrans {\map {f_a} t}
+      | r = \laptrans a
+      | c = {{Defof|Constant Mapping}}
+}}
+{{eqn | r = a \, \laptrans 1
+      | c = [[Linear Combination of Laplace Transforms]]
+}}
+{{eqn | r = a \int_0^{\to +\infty} e^{-s t} \rd t
+      | c = {{Defof|Laplace Transform}}
+}}
+{{eqn | r = \intlimits {- \frac a s e^{-st} } {t \mathop = 0} {t \mathop \to +\infty}
+}}
+{{eqn | r = 0 - \paren {- \frac a s}
+      | c = [[Complex Exponential Tends to Zero]], [[Exponential of Zero]]
+}}
+{{eqn | r = \frac a s
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \laptrans {\map {f_a} t}
+      | r = \laptrans a
+      | c = {{Defof|Constant Mapping}}
+}}
+{{eqn | r = a \, \laptrans 1
+      | c = [[Linear Combination of Laplace Transforms]]
+}}
+{{eqn | r = a \frac 1 s
+      | c = [[Laplace Transform of 1]]
+}}
+{{eqn | r = \frac a s
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Laplace Transform of Identity Mapping}
+Tags: Laplace Transforms, Identity Mappings, Laplace Transform of Identity Mapping
+
+\begin{theorem}
+Let $\laptrans f$ denote the [[Definition:Laplace Transform|Laplace transform]] of a [[Definition:Function|function]] $f$.
+Let $\map {I_\R} t$ denote the [[Definition:Identity Mapping|identity mapping]] on $\R$ for $t > 0$.
+Then:
+:$\laptrans {\map {I_\R} t} = \dfrac 1 {s^2}$
+for $\map \Re s > 0$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \laptrans {\map {I_\R} t}
+      | r = \laptrans t
+      | c = {{Defof|Identity Mapping}}
+}}
+{{eqn | r = \int_0^{\to +\infty} t e^{-st} \rd t
+      | c = {{Defof|Laplace Transform}}
+}}
+{{end-eqn}}
+From [[Integration by Parts]]:
+:$\displaystyle \int f g' \rd t = f g - \int f'g \rd t$
+Here:
+{{begin-eqn}}
+{{eqn | l = f
+      | r = t
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = f'
+      | r = 1
+      | c = [[Derivative of Identity Function]]
+}}
+{{eqn | l = g'
+      | r = e^{-st}
+}}
+{{eqn | ll= \leadsto
+      | l = g
+      | r = -\frac 1 s e^{-s t}
+      | c = [[Primitive of Exponential Function]]
+}}
+{{end-eqn}}
+So: 
+{{begin-eqn}}
+{{eqn | l = \int t e^{-s t} \rd t
+      | r = -\frac t s e^{-s t} - \frac 1 s \int e^{-s t} \rd t
+}}
+{{eqn | r = -\frac t s e^{-s t} - \frac 1 {s^2} e^{-s t}
+      | c = [[Primitive of Exponential Function]]
+}}
+{{end-eqn}}
+Evaluating at $t = 0$ and $t \to +\infty$:
+{{begin-eqn}}
+{{eqn | l = \laptrans t
+      | r = \intlimits {-\frac t s e^{-s t} - \frac 1 {s^2} e^{-s t} } {t \mathop = 0} {t \mathop \to +\infty}
+}}
+{{eqn | r = -\frac 1 s \lim_{t \mathop \to +\infty} \frac t { e^{s t} } - \paren {0 - \frac 1 {s^2} }
+      | c = [[Exponential of Zero and One]], [[Exponent Combination Laws/Negative Power|Exponent Combination Laws: Negative Power]]
+}}
+{{eqn | r = 0 + \frac 1 {s^2}
+      | c = [[Limit at Infinity of Polynomial over Complex Exponential]]
+}}
+{{eqn | r = \frac 1 {s^2}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Laplace Transform of Positive Integer Power}
+Tags: Laplace Transforms, Laplace Transform of Positive Integer Power
+
+\begin{theorem}
+Let $\laptrans f$ denote the [[Definition:Laplace Transform|Laplace transform]] of a [[Definition:Function|function]] $f$.
+Let $t^n: \R \to \R$ be [[Definition:Integer Power|$t$ to the $n$th power]] for some $n \in \N_{\ge 0}$.
+Then:
+:$\laptrans {t^n} = \dfrac {n!} { s^{n + 1} }$
+for $\map \Re s > 0$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \laptrans {t^n}
+      | r = \int_0^{\to +\infty} t^n e^{-s t} \rd t
+      | c = {{Defof|Laplace Transform}}
+}}
+{{eqn | r = \intlimits {\frac {e^{-s t} } {-s} \sum_{k \mathop = 0}^n \paren {\paren {-1}^k \frac {n^{\underline k} t^{n - k} } {\paren {-s}^k} } } {t \mathop = 0} {t \mathop \to +\infty}
+      | c = [[Primitive of Power of x by Exponential of a x|Primitive of $x^n e^{a x}$]]
+}}
+{{eqn | r = \sum_{k \mathop = 0}^n \paren {0 - \frac 1 {-s} \paren {-1}^k \frac {n^{\underline k} 0^{n - k} } {\paren {-s}^k} }
+      | c = [[Exponential Tends to Zero and Infinity]], [[Exponent Combination Laws/Negative Power|Exponent Combination Laws: Negative Power]]
+}}
+{{eqn | r = \sum_{k \mathop = 0}^n \paren {\dfrac 1 {s^{k + 1} } n^{\underline k} 0^{n - k} }
+      | c = simplifying
+}}
+{{eqn | r = \dfrac 1 {s^{k + 1} } n^{\underline n}
+      | c = $0^{n - k} = 0$ for all terms except $n = k$, when $0^{n - n} = 1$
+}}
+{{eqn | r = \dfrac {n!} {s^{k + 1} }
+      | c = [[Integer to Power of Itself Falling is Factorial]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Function under its Derivative}
+Tags: Primitives
+
+\begin{theorem}
+Let $f$ be a [[Definition:Real Function|real function]] which is [[Definition:Integrable Function|integrable]].
+Then:
+:$\ds \int \frac {\map {f'} x} {\map f x} \rd x = \ln \size {\map f x} + C$
+where $C$ is an [[Definition:Arbitrary Constant (Calculus)|arbitrary constant]].
+\end{theorem}
+
+\begin{proof}
+By [[Integration by Substitution]] (with appropriate renaming of variables):
+:$\ds \int \map g u \rd u = \int \map g {\map f x} \map {f'} x \rd x$
+Then:
+{{begin-eqn}}
+{{eqn | l = \int \frac {\map {f'} x} {\map f x} \rd x
+      | r = \int \frac {\d u} u
+      | c = putting $\map g u := \dfrac 1 u$
+}}
+{{eqn | r = \ln \size u + C
+      | c = [[Primitive of Reciprocal]]
+}}
+{{eqn | r = \ln \size {\map f x} + C
+      | c = Definition of $u$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Exponential Function/General Result}
+Tags: Primitive of Exponential Function
+
+\begin{theorem}
+Let $a \in \R_{>0}$ be a [[Definition:Constant|constant]] such that $a \ne 1$.
+Then:
+:$\ds \int a^x \rd x = \frac {a^x} {\ln a} + C$
+where $C$ is an [[Definition:Arbitrary Constant (Calculus)|arbitrary constant]].
+\end{theorem}
+
+\begin{proof}
+Let $u = x \ln a$.
+{{begin-eqn}}
+{{eqn | l = \int a^x \rd x
+      | r = \int \map \exp {x \ln a} \rd x
+      | c = {{Defof|Power to Real Number}}
+}}
+{{eqn | r = \frac 1 {\ln a} \int \map \exp u \rd u
+      | c = [[Primitive of Function of Constant Multiple]]
+}}
+{{eqn | r = \frac {\map \exp u} {\ln a} + C
+      | c = [[Primitive of Exponential Function]]
+}}
+{{eqn | r = \frac {\map \exp {x \ln a} } {\ln a} + C
+      | c = Definition of $u$
+}}
+{{eqn | r = \frac {a^x} {\ln a} + C
+      | c = {{Defof|Power to Real Number}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Tangent Function/Secant Form}
+Tags: Primitive of Tangent Function
+
+\begin{theorem}
+:$\ds \int \tan x \rd x = \ln \size {\sec x} + C$
+where $\sec x$ is defined.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int \tan x \rd x
+      | r = -\ln \size {\cos x} + C
+      | c = [[Primitive of Tangent Function/Cosine Form|Primitive of $\tan x$: Cosine Form]]
+}}
+{{eqn | r = \ln \size {\frac 1 {\cos x} } + C
+      | c = [[Logarithm of Reciprocal]]
+}}
+{{eqn | r = \ln \size {\sec x} + C
+      | c = [[Secant is Reciprocal of Cosine]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Secant Function/Tangent plus Angle Form}
+Tags: Primitive of Secant Function
+
+\begin{theorem}
+:$\ds \int \sec x \rd x = \ln \size {\map \tan {\frac x 2 + \frac \pi 4} } + C$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int \sec x \rd x
+      | r = \ln \size {\tan x + \sec x} + C
+      | c = [[Primitive of Secant Function/Secant plus Tangent Form|Primitive of $\sec x$: Secant plus Tangent Form]]
+}}
+{{eqn | r = \ln \size {\map \tan {\frac x 2 + \frac \pi 4} } + C
+      | c = [[Tangent of Half Angle plus Quarter Pi|Tangent of Half Angle plus $\dfrac \pi 4$]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Exponential of x not less than 1+x}
+Tags: Exponential Function, Inequalities
+
+\begin{theorem}
+:$e^x \ge 1 + x$
+for all $x \in \R$.
+\end{theorem}
+
+\begin{proof}
+For $x > - 1$:
+{{begin-eqn}}
+{{eqn | l = e^x
+      | r = \lim_{n \mathop \to \infty} \left({1 + \frac x n}\right)^n
+      | c = {{Defof|Real Exponential Function}}
+}}
+{{eqn | o = \ge
+      | r = \lim_{n \mathop \to \infty} \left({1 + x}\right)
+      | c = [[Bernoulli's Inequality]]
+}}
+{{eqn | r = 1 + x
+      | c = 
+}}
+{{end-eqn}}
+as required.
+For $x \le -1$, the inequality follows from the fact that $e^x$ is [[Definition:Positive Real Function|positive]] for all $x$.
+{{qed}}
+[[Category:Exponential Function]]
+[[Category:Inequalities]]
+hqlcsp872ooy00iyqpsw50x7c4n1qxm
+\end{proof}<|endoftext|>
+\section{Primitive of Cosecant Function/Cosecant minus Cotangent Form}
+Tags: Primitive of Cosecant Function
+
+\begin{theorem}
+:$\ds \int \csc x \rd x = \ln \size {\csc x - \cot x} + C$
+where $\csc x - \cot x \ne 0$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int \csc x \rd x
+      | r = \ln \size {\tan \frac x 2} + C
+      | c = [[Primitive of Cosecant Function/Tangent Form|Primitive of $\csc x$: Tangent Form]]
+}}
+{{eqn | r = \ln \size {\frac {1 - \cos x} {\sin x} } + C
+      | c = [[Half Angle Formulas/Tangent/Corollary 2|Half Angle Formula for Tangent: Corollary 2]]
+}}
+{{eqn | r = \ln \size {\frac 1 {\sin x} - \frac {\cos x} {\sin x} } + C
+      | c = 
+}}
+{{eqn | r = \ln \size {\csc x - \cot x} + C
+      | c = {{Defof|Cosecant}} and {{Defof|Cotangent}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Cosecant Function/Tangent Form}
+Tags: Primitive of Cosecant Function
+
+\begin{theorem}
+:$\ds \int \csc x \rd x = \ln \size {\tan \frac x 2} + C$
+where $\tan \dfrac x 2 \ne 0$.
+\end{theorem}<|endoftext|>
+\section{Primitive of Square of Secant Function}
+Tags: Primitives involving Secant Function, Tangent Function
+
+\begin{theorem}
+:$\ds \int \sec^2 x \rd x = \tan x + C$
+where $C$ is an [[Definition:Arbitrary Constant (Calculus)|arbitrary constant]].
+\end{theorem}
+
+\begin{proof}
+From [[Derivative of Tangent Function]]:
+:$\map {\dfrac \d {\d x} } {\tan x} = \sec^2 x$
+The result follows from the definition of [[Definition:Primitive (Calculus)|primitive]].
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Square of Cosecant Function}
+Tags: Primitives involving Cosecant Function, Cotangent Function
+
+\begin{theorem}
+:$\ds \int \csc^2 x \rd x = -\cot x + C$
+where $C$ is an [[Definition:Arbitrary Constant (Calculus)|arbitrary constant]].
+\end{theorem}
+
+\begin{proof}
+From [[Derivative of Cotangent Function]]:
+:$\dfrac \d {\d x} \cot x = -\csc^2 x$
+The result follows from the definition of [[Definition:Primitive (Calculus)|primitive]].
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Square of Tangent Function}
+Tags: Primitives involving Tangent Function, Primitive of Square of Tangent Function
+
+\begin{theorem}
+:$\displaystyle \int \tan^2 x \rd x = \tan x - x + C$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int \tan^2 x \ \mathrm d x
+      | r = \int \left({\sec^2 x - 1}\right) \ \mathrm d x
+      | c = [[Difference of Squares of Secant and Tangent]]
+}}
+{{eqn | r = \int \sec^2 x \ \mathrm d x + \int \left({-1}\right) \ \mathrm d x
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \tan x + C + \int \left({-1}\right) \ \mathrm d x
+      | c = [[Primitive of Square of Secant Function]]
+}}
+{{eqn | r = \tan x - x + C
+      | c = [[Primitive of Constant]]
+}}
+{{end-eqn}}
+{{Qed}}
+\end{proof}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn	| l = I_n
+	| r = \int \tan^n\left({x}\right) \mathrm d x
+}}
+{{eqn	| r = \frac {\tan^{n - 1} \left({x}\right)} {n - 1} - I_{n - 2} 
+	| c = [[Reduction Formula for Integral of Power of Tangent]]
+}}
+{{eqn	| l = I_0
+	| r = \int \left(\tan\left({x}\right)\right)^0 \mathrm d x
+}}
+{{eqn	| r = \int \mathrm d x
+	| c = 
+}}
+{{eqn	| r = x + C
+	| c = [[Primitive of Constant]]
+}}
+{{eqn	| ll= \leadsto
+	| l = I_2
+	| r = \tan\left({x}\right) - x + C'
+	| c = setting $n = 2$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Square of Cotangent Function}
+Tags: Primitives involving Cotangent Function
+
+\begin{theorem}
+:$\displaystyle \int \cot^2 x \rd x = -\cot x - x + C$
+where $C$ is an [[Definition:Arbitrary Constant (Calculus)|arbitrary constant]].
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int \cot^2 x \rd x
+      | r = \int \paren {\csc^2 x - 1} \rd x
+      | c = [[Difference of Squares of Cosecant and Cotangent]]
+}}
+{{eqn | r = \int \csc^2 x \rd x + \int \paren {-1} \rd x
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = -\cot x + C + \int \paren {-1} \rd x
+      | c = [[Primitive of Square of Cosecant Function]]
+}}
+{{eqn | r = -\cot x - x + C
+      | c = [[Primitive of Constant]]
+}}
+{{end-eqn}}
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Square of Sine Function}
+Tags: Primitives involving Sine Function
+
+\begin{theorem}
+:$\ds \int \sin^2 x \rd x = \frac x 2 - \frac {\sin 2 x} 4 + C$
+where $C$ is an [[Definition:Arbitrary Constant (Calculus)|arbitrary constant]].
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int \sin^2 x \rd x
+      | r = \int \paren {\frac {1 - \cos 2 x} 2} \rd x
+      | c = [[Square of Sine]]
+}}
+{{eqn | r = \frac 1 2 \int \d x - \frac 1 2 \int \cos 2 x \rd x
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \frac x 2 - \frac 1 2 \int \cos 2 x \rd x + C
+      | c = [[Primitive of Constant]]
+}}
+{{eqn | r = \frac x 2 - \frac 1 2 \paren {\frac {\sin 2 x} 2} + C
+      | c = [[Primitive of Cosine of a x|Primitive of $\cos a x$]]
+}}
+{{eqn | r = \frac x 2 - \frac {\sin 2 x} 4 + C
+      | c = rearranging
+}}
+{{end-eqn}}
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Square of Cosine Function}
+Tags: Primitive of Square of Cosine Function
+
+\begin{theorem}
+:$\ds \int \cos^2 x \rd x = \frac x 2 + \frac {\sin 2 x} 4 + C$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn	| l = I_n
+	| r = \int \cos^n x \rd x
+}}
+{{eqn	| r = \dfrac {\cos^{n - 1} x \sin x} n + \dfrac {n - 1} n I_{n-2}
+	| c = [[Reduction Formula for Integral of Power of Cosine]]
+}}
+{{eqn	| l = I_0
+	| r = \int \left({\cos x}\right)^0 \rd x
+	| c = 
+}}
+{{eqn	| r = \int \rd x
+	| c = 
+}}
+{{eqn	| r = x + C
+	| c = [[Primitive of Constant]]
+}}
+{{eqn	| ll= \leadsto
+	| l  = I_2
+	| r  = \frac {\cos x \sin x} 2 + \frac x 2 + \frac C 2
+	| c  = setting $n = 2$
+}}
+{{eqn	| r = \frac {\sin 2 x} 4 + \frac x 2 + C'
+	| c = [[Double Angle Formula for Sine]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int \cos^2 x \rd x
+      | r = \int \paren {\frac {1 + \cos 2 x} 2} \rd x
+      | c = [[Square of Cosine]]
+}}
+{{eqn | r = \int \frac 1 2 \rd x + \int \paren {\frac {\cos 2 x} 2} \rd x
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \frac x 2 + C + \int \paren {\frac {\cos 2 x} 2} \rd x
+      | c = [[Primitive of Constant]]
+}}
+{{eqn | r = \frac x 2 + C + \frac 1 2 \int \cos 2 x \rd x
+      | c = [[Primitive of Constant Multiple of Function]]
+}}
+{{eqn | r = \frac x 2 + \frac 1 2 \paren {\frac {\sin 2 x} 2} + C
+      | c = [[Primitive of Function of Constant Multiple]] and [[Primitive of Cosine Function]]
+}}
+{{eqn | r = \frac x 2 + \frac {\sin 2 x} 4 + C
+      | c = [[Primitive of Function of Constant Multiple]] and [[Primitive of Cosine Function]]
+}}
+{{end-eqn}}
+{{Qed}}
+\end{proof}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int \cos^2 x \rd x 
+      | r = \frac 1 4 \int \left({e^{i x} + e^{-i x} }\right)^2 \rd x
+      | c = [[Cosine Exponential Formulation]]
+}}
+{{eqn | r = \frac 1 4 \int \left({e^{2 i x} + 2 + e^{-2 i x} }\right) \rd x
+}}
+{{eqn | r = \frac 1 4 \left({\frac{e^{2 i x} - e^{-2 i x} } {2 i} + 2 x}\right) + C
+      | c = [[Primitive of Exponential of a x|Primitive of $e^{a x}$]], [[Primitive of Constant]]
+}}
+{{eqn | r = \frac {\sin 2 x} 4 + \frac x 2 + C
+      | c = [[Sine Exponential Formulation]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Product of Secant and Tangent}
+Tags: Primitives involving Tangent Function, Primitives involving Secant Function
+
+\begin{theorem}
+:$\ds \int \sec x \tan x \rd x = \sec x + C$
+where $C$ is an [[Definition:Arbitrary Constant (Calculus)|arbitrary constant]].
+\end{theorem}
+
+\begin{proof}
+From [[Derivative of Secant Function]]:
+:$\dfrac \d {\d x} \sec x = \sec x \tan x$
+The result follows from the definition of [[Definition:Primitive (Calculus)|primitive]].
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Product of Cosecant and Cotangent}
+Tags: Primitives involving Cosecant Function, Primitives involving Cotangent Function
+
+\begin{theorem}
+:$\ds \int \csc x \cot x \rd x = -\csc x + C$
+where $C$ is an [[Definition:Arbitrary Constant (Calculus)|arbitrary constant]].
+\end{theorem}
+
+\begin{proof}
+From [[Derivative of Cosecant Function]]:
+:$\dfrac \d {\d x} \csc x = -\csc x \cot x$
+The result follows from the definition of [[Definition:Primitive (Calculus)|primitive]].
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Hyperbolic Sine Function}
+Tags: Primitives of Hyperbolic Functions, Hyperbolic Sine Function, Primitives involving Hyperbolic Sine Function
+
+\begin{theorem}
+:$\ds \int \sinh x \rd x = \cosh x + C$
+\end{theorem}
+
+\begin{proof}
+From [[Derivative of Hyperbolic Cosine]]:
+:$\map {\dfrac \d {\d x} } {\cosh x} = \sinh x$
+The result follows from the definition of [[Definition:Primitive (Calculus)|primitive]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Hyperbolic Cosine Function}
+Tags: Primitives of Hyperbolic Functions, Hyperbolic Cosine Function, Primitives involving Hyperbolic Cosine Function
+
+\begin{theorem}
+:$\displaystyle \int \cosh x \rd x = \sinh x + C$
+\end{theorem}
+
+\begin{proof}
+From [[Derivative of Hyperbolic Sine]]:
+:$\map {\dfrac \d {\d x} } {\sinh x} = \cosh x$
+The result follows from the definition of [[Definition:Primitive (Calculus)|primitive]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Hyperbolic Tangent Function}
+Tags: Primitives of Hyperbolic Functions, Hyperbolic Tangent Function, Primitive of Hyperbolic Tangent Function
+
+\begin{theorem}
+:$\ds \int \tanh x \rd x = \map \ln {\cosh x} + C$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn	| l = \int \tanh x \rd x
+	| r = -i \int \tan i x \rd x
+	| c = [[Hyperbolic Tangent in terms of Tangent]]
+}}
+{{eqn	| r = -\int \tan i x \rd \paren {i x}
+	| c = [[Primitive of Function of Constant Multiple]]
+}}
+{{eqn	| r = \ln \cmod {\cos i x} + C
+	| c = [[Primitive of Tangent Function/Cosine Form|Primitive of $\tan x$: Cosine Form]]
+}}
+{{eqn	| r = \ln \cmod {\cosh x} + C
+	| c = {{Defof|Hyperbolic Cosine}}
+}}
+{{eqn | r = \map \ln {\cosh x} + C
+      | c = [[Graph of Hyperbolic Cosine Function]]: $\cosh x > 0$ for all $x$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int \tanh x \rd x
+      | r = \int \frac {\sinh x} {\cosh x} \rd x
+      | c = {{Defof|Hyperbolic Tangent}}
+}}
+{{eqn | r = \int \frac {\paren {\cosh x}'} {\cosh x} \rd x
+      | c = [[Derivative of Hyperbolic Cosine]]
+}}
+{{eqn | r = \ln \size {\cosh x} + C
+      | c = [[Primitive of Function under its Derivative]]
+}}
+{{eqn | r = \map \ln {\cosh x} + C
+      | c = [[Graph of Hyperbolic Cosine Function]]: $\cosh x > 0$ for all $x$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Hyperbolic Cotangent Function}
+Tags: Primitives of Hyperbolic Functions, Hyperbolic Cotangent Function, Primitives involving Hyperbolic Cotangent Function
+
+\begin{theorem}
+:$\ds \int \coth x \rd x = \ln \size {\sinh x} + C$
+where $\sinh x \ne 0$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int \coth x \rd x
+      | r = \int \frac {\cosh x} {\sinh x} \rd x
+      | c = {{Defof|Hyperbolic Cotangent}}
+}}
+{{eqn | r = \int \frac {\paren {\sinh x}'} {\sinh x} \rd x
+      | c = [[Derivative of Hyperbolic Sine]]
+}}
+{{eqn | r = \ln \size {\sinh x} + C
+      | c = [[Primitive of Function under its Derivative]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Limit of Tan X over X}
+Tags: Tangent Function, Limit of Tan X over X
+
+\begin{theorem}
+:$\displaystyle \lim_{x \mathop \to 0} \frac {\tan x} x = 1$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn |l = \lim_{x \mathop \to 0} \frac {\tan x} x 
+      |r = \lim_{x \mathop \to 0} \frac 1 {\cos x} \frac {\sin x} x
+      |c = {{Defof|Tangent Function}}
+}}
+{{eqn |r = \lim_{x \mathop \to 0} \frac 1 {\cos x} \lim_{x \mathop \to 0} \frac {\sin x} x
+      |c = [[Product Rule for Limits of Functions]]
+}}
+{{eqn |r = \lim_{x \mathop \to 0} \frac {\sin x} x 
+      |c = [[Cosine of Zero is One]]
+}}
+{{eqn |r = 1
+      |c = [[Limit of Sine of X over X]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+By [[L'Hôpital's Rule]]:
+{{begin-eqn}}
+{{eqn |l = \lim_{x \mathop \to 0} \frac {\tan x} x 
+      |r = \lim_{x \mathop \to 0} \frac {\sec^2 x} 1
+      |c = [[Derivative of Tangent Function]]
+}}
+{{eqn |r = 1
+      |c = [[Secant of Zero]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Euler Formula for Sine Function/Real Numbers}
+Tags: Euler Formula for Sine Function
+
+\begin{theorem}
+:$\displaystyle \sin x = x \prod_{n \mathop = 1}^\infty \paren {1 - \frac {x^2} {n^2 \pi^2} }$
+for all $x \in \R$.
+\end{theorem}<|endoftext|>
+\section{Euler Formula for Sine Function/Complex Numbers}
+Tags: Euler Formula for Sine Function
+
+\begin{theorem}
+:$\displaystyle \sin z = z \prod_{n \mathop = 1}^\infty \left({1 - \frac {z^2} {n^2 \pi^2}}\right)$
+for all $z \in \C$.
+\end{theorem}<|endoftext|>
+\section{Logarithm of Reciprocal}
+Tags: Logarithms, Reciprocals
+
+\begin{theorem}
+:$\map {\log_b} {\dfrac 1 x} = -\log_b x$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map {\log_b} {\dfrac 1 x}
+      | r = \log_b 1 - \log_b x
+      | c = [[Difference of Logarithms]]
+}}
+{{eqn | r = 0 - \log_b x
+      | c = [[Logarithm of 1 is 0]]
+}}
+{{end-eqn}}
+Hence the result.
+{{qed}}
+[[Category:Logarithms]]
+[[Category:Reciprocals]]
+kgfbjw7h1n6cupqiu7qtvossewt0xp8
+\end{proof}<|endoftext|>
+\section{Primitive of Hyperbolic Secant Function/Arcsine Form}
+Tags: Primitive of Hyperbolic Secant Function
+
+\begin{theorem}
+:$\ds \int \sech x \rd x = \map \arcsin {\tanh x} + C$
+\end{theorem}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = \int \sech x \rd x
+      | r = \int \frac {\sech^2 x} {\sech x} \rd x
+      | c = 
+}}
+{{eqn | r = \int \frac {\sech^2 x} {\sqrt {1 - \tanh^2 x} } \rd x
+      | c = [[Sum of Squares of Hyperbolic Secant and Tangent]]
+}}
+{{end-eqn}}
+Let:
+{{begin-eqn}}
+{{eqn | l = u
+      | r = \tanh x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = u'
+      | r = \sech^2 x
+      | c = [[Derivative of Hyperbolic Tangent]]
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = \int \sech x \rd x
+      | r = \int \frac {\d u} {\sqrt {1 - u^2} }
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \map \arcsin u + C
+      | c = [[Primitive of Reciprocal of Root of a squared minus x squared/Arcsine Form|Primitive of $\dfrac 1 {\sqrt {a^2 - x^2} }$: Arcsine Form]]
+}}
+{{eqn | r = \map \arcsin {\tanh x} + C
+      | c = Definition of $u$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Hyperbolic Secant Function/Arctangent of Exponential Form}
+Tags: Primitive of Hyperbolic Secant Function
+
+\begin{theorem}
+:$\ds \int \sech x \rd x = 2 \map \arctan {e^x} + C$
+\end{theorem}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = \int \sech x \rd x
+      | r = \int \frac 2 {e^x + e^{-x} } \rd x
+      | c = {{Defof|Hyperbolic Secant|index = 1}}
+}}
+{{eqn | r = \int \frac {2 e^x} {e^{2 x} + 1} \rd x
+      | c = multiplying [[Definition:Numerator|top]] and [[Definition:Denominator|bottom]] by $e^x$
+}}
+{{end-eqn}}
+Let:
+{{begin-eqn}}
+{{eqn | l = u
+      | r = e^x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = u'
+      | r = e^x
+      | c = [[Derivative of Exponential Function]]
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = \int \sech x \rd x
+      | r = \int \frac {2 \rd u} {u^2 + 1}
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = 2 \arctan u + C
+      | c = [[Primitive of Reciprocal of x squared plus a squared/Arctangent Form|Primitive of $\dfrac 1 {x^2 + a^2}$: Arctangent Form]]
+}}
+{{eqn | r = 2 \map \arctan {e^x} + C
+      | c = Definition of $u$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Hyperbolic Cosecant Function/Logarithm Form}
+Tags: Primitive of Hyperbolic Cosecant Function
+
+\begin{theorem}
+:$\displaystyle \int \csch x \rd x = -\ln \size {\csch x + \coth x} + C$
+where $\csch x + \coth x \ne 0$.
+\end{theorem}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = u
+      | r = \coth x + \csch x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d u} {\d x}
+      | r = \frac \d {\d x} \coth x + \frac \d {\d x} \csch x 
+      | c = [[Linear Combination of Derivatives]]
+}}
+{{eqn | r = -\csch^2 x + \frac \d {\d x} \csch x 
+      | c = [[Derivative of Hyperbolic Cotangent]]
+}}
+{{eqn | r = -\csch^2 x - \csch x \coth x 
+      | c = [[Derivative of Hyperbolic Cosecant]]
+}}
+{{eqn | r = -\csch x \left({\csch x + \coth x}\right)
+      | c = factorising
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = \int \csch x \rd x
+      | r = \int \frac {\csch x \paren {\csch x + \coth x} } {\csch x + \coth x} \rd x
+      | c = multiplying [[Definition:Numerator|top]] and [[Definition:Denominator|bottom]] by $\csch x + \coth x$
+}}
+{{eqn | r = -\int \frac {-\csch x \paren {\csch x + \coth x} } {\csch x + \coth x} \rd x
+      | c = [[Primitive of Constant Multiple of Function]]
+}}
+{{eqn | r = -\ln \size {\csch x + \coth x} + C
+      | c = [[Primitive of Function under its Derivative]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Hyperbolic Cosecant Function/Hyperbolic Tangent Form}
+Tags: Primitive of Hyperbolic Cosecant Function
+
+\begin{theorem}
+:$\ds \int \csch x \rd x = \ln \size {\tanh \frac x 2} + C$
+where $\tanh \dfrac x 2 \ne 0$.
+\end{theorem}<|endoftext|>
+\section{Primitive of Hyperbolic Cosecant Function/Inverse Hyperbolic Cotangent Form}
+Tags: Primitive of Hyperbolic Cosecant Function, Primitives involving Hyperbolic Cotangent Function
+
+\begin{theorem}
+:$\displaystyle \int \csch x \rd x = -2 \map {\coth^{-1} } {e^x} + C$
+\end{theorem}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = \int \csch x \rd x
+      | r = \int \frac 2 {e^x - e^{-x} } \rd x
+      | c = {{Defof|Hyperbolic Cosecant}}
+}}
+{{eqn | r = \int \frac {2 e^x} {e^{2 x} - 1} \rd x
+      | c = multiplying [[Definition:Numerator|top]] and [[Definition:Denominator|bottom]] by $e^x$
+}}
+{{end-eqn}}
+Let:
+{{begin-eqn}}
+{{eqn | l = u
+      | r = e^x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = u'
+      | r = e^x
+      | c = [[Derivative of Exponential Function]]
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = \int \csch x \rd x
+      | r = \int \frac {2 \rd u} {u^2 - 1}
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = -2 \coth^{-1} u + C
+      | c = [[Primitive of Reciprocal of x squared minus a squared/Inverse Hyperbolic Cotangent Form|Primitive of Reciprocal of $x^2 - a^2$: $\coth^{-1}$ form]]
+}}
+{{eqn | r = -2 \map {\coth^{-1} } {e^x} + C
+      | c = Definition of $u$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Square of Hyperbolic Secant Function}
+Tags: Primitives of Hyperbolic Functions, Hyperbolic Secant Function, Hyperbolic Tangent Function
+
+\begin{theorem}
+:$\ds \int \sech^2 x \rd x = \tanh x + C$
+where $C$ is an [[Definition:Arbitrary Constant (Calculus)|arbitrary constant]].
+\end{theorem}
+
+\begin{proof}
+From [[Derivative of Hyperbolic Tangent]]:
+:$\map {\dfrac \d {\d x} } {\tanh x} = \sech^2 x$
+The result follows from the definition of [[Definition:Primitive (Calculus)|primitive]].
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Square of Hyperbolic Cosecant Function}
+Tags: Primitives of Hyperbolic Functions, Hyperbolic Cosecant Function, Hyperbolic Cotangent Function
+
+\begin{theorem}
+:$\ds \int \csch^2 x \rd x = -\coth x + C$
+where $C$ is an [[Definition:Arbitrary Constant (Calculus)|arbitrary constant]].
+\end{theorem}
+
+\begin{proof}
+From [[Derivative of Hyperbolic Cotangent]]:
+:$\map {\dfrac \d {\d x} } {\coth x} = -\csch^2 x$
+The result follows from the definition of [[Definition:Primitive (Calculus)|primitive]].
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Square of Hyperbolic Tangent Function}
+Tags: Primitives of Hyperbolic Functions, Hyperbolic Tangent Function
+
+\begin{theorem}
+:$\displaystyle \int \tanh^2 x \rd x = x - \tanh x + C$
+where $C$ is an [[Definition:Arbitrary Constant (Calculus)|arbitrary constant]].
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int \tanh^2 x \rd x
+      | r = \int \paren {1 - \sech^2 x} \rd x
+      | c = [[Sum of Squares of Hyperbolic Secant and Tangent]]
+}}
+{{eqn | r = \int 1 \rd x - \int \sech^2 x \rd x
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \int 1 \rd x - \tanh x + C
+      | c = [[Primitive of Square of Hyperbolic Secant Function]]
+}}
+{{eqn | r = x - \tanh x + C
+      | c = [[Primitive of Constant]]
+}}
+{{end-eqn}}
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Square of Hyperbolic Cotangent Function}
+Tags: Primitives of Hyperbolic Functions, Hyperbolic Cotangent Function
+
+\begin{theorem}
+:$\displaystyle \int \coth^2 x \rd x = x - \coth x + C$
+where $C$ is an [[Definition:Arbitrary Constant (Calculus)|arbitrary constant]].
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int \coth^2 x \rd x
+      | r = \int \paren {1 + \csch^2 x} \rd x
+      | c = [[Difference of Squares of Hyperbolic Cotangent and Cosecant]]
+}}
+{{eqn | r = \int 1 \rd x + \int \csch^2 x \rd x
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \int 1 \rd x - \coth x + C
+      | c = [[Primitive of Square of Hyperbolic Cosecant Function]]
+}}
+{{eqn | r = x - \coth x + C
+      | c = [[Primitive of Constant]]
+}}
+{{end-eqn}}
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Heaviside Step Function is Piecewise Continuous}
+Tags: Heaviside Step Function, Continuous Functions
+
+\begin{theorem}
+Let $c \ge 0$ be a [[Definition:Constant|constant]] [[Definition:Real Number|real number]].
+The [[Definition:Heaviside Step Function|Heaviside step function]]:
+:$\mu_c \left({t}\right) = \begin{cases}
+1 & : t > c \\
+0 & : t < c
+\end{cases}$
+is [[Definition:Piecewise Continuous Function|piecewise continuous]] for any [[Definition:Closed Real Interval|interval]] of the form:
+:$\left[{c - M \,.\,.\, c + M}\right]$
+where $M > 0$ is some [[Definition:Arbitrarily Large|arbitrarily large]] [[Definition:Constant|constant]].
+\end{theorem}
+
+\begin{proof}
+Let the [[Definition:Finite Set|finite]] [[Definition:Subdivision (Real Analysis)|subdivision]] of $\mu_c$ be:
+:$\left\{{c - M, c, c + M}\right\}$
+Let $\epsilon > 0$.
+For $t \to \left({c - M}\right)^+$, choose $\delta$ to be any number at all, because:
+:$c - M < t < c - M + \delta \implies \left\vert{ \mu_c \left({t}\right) - 0 }\right\vert = \left\vert{ 0 - 0 }\right\vert < \epsilon$
+holds, from [[True Statement is implied by Every Statement]].
+The case for $t \to c^-$ is proved similarly.
+For $t \to c^+$, again, choose $\delta$ to be any number at all, because:
+:$c - \delta < t < c \implies \left\vert{ \mu_c \left({t}\right) - 1 }\right\vert = \left\vert{ 1 - 1 }\right\vert < \epsilon$
+holds, from [[True Statement is implied by Every Statement]].
+The case for $t \to \left({c + M}\right)^-$ is proved similarly.
+For all other $t$, we invoke [[Continuity of Heaviside Step Function]].
+{{qed}}
+[[Category:Heaviside Step Function]]
+[[Category:Continuous Functions]]
+f1rnjh4xr4nyczh29qeqxrivz84a20x
+\end{proof}<|endoftext|>
+\section{Euler Formula for Sine Function/Complex Numbers/Proof 1/Lemma 1}
+Tags: Hyperbolic Sine Function, Euler Formula for Sine Function
+
+\begin{theorem}
+The function:
+:$\dfrac {\sinh x} x$ 
+is [[Definition:Increasing Real Function|increasing]] for [[Definition:Positive Real Number|positive real]] $x$.
+\end{theorem}
+
+\begin{proof}
+Let $\map f x = \dfrac {\sinh x} x$.
+By [[Quotient Rule for Derivatives]] and [[Derivative of Hyperbolic Sine]]:
+:$\map {f'} x = \dfrac {x \cosh x - \sinh x} {x^2}$
+From [[Hyperbolic Tangent Less than X]], we have $\tanh x \le x$ for $x \ge 0$.
+Since $\cosh x \ge 0$, we can rearrange to get $x \cosh x - \sinh x \ge 0$.
+Since $x^2 \ge 0$, we have $\map {f'} x \ge 0$ for $x \ge 0$.
+So by [[Derivative of Monotone Function]] it follows that $\map f x$ is [[Definition:Increasing Real Function|increasing]] for $x \ge 0$.
+{{qed}}
+[[Category:Hyperbolic Sine Function]]
+[[Category:Euler Formula for Sine Function]]
+pj2z8lxqa3jxmucl8rypl8bqmyxs3ss
+\end{proof}<|endoftext|>
+\section{Hyperbolic Tangent Less than X}
+Tags: Hyperbolic Tangent Function
+
+\begin{theorem}
+:$\tanh x \le x$
+for $x \ge 0$.
+\end{theorem}
+
+\begin{proof}
+Let $\map f x = x - \tanh x$.
+By [[Derivative of Hyperbolic Tangent]]:
+:$\map {f'} x = 1 - \sech^2 x$
+Since $\cosh x \ge 1$ for all $x \in \R$, we can deduce that:
+:$\map {f'} x \ge 0$
+From [[Derivative of Monotone Function]], $\map f x$ is [[Definition:Increasing Real Function|increasing]].
+By definition of [[Definition:Hyperbolic Tangent|hyperbolic tangent]]:
+:$\map f x = 0$
+It follows that $\map f x \ge 0$ for $x \ge 0$.
+Hence the result.
+{{qed}}
+[[Category:Hyperbolic Tangent Function]]
+ryxjy4bcq7w21sw5ohci51xqz8sts03
+\end{proof}<|endoftext|>
+\section{Primitive of Square of Hyperbolic Sine Function}
+Tags: Primitives of Hyperbolic Functions, Hyperbolic Sine Function
+
+\begin{theorem}
+:$\displaystyle \int \sinh^2 x \rd x = \frac {\sinh 2 x} 4 - \frac x 2 + C$
+where $C$ is an [[Definition:Arbitrary Constant (Calculus)|arbitrary constant]].
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int \sinh^2 x \rd x
+      | r = \int \paren {\frac {\cosh 2 x - 1} 2} \rd x
+      | c = [[Square of Hyperbolic Sine]]
+}}
+{{eqn | r = \int \paren {\frac {\cosh 2 x} 2} \rd x - \int \frac 1 2 \rd x
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \int \paren {\frac {\cosh 2 x} 2} \rd x - \frac x 2 + C
+      | c = [[Primitive of Constant]]
+}}
+{{eqn | r = \frac 1 2 \paren {\frac {\sinh 2 x} 2} - \frac x 2 + C
+      | c = [[Primitive of Function of Constant Multiple]] and [[Primitive of Hyperbolic Cosine Function]]
+}}
+{{eqn | r = \frac {\sinh 2 x} 4 - \frac x 2 + C
+      | c = rearranging
+}}
+{{end-eqn}}
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Square of Hyperbolic Cosine Function}
+Tags: Primitives of Hyperbolic Functions, Hyperbolic Cosine Function
+
+\begin{theorem}
+:$\displaystyle \int \cosh^2 x \rd x = \frac {\sinh 2 x} 4 + \frac x 2 + C$
+where $C$ is an [[Definition:Arbitrary Constant (Calculus)|arbitrary constant]].
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int \cosh^2 x \rd x
+      | r = \int \paren {\frac {\cosh 2 x + 1} 2} \rd x
+      | c = [[Square of Hyperbolic Cosine]]
+}}
+{{eqn | r = \int \paren {\frac {\cosh 2 x} 2} \rd x + \int \frac 1 2 \rd x
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \int \paren {\frac {\cosh 2 x} 2} \rd x + \frac x 2 + C
+      | c = [[Primitive of Constant]]
+}}
+{{eqn | r = \frac 1 2 \paren {\frac {\map \sinh {2 x} } 2} + \frac x 2 + C
+      | c = [[Primitive of Function of Constant Multiple]] and [[Primitive of Hyperbolic Cosine Function]]
+}}
+{{eqn | r = \frac {\sinh 2 x} 4 + \frac x 2 + C
+      | c = rearranging
+}}
+{{end-eqn}}
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Product of Hyperbolic Secant and Tangent}
+Tags: Primitives of Hyperbolic Functions, Hyperbolic Secant Function, Hyperbolic Tangent Function
+
+\begin{theorem}
+:$\ds \int \sech x \tanh x \rd x = -\sech x + C$
+where $C$ is an [[Definition:Arbitrary Constant (Calculus)|arbitrary constant]].
+\end{theorem}
+
+\begin{proof}
+From [[Derivative of Hyperbolic Secant]]:
+:$\dfrac \d {\d x} \sech x = -\sech x \tanh x$
+The result follows from the definition of [[Definition:Primitive (Calculus)|primitive]].
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Product of Hyperbolic Cosecant and Cotangent}
+Tags: Primitives of Hyperbolic Functions, Hyperbolic Cosecant Function, Hyperbolic Cotangent Function
+
+\begin{theorem}
+:$\ds \int \csch x \coth x \rd x = -\csch x + C$
+where $C$ is an [[Definition:Arbitrary Constant (Calculus)|arbitrary constant]].
+\end{theorem}
+
+\begin{proof}
+From [[Derivative of Hyperbolic Cosecant]]:
+:$\dfrac \d {\d x} \csch x = -\csch x \coth x$
+The result follows from the definition of [[Definition:Primitive (Calculus)|primitive]].
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Reciprocal of x squared minus a squared}
+Tags: Primitives involving x squared minus a squared, Primitive of Reciprocal of x squared minus a squared, Primitives involving Reciprocals
+
+\begin{theorem}
+[[File:Primitive-of-1-over-x-squared-minus-a-squared.png|thumb|right|600px|$\color {blue} {\dfrac 1 {x^2 - a^2} } \qquad \color {green} {-\dfrac 1 a \tanh^{-1} \dfrac x a} \qquad \color {red} {-\dfrac 1 a \coth^{-1} \dfrac x a}$]]
+Let $a \in \R_{>0}$ be a [[Definition:Strictly Positive Real Number|strictly positive real]] [[Definition:Constant|constant]].
+\end{theorem}<|endoftext|>
+\section{Primitive of Reciprocal of x squared minus a squared/Logarithm Form 2}
+Tags: Primitive of Reciprocal of x squared minus a squared
+
+\begin{theorem}
+:$\ds \int \frac {\d x} {x^2 - a^2} = \frac 1 {2 a} \ln \size {\frac {x - a} {x + a} } + C$
+\end{theorem}
+
+\begin{proof}
+From the [[Primitive of Reciprocal of x squared minus a squared/Logarithm Form 1|$1$st logarithm form]]:
+:$\ds \int \frac {\d x} {x^2 - a^2} = \begin {cases} \dfrac 1 {2 a} \map \ln {\dfrac {a - x} {a + x} } + C & : \size x < a\\
+& \\
+ \dfrac 1 {2 a} \map \ln {\dfrac {x - a} {x + a} } + C & : \size x > a \\
+& \\
+\text {undefined} & : \size x = a \end {cases}$
+From [[Primitive of Reciprocal of x squared minus a squared/Logarithm Form/Lemma|Primitive of Reciprocal of a squared minus x squared: Logarithm Form: Lemma]]:
+:$\map \ln {\dfrac {x - a} {x + a} }$ is defined {{iff}} $\size x > a$
+:$\map \ln {\dfrac {a - x} {a + x} }$ is defined {{iff}} $\size x < a$
+Let $\size x > a$.
+Then $\map \ln {\dfrac {x - a} {x + a} }$ is defined.
+We have that:
+{{begin-eqn}}
+{{eqn | l = \dfrac {x - a} {x + a}
+      | o = >
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \dfrac {x - a} {x + a}
+      | r = \size {\dfrac {x - a} {x + a} }
+      | c = 
+}}
+{{end-eqn}}
+So the result holds for $\size x > a$.
+Let $\size x < a$.
+Then $\map \ln {\dfrac {a - x} {a + x} }$ is defined.
+We have:
+We have that:
+{{begin-eqn}}
+{{eqn | l = \dfrac {a - x} {a + x}
+      | r = -\dfrac {x - a} {x + a}
+      | c = 
+}}
+{{eqn | o = <
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \dfrac {a - x} {a + x}
+      | r = \size {\dfrac {x - a} {x + a} }
+      | c = 
+}}
+{{end-eqn}}
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Reciprocal of a squared minus x squared/Logarithm Form 1}
+Tags: Primitive of Reciprocal of a squared minus x squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {\d x} {a^2 - x^2} = \begin {cases} \dfrac 1 {2 a} \map \ln {\dfrac {a + x} {a - x} } + C & : \size x < a\\
+& \\
+ \dfrac 1 {2 a} \map \ln {\dfrac {x + a} {x - a} } + C & : \size x > a \\
+& \\
+\text {undefined} & : \size x = a \end {cases}$
+\end{theorem}
+
+\begin{proof}
+=== [[Primitive of Reciprocal of a squared minus x squared/Logarithm Form 1/size of x less than a|Case where $\size x < a$]] ===
+{{:Primitive of Reciprocal of a squared minus x squared/Logarithm Form 1/size of x less than a}}
+=== [[Primitive of Reciprocal of a squared minus x squared/Logarithm Form 1/size of x greater than a|Case where $\size x > a$]] ===
+{{:Primitive of Reciprocal of a squared minus x squared/Logarithm Form 1/size of x greater than a}}
+\end{proof}<|endoftext|>
+\section{Primitive of Reciprocal of a squared minus x squared}
+Tags: Primitives involving a squared minus x squared, Primitive of Reciprocal of a squared minus x squared, Primitives involving Reciprocals
+
+\begin{theorem}
+[[File:Primitive-of-1-over-a-squared-minus-x-squared.png|thumb|right|600px|$\color {blue} {\dfrac 1 {a^2 - x^2} } \qquad \color {red} {\dfrac 1 a \tanh^{-1} \dfrac x a} \qquad \color {green} {\dfrac 1 a \coth^{-1} \dfrac x a}$]]
+Let $a \in \R_{>0}$ be a [[Definition:Strictly Positive Real Number|strictly positive real]] [[Definition:Constant|constant]].
+\end{theorem}<|endoftext|>
+\section{Primitive of Reciprocal of Root of x squared plus a squared/Logarithm Form}
+Tags: Primitive of Reciprocal of Root of x squared plus a squared
+
+\begin{theorem}
+:$\ds \int \frac {\d x} {\sqrt {x^2 + a^2} } = \map \ln {x + \sqrt {x^2 + a^2} } + C$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int \frac {\d x} {\sqrt {x^2 + a^2} }
+      | r = \sinh^{-1} {\frac x a} + C
+      | c = [[Primitive of Reciprocal of Root of x squared plus a squared/Inverse Hyperbolic Sine Form|Primitive of $\dfrac 1 {\sqrt {x^2 - a^2} }$ in $\sinh^{-1}$ form]]
+}}
+{{eqn | r = \map \ln {x + \sqrt {x^2 + a^2} } - \ln a + C
+      | c = [[Inverse Hyperbolic Sine of x over a in Logarithm Form|$\sinh^{-1} {\dfrac x a}$ in Logarithm Form]]
+}}
+{{eqn | r = \map \ln {x + \sqrt {x^2 + a^2} } + C
+      | c = subsuming $-\ln a$ into [[Definition:Arbitrary Constant (Calculus)|arbitrary constant]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Limit at Infinity of Polynomial over Complex Exponential}
+Tags: Limits of Functions, Polynomial Theory, Exponential Function
+
+\begin{theorem}
+Let $n \in \N$.
+Let $\map {P_n} x$ be a [[Definition:Real Polynomial Function|real polynomial]], of [[Definition:Degree of Polynomial|degree]] $n$.
+Let $e^z$ be the [[Definition:Complex Exponential Function|complex exponential function]], where $z = x + i y$.
+Let $a \in \R_{>0}$.
+Then:
+:$\displaystyle \lim_{x \mathop \to +\infty} \frac {\map {P_n} x} {e^{a z} } = 0$
+\end{theorem}
+
+\begin{proof}
+Let $\epsilon > 0$.
+By the definition of [[Definition:Limit at Infinity|limits at infinity]], we need to show that there is some $M \in \R$ such that:
+:$\displaystyle x > M \implies \size {\frac {\map {P_n} x} {e^{a z} } - 0} < \epsilon$
+But:
+{{begin-eqn}}
+{{eqn | l = \size {\frac {\map {P_n} x}{e^{a z} } - 0}
+      | r = \frac {\size {\map {P_n} x} } {\size {e^{a z} } }
+      | c = [[Complex Modulus of Product of Complex Numbers|Modulus of Product]]
+}}
+{{eqn | r = \frac {\size {\map {P_n} x} } {\size {e^{a x \mathop + i a y} } }
+}}
+{{eqn | r = \frac {\size {\map {P_n} x} } {\size {e^{a x} } }
+      | c = [[Modulus of Exponential is Modulus of Real Part]]
+}}
+{{end-eqn}}
+This means it is sufficient to find an $M \in \R$ such that:
+{{begin-eqn}}
+{{eqn | l = \frac {\size {\map {P_n} x} } {\size {e^{a x} } }
+      | o = <
+      | r = \epsilon
+}}
+{{end-eqn}}
+{{questionable|[[Exponential Dominates Polynomial]] doesn't say anything about the quotient. (Proof corrected Nov. 2016.)}}
+Let $\displaystyle \map {P_n} x = \sum_{j \mathop = 0}^n \paren {a_j x^j}$ where $a_j \in \R$ for every $j$ in $\set {0, \ldots, n}$.
+Let $\displaystyle M' = \frac 1 \epsilon \sum_{j \mathop = 0}^n \size {a_j}$.
+We observe that $M' \ge 0$.
+Also, for every $x > M'$:
+:$x > 0$ as $M' \ge 0$
+We have for every $x > M'$:
+{{begin-eqn}}
+{{eqn | l = M'
+      | o = <
+      | r = x
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \frac 1 \epsilon \sum_{j \mathop = 0}^n \size {a_j}
+      | o = <
+      | r = x
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \frac 1 x \sum_{j \mathop = 0}^n \size {a_j}
+      | o = <
+      | r = \epsilon
+      | c = as $x \not = 0$ since $x > 0$
+}}
+{{end-eqn}}
+We have that for every $k \in \N$ there exists $N_k \in \N$ such that:
+:$x^k < e^{a x}$ for all $x > N_k$ by [[Exponential Dominates Polynomial]]
+Let $N = \map \max {N_1, \ldots, N_{n + 1} }$.
+We have for every $k \in \set {1, \ldots, n + 1}$:
+{{begin-eqn}}
+{{eqn | l = x^k
+      | o = <
+      | r = e^{a x}
+      | c = for all $x > N$
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \frac {x^k} {e^{a x} }
+      | o = <
+      | r = 1
+      | c = for all $x > N$
+}}
+{{end-eqn}}
+Let $M = \map \max {M', N}$.
+We get for every $x > M$:
+{{begin-eqn}}
+{{eqn | l = \frac {\size {\map {P_n} x} } {\size {e^{a x} } }
+      | r = \frac {\size {\sum_{j \mathop = 0}^n \paren {a_j x^j} } } {e^{a x} }
+}}
+{{eqn | o = \le
+      | r = \frac {\sum_{j \mathop = 0}^n \size {a_j x^j} } {e^{a x} }
+}}
+{{eqn | r = \frac {\sum_{j \mathop = 0}^n \paren {\size {a_j}  x^j} } {e^{a x} }
+      | c = as $x^j > 0$ since $x > 0$
+}}
+{{eqn | r = \frac 1 x \frac {\sum_{j \mathop = 0}^n \paren {\size {a_j} x^\paren {j + 1} } } {e^{a x} }
+      | c = as $x \not = 0$ since $x > 0$
+}}
+{{eqn | r = \frac 1 x \sum_{j \mathop = 0}^n \paren {\size {a_j} \frac {x^\paren {j + 1} } {e^{a x} } }
+}}
+{{eqn | o = \le
+      | r = \frac 1 x \sum_{j \mathop = 0}^n \size {a_j} 1
+      | c = as $\dfrac {x^\paren {j + 1} } {e^{a x} } < 1$ for every $j$ in $\set {0, \ldots, n}$
+}}
+{{eqn | o = <
+      | r = \epsilon
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Limits of Functions]]
+[[Category:Polynomial Theory]]
+[[Category:Exponential Function]]
+3r6mkfmu8w1d1jzcycs88kojf9c27bo
+\end{proof}<|endoftext|>
+\section{Primitives of Trigonometric Functions}
+Tags: Primitives, Primitives of Trigonometric Functions
+
+\begin{theorem}
+This page gathers together [[Definition:Primitive (Calculus)|primitives]] of [[Definition:Trigonometric Function|trigonometric functions]].
+In the below, $C$ is an [[Definition:Arbitrary Constant (Calculus)|arbitrary constant]] throughout.
+\end{theorem}<|endoftext|>
+\section{Primitives of Hyperbolic Functions}
+Tags: Primitives, Primitives of Hyperbolic Functions
+
+\begin{theorem}
+This page gathers together [[Definition:Primitive (Calculus)|primitives]] of [[Definition:Hyperbolic Function|hyperbolic functions]].
+In the below, $C$ is an [[Definition:Arbitrary Constant (Calculus)|arbitrary constant]] throughout.
+\end{theorem}<|endoftext|>
+\section{Derivatives of Trigonometric Functions}
+Tags: Derivatives, Derivatives of Trigonometric Functions
+
+\begin{theorem}
+This page gathers together [[Definition:Derivative|derivatives]] of [[Definition:Trigonometric Function|trigonometric functions]].
+\end{theorem}<|endoftext|>
+\section{Derivatives of Inverse Trigonometric Functions}
+Tags: Derivatives, Derivatives of Inverse Trigonometric Functions
+
+\begin{theorem}
+This page gathers together [[Definition:Derivative|derivatives]] of [[Definition:Inverse Trigonometric Function|inverse trigonometric functions]].
+\end{theorem}<|endoftext|>
+\section{Derivatives of Hyperbolic Functions}
+Tags: Derivatives, Derivatives of Hyperbolic Functions
+
+\begin{theorem}
+This page gathers together [[Definition:Derivative|derivatives]] of [[Definition:Hyperbolic Function|hyperbolic functions]].
+Let $u$ be a [[Definition:Differentiable Real Function|differentiable real function]] of $x$.
+\end{theorem}<|endoftext|>
+\section{Derivatives of Inverse Hyperbolic Functions}
+Tags: Derivatives, Derivatives of Inverse Hyperbolic Functions
+
+\begin{theorem}
+This page gathers together [[Definition:Derivative|derivatives]] of [[Definition:Inverse Hyperbolic Function|inverse hyperbolic functions]].
+\end{theorem}<|endoftext|>
+\section{Primitive of Reciprocal of Root of x squared minus a squared/Logarithm Form}
+Tags: Primitive of Reciprocal of Root of x squared minus a squared
+
+\begin{theorem}
+:$\ds \int \frac {\d x} {\sqrt {x^2 - a^2} } = \ln \size {x + \sqrt {x^2 - a^2} } + C$
+\end{theorem}
+
+\begin{proof}
+We have that $\sqrt {x^2 - a^2}$ is defined only when $x^2 > a^2$, that is, either:
+:$x > a$
+or:
+:$x < -a$
+where it is assumed that $a > 0$.
+Consider the [[Definition:Arcsecant|arcsecant]] substitution:
+:$u = \arcsec {\dfrac x a}$
+which is defined for all $x$ such that $\size {\dfrac x a} \ge 1$.
+That is:
+:$\size x \ge a$
+Hence from [[Shape of Secant Function]], this substitution is valid for all for all $x$ such that $\size {\dfrac x a} > 1$.
+{{begin-eqn}}
+{{eqn | l = u
+      | r = \arcsec {\frac x a}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x
+      | r = a \sec u
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d x} {\d u}
+      | r = a \sec u \tan u
+      | c = [[Derivative of Secant Function]]
+}}
+{{end-eqn}}
+Let $x > a$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \int \frac 1 {\sqrt {x^2 - a^2} } \rd x
+      | r = \int \frac {a \sec u \tan u} {\sqrt {a^2 \sec^2 u - a^2} } \rd u
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \frac a a \int \frac {\sec u \tan u} {\sqrt {\sec^2 u - 1} } \rd u
+      | c = simplifying
+}}
+{{eqn | r = \int \sec u \frac {\tan u} {\sqrt {\tan^2 u} } \rd u
+      | c = [[Difference of Squares of Secant and Tangent]]
+}}
+{{eqn | r = \int \sec u \rd u
+      | c = as $x > 1$ both $\sec u$ and $\tan u$ are greater than $0$
+}}
+{{eqn | r = \ln \size {\sec u + \tan u} + C
+      | c = [[Primitive of Secant Function/Secant plus Tangent Form|Primitive of $\sec x$: Secant plus Tangent Form]]
+}}
+{{eqn | r = \map \ln {\sec u + \tan u} + C
+      | c = as both $\sec u$ and $\tan u$ are greater than $0$
+}}
+{{eqn | r = \map \ln {\frac x a + \sqrt {\frac {x^2} {a^2} - 1} } + C
+      | c = substituting back: $\sec u = \dfrac x a$ and [[Difference of Squares of Secant and Tangent]]
+}}
+{{eqn | r = \map \ln {\frac x a + \sqrt {\frac {x^2 - a^2} {a^2} } } + C
+      | c = rearrangement
+}}
+{{eqn | r = \map \ln {\frac {x + \sqrt {x^2 - a^2} } a} + C
+      | c = rearrangement
+}}
+{{eqn | r = \map \ln {x + \sqrt {x^2 - a^2} - \ln a} + C
+      | c = [[Difference of Logarithms]]
+}}
+{{eqn | r = \map \ln {x + \sqrt {x^2 - a^2} } + C
+      | c = subsuming $-\ln a$ into [[Definition:Arbitrary Constant (Calculus)|constant]]
+}}
+{{eqn | r = \ln \size {x + \sqrt {x^2 - a^2} } + C
+      | c = as $x + \sqrt {x^2 - a^2} > 0$ for $x > a$
+}}
+{{end-eqn}}
+Now suppose $x < -a$.
+Let $z = -x$.
+Then:
+:$\d x = -\d z$
+and we then have:
+{{begin-eqn}}
+{{eqn | l = \int \frac {\d x} {\sqrt {x^2 - a^2} }
+      | r = \int \frac {-\d z} {\sqrt {\paren {-z}^2 - a^2} }
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = -\int \frac {\d z} {\sqrt {z^2 - a^2} }
+      | c = simplifying
+}}
+{{eqn | r = -\map \ln {z + \sqrt {z^2 - a^2} } + C
+      | c = from above
+}}
+{{eqn | r = \map \ln {z - \sqrt {z^2 - a^2} } - \map \ln {a^2} + C
+      | c = [[Negative of Logarithm of x plus Root x squared minus a squared|Negative of $\map \ln {z + \sqrt {z^2 - a^2} }$]]
+}}
+{{eqn | r = \map \ln {z - \sqrt {z^2 - a^2} } + C
+      | c = subsuming $-\map \ln {a^2}$ into [[Definition:Arbitrary Constant (Calculus)|constant]]
+}}
+{{eqn | r = \map \ln {-x - \sqrt {\paren {-x}^2 - a^2} } + C
+      | c = substituting back for $x$
+}}
+{{eqn | r = \map \ln {-\paren {x + \sqrt {x^2 - a^2} } } + C
+      | c = simplifying
+}}
+{{eqn | r = \ln \size {x + \sqrt {x^2 - a^2} } + C
+      | c = as $x + \sqrt {x^2 - a^2} < 0$ for $x < -a$
+}}
+{{end-eqn}}
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Reciprocal of x by Root of x squared minus a squared}
+Tags: Expressions whose Primitives are Inverse Trigonometric Functions, Primitives involving Root of x squared minus a squared, Arcsecant Function, Primitives involving Reciprocals
+
+\begin{theorem}
+:$\displaystyle \int \frac {\d x} {x \sqrt {x^2 - a^2} } = \frac 1 a \arcsec \size {\frac x a} + C$
+for $\size x > a$.
+\end{theorem}
+
+\begin{proof}
+We have that $\sqrt {x^2 - a^2}$ is defined only when $x^2 > a^2$, that is, either:
+:$x > a$
+or:
+:$x < -a$
+where it is assumed that $a > 0$.
+Consider the [[Definition:Arcsecant|arcsecant]] substitution:
+:$u = \arcsec {\dfrac x a}$
+which is defined for all $x$ such that $\size {\dfrac x a} \ge 1$.
+That is:
+:$\size x \ge a$
+and it is seen that $u = \arcsec {\dfrac x a}$ is defined over the whole domain of the integrand.
+Hence:
+{{begin-eqn}}
+{{eqn | l = u
+      | r = \arcsec {\frac x a}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x
+      | r = a \sec u
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d x} {\d u}
+      | r = a \sec u \tan u
+      | c = [[Derivative of Secant Function]]
+}}
+{{end-eqn}}
+Let $x > a$.
+{{begin-eqn}}
+{{eqn | l = \int \frac {\d x} {x \sqrt {x^2 - a^2} }
+      | r = \int \frac {a \sec u \tan u} {a \sec u \sqrt {a^2 \sec^2 u - a^2} } \rd u
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \frac a {a^2} \int \frac {\sec u \tan u} {\sec u \sqrt {\sec^2 u - 1} } \rd u
+      | c = [[Primitive of Constant Multiple of Function]]
+}}
+{{eqn | r = \frac 1 a \int \frac {\sec u \tan u} {\sec u \tan u} \rd u
+      | c = [[Difference of Squares of Secant and Tangent]]
+}}
+{{eqn | r = \frac 1 a \int 1 \rd u
+}}
+{{eqn | r = \frac 1 a u + C
+      | c = [[Integral of Constant]]
+}}
+{{eqn | r = \frac 1 a \arcsec {\frac x a} + C
+      | c = Definition of $u$
+}}
+{{eqn | r = \frac 1 a \arcsec {\frac {\size x} a} + C
+      | c = {{Defof|Absolute Value}}: $\size x = x$ for $x > 0$
+}}
+{{end-eqn}}
+Now suppose $x < -a$.
+Let $z = -x$.
+Then:
+:$\d x = -\d z$
+and we then have:
+{{begin-eqn}}
+{{eqn | l = \int \frac {\d x} {x \sqrt {x^2 - a^2} }
+      | r = \int \frac {-\d z} {\paren {-z} \sqrt {\paren {-z}^2 - a^2} }
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \int \frac {\d z} {z \sqrt {z^2 - a^2} }
+      | c = simplifying
+}}
+{{eqn | r = \frac 1 a \arcsec {\frac z a} + C
+      | c = from above
+}}
+{{eqn | r = \frac 1 a \arcsec {\frac {\paren {-x} } a} + C
+      | c = substituting back for $x$
+}}
+{{eqn | r = \frac 1 a \arcsec {\frac {\size x} a} + C
+      | c = {{Defof|Absolute Value}}: $\size x = -x$ for $x < 0$
+}}
+{{end-eqn}}
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Reciprocal of x by Root of x squared plus a squared/Logarithm Form}
+Tags: Primitive of Reciprocal of x by Root of x squared plus a squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {\d x} {x \sqrt {x^2 + a^2} } = -\frac 1 a \map \ln {\frac {a + \sqrt {x^2 + a^2} } x} + C$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int \frac {\d x} {x \sqrt {x^2 + a^2} }
+      | r = -\frac 1 a \csch^{-1} {\frac x a} + C
+      | c = [[Primitive of Reciprocal of x by Root of x squared plus a squared/Inverse Hyperbolic Cosecant Form|Primitive of Reciprocal of $x \sqrt {x^2 + a^2}$: $\csch^{-1}$ form]]
+}}
+{{eqn | r = -\frac 1 a \map \ln {\frac {a + \sqrt{a^2 + x^2} } x} + C
+      | c = [[Inverse Hyperbolic Cosecant of x over a in Logarithm Form|$\csch^{-1} {\dfrac x a}$ in Logarithm Form]]
+}}
+{{end-eqn}}
+{{qed}}
+{{questionable|Negative $x$ not taken into account. The result should end up being absolute valued.}}
+\end{proof}<|endoftext|>
+\section{Primitive of Reciprocal of x by Root of a squared minus x squared/Logarithm Form}
+Tags: Primitive of Reciprocal of x by Root of a squared minus x squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {\d x} {x \sqrt {a^2 - x^2} } = -\frac 1 a \map \ln {\frac {a + \sqrt {a^2 - x^2} } x} + C$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int \frac {\d x} {x \sqrt {a^2 - x^2} }
+      | r = -\frac 1 a \sech^{-1} {\frac x a} + C
+      | c = [[Primitive of Reciprocal of x by Root of a squared minus x squared/Inverse Hyperbolic Secant Form|Primitive of Reciprocal of $x \sqrt {a^2 - x^2}$: $\sech^{-1}$ form]]
+}}
+{{eqn | r = -\frac 1 a \map \ln {\frac {1 + \sqrt {1 - \paren {\frac x a}^2} } {\frac x a} } + C
+      | c = {{Defof|Inverse Hyperbolic Secant|subdef = Real|index = 2}}
+}}
+{{eqn | r = -\frac 1 a \map \ln {\frac {a + a \sqrt {1 - \paren {\frac x a}^2} } x} + C
+      | c = multiplying [[Definition:Numerator|top]] and [[Definition:Denominator|bottom]] by $a$
+}}
+{{eqn | r = -\frac 1 a \map \ln {\frac {a + \sqrt {a^2 - a^2 \paren {\frac x a}^2} } x} + C
+      | c = moving $a$ within the [[Definition:Square Root|square root]]
+}}
+{{eqn | r = -\frac 1 a \map \ln {\frac {a + \sqrt {a^2 - x^2} } x} + C
+      | c = simplifying
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Generalized Integration by Parts}
+Tags: Integral Calculus
+
+\begin{theorem}
+Let $\map f x, \map g x$ be [[Definition:Real Function|real functions]] which are [[Definition:Integrable Function|integrable]] and at least [[Definition:Nth Derivative|$n$ times differentiable]].
+Then:
+{{begin-eqn}}
+{{eqn | l = \int f^{\paren n} g \rd x
+      | r = \sum_{j \mathop = 0}^{n - 1} \paren {-1}^j f^{\paren {n - j - 1} } g^{\paren j} + \paren {-1}^n \int f g^{\paren n} \rd x
+      | c = 
+}}
+{{eqn | r = f^{\paren {n - 1} } g - f^{\paren {n - 2} } g' + f^{\paren {n - 3} } g'' - \cdots + \paren {-1}^n \int f g^{\paren n} \rd x
+      | c = 
+}}
+{{end-eqn}}
+where $f^{\paren n}$ denotes the [[Definition:Nth Derivative|$n$th derivative]] of $f$.
+\end{theorem}
+
+\begin{proof}
+Proof by [[Principle of Mathematical Induction|induction]]:
+For all $n \in \N_{> 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
+:$\displaystyle \int f^{\paren n} g \rd x = \sum_{j \mathop = 0}^{n - 1} \paren {-1}^j f^{\paren {n - j - 1} } g^\paren j + \paren {-1}^n \int f g^{\paren n} \rd x$
+=== Basis for the Induction ===
+$\map P 1$ is the case:
+:$\displaystyle \int f' g \rd x = f g - \int f g' \rd x$
+which is proved in [[Integration by Parts]].
+This is our [[Definition:Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
+So this is our [[Definition:Induction Hypothesis|induction hypothesis]]:
+:$\displaystyle \int f^{\paren k} g \rd x = \sum_{j \mathop = 0}^{k - 1} \paren {-1}^j f^{\paren {k - j - 1} } g^{\paren j} + \paren {-1}^k \int f g^{\paren k} \rd x$
+Then we need to show:
+:$\displaystyle \int f^{\paren {k + 1} } g \rd x = \sum_{j \mathop = 0}^k \paren {-1}^j f^{\paren {k - j} } g^{\paren j} + \paren {-1}^{k + 1} \int f g^{\paren {k + 1} } \rd x$
+=== Induction Step ===
+This is our [[Definition:Induction Step|induction step]]:
+{{begin-eqn}}
+{{eqn | l = \int f^{\paren {k + 1} } g \rd x
+      | r = \int \paren {f^{\paren k} }' g \rd x
+      | c = 
+}}
+{{eqn | r = f^{\paren k } g - \int f^{\paren k} g' \rd x
+      | c = [[Generalized Integration by Parts#Basis for the Induction|Basis for the Induction]]
+}}
+{{eqn | r = f^{\paren k} g - \paren {\sum_{j \mathop = 0}^{k - 1} \paren {-1}^j f^{\paren {k - j - 1} } g^\paren {j + 1} + \paren {-1}^k \int f g^{\paren {k + 1} } \rd x}
+      | c = [[Generalized Integration by Parts#Induction Hypothesis|Induction Hypothesis]]
+}}
+{{eqn | r = f^{\paren k} g + \paren {\sum_{j \mathop = 0}^{k - 1} \paren {-1}^{j + 1} f^{\paren {k - \paren {j + 1} } } g^{\paren {j + 1} } + \paren {-1}^{k + 1} \int f g^{\paren {k + 1} } \rd x}
+      | c = moving $-1$ into the parenthesis
+}}
+{{eqn | r = f^{\paren k} g + \paren {\sum_{j \mathop = 1}^k \paren {-1}^j f^{\paren {k - j} } g^{\paren j} + \paren {-1}^{k + 1} \int f g^{\paren {k + 1} } \rd x}
+      | c = substituting $j$ for $j + 1$
+}}
+{{eqn | r = \sum_{j \mathop = 0}^k \paren {-1}^j f^{\paren {k - j} } g^{\paren j} + \paren {-1}^{k + 1} \int f g^{\paren {k + 1} } \rd x
+      | c = as $f^{\paren k} g$ is the $0$th element of the [[Definition:Summation|summation]]
+}}
+{{end-eqn}}
+So $\map P k \implies \map P {k + 1}$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\displaystyle \forall n \in \N_{>0}: \int f^{\paren n} g \rd x = \sum_{j \mathop = 0}^{n - 1} \paren {-1}^j f^{\paren {n - j - 1} } g^{\paren j} + \paren {-1}^n \int f g^{\paren n} \rd x$
+assuming that $f$ and $g$ are [[Definition:Nth Derivative|sufficiently differentiable]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Function of a x + b}
+Tags: Integral Substitutions, Primitives involving a x + b
+
+\begin{theorem}
+:$\displaystyle \int \map F {a x + b} \rd x = \frac 1 a \int \map F u \rd u$
+where $u = a x + b$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = u 
+      | r = a x + b
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d u} {\d x}
+      | r = a
+      | c = [[Derivative of Function of Constant Multiple/Corollary|Derivative of Function of Constant Multiple: Corollary]]
+}}
+{{eqn | ll= \leadsto
+      | l = \int \map F {a x + b} \rd x
+      | r = \int \frac {\map F u} a \d u
+      | c = [[Primitive of Composite Function]]
+}}
+{{eqn | r = \frac 1 a \int \map F u \rd u
+      | c = [[Primitive of Constant Multiple of Function]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Function of Root of a x + b}
+Tags: Integral Substitutions, Primitives involving Root of a x + b
+
+\begin{theorem}
+:$\displaystyle \int F \left({\sqrt {a x + b}}\right) \ \mathrm d x = \frac 2 a \int u \ F \left({u}\right) \ \mathrm d u$
+where $u = \sqrt {a x + b}$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = u 
+      | r = \sqrt {a x + b}
+      | c = 
+}}
+{{eqn | l = u 
+      | r = \left({a x + b}\right)^{1/2}
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \frac {\mathrm d u} {\mathrm d x}
+      | r = \frac 1 2 \left({a x + b}\right)^{-1/2} \frac {\mathrm d} {\mathrm d x} \left({a x + b}\right)
+      | c = [[Chain Rule for Derivatives]], [[Power Rule for Derivatives]]
+}}
+{{eqn | r = \frac 1 {2 u} \frac {\mathrm d} {\mathrm d x} \left({a x + b}\right)
+      | c = substituting for $u$
+}}
+{{eqn | ll= \implies
+      | l = \frac {\mathrm d u} {\mathrm d x}
+      | r = \frac a {2 u}
+      | c = [[Derivative of Function of Constant Multiple/Corollary|Derivative of Function of Constant Multiple: Corollary]]
+}}
+{{eqn | ll= \implies
+      | l = \int F \left({\sqrt {a x + b} }\right) \ \mathrm d x
+      | r = \int \frac {2 u} a F \left({u}\right) \ \mathrm d u
+      | c = [[Primitive of Composite Function]]
+}}
+{{eqn | r = \frac 1 a \int u \ F \left({u}\right) \ \mathrm d u
+      | c = [[Primitive of Constant Multiple of Function]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Derivative of Nth Root}
+Tags: Derivatives
+
+\begin{theorem}
+Let $n \in \N_{>0}$.
+Let $f: \R \to \R$ be the [[Definition:Real Function|real function]] defined as $f \left({x}\right) = \sqrt [n] x$.
+Then:
+:$f' \left({x}\right) = \dfrac 1 {n \left({\sqrt [n] x}\right)^{n-1} }$
+everywhere that $f \left({x}\right) = \sqrt [n] x$ is defined.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = f \left({x}\right)
+      | r = \sqrt [n] x
+      | c = 
+}}
+{{eqn | r = x^{1 / n}
+      | c = Definition of [[Definition:Root (Analysis)|$n$th Root]]
+}}
+{{eqn | ll= \implies
+      | l = f' \left({x}\right)
+      | r = \frac 1 n x^{\left({1 / n}\right) - 1}
+      | c = [[Power Rule for Derivatives]]
+}}
+{{eqn | r = \frac 1 n x^{\left({1 / n}\right)\left({1 - n}\right)}
+      | c = rearrangement
+}}
+{{eqn | r = \frac 1 {n x^{\left({1 / n}\right) \left({n - 1}\right)} }
+      | c = [[Exponent Combination Laws/Negative Power|Exponent Combination Laws: Negative Power]]
+}}
+{{eqn | r = \frac 1 {n \left({\sqrt [n] x}\right)^{n - 1} }
+      | c = Definition of [[Definition:Root (Analysis)|$n$th Root]]
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Derivatives]]
+igt3uezk8pi0aw2x6m1ojrw2wb2p5ll
+\end{proof}<|endoftext|>
+\section{Primitive of Function of Nth Root of a x + b}
+Tags: Integral Substitutions
+
+\begin{theorem}
+:$\displaystyle \int F \left({\sqrt [n] {a x + b}}\right) \ \mathrm d x = \frac n a \int u^{n-1} F \left({u}\right) \ \mathrm d u$
+where $u = \sqrt [n] {a x + b}$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = u 
+      | r = \sqrt [n] {a x + b}
+      | c = 
+}}
+{{eqn | l = u 
+      | r = \left({a x + b}\right)^{1/n}
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \frac {\mathrm d u} {\mathrm d x}
+      | r = \frac 1 {n \left({\sqrt [n] {a x + b} }\right)^{n - 1} } \frac {\mathrm d} {\mathrm d x} \left({a x + b}\right)
+      | c = [[Chain Rule for Derivatives]], [[Derivative of Nth Root]]
+}}
+{{eqn | r = \frac 1 {n u^{n - 1} } \frac {\mathrm d} {\mathrm d x} \left({a x + b}\right)
+      | c = substituting for $u$
+}}
+{{eqn | r = \frac a {n u^{n - 1} }
+      | c = [[Derivative of Function of Constant Multiple/Corollary|Derivative of Function of Constant Multiple: Corollary]]
+}}
+{{eqn | ll= \implies
+      | l = \int F \left({\sqrt [n] {a x + b} }\right) \ \mathrm d x
+      | r = \int \frac {n u^{n - 1} } a F \left({u}\right) \ \mathrm d u
+      | c = [[Primitive of Composite Function]]
+}}
+{{eqn | r = \frac n a \int u^{n - 1} F \left({u}\right) \ \mathrm d u
+      | c = [[Primitive of Constant Multiple of Function]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Function of Root of a squared minus x squared}
+Tags: Integral Substitutions
+
+\begin{theorem}
+:$\displaystyle \int F \left({\sqrt {a^2 - x^2}}\right) \ \mathrm d x = a \int \cos u \ F \left({a \cos u}\right) \ \mathrm d u$
+where $x = a \sin u$.
+\end{theorem}
+
+\begin{proof}
+First note that:
+{{begin-eqn}}
+{{eqn | l = x 
+      | r = a \sin u
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \sqrt {a^2 - x^2}
+      | r = \sqrt {a^2 - \left({a \sin u}\right)^2}
+      | c =
+}}
+{{eqn | r = a \sqrt {1 - \sin^2 u}
+      | c = taking $a$ outside the square root
+}}
+{{eqn | n = 1
+      | r = a \cos u
+      | c = [[Sum of Squares of Sine and Cosine]]
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = x 
+      | r = a \sin u
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \frac{\mathrm d x} {\mathrm d u}
+      | r = a \cos u
+      | c = [[Derivative of Sine Function]]
+}}
+{{eqn | ll= \implies
+      | l = \int F \left({\sqrt {a^2 - x^2} }\right) \ \mathrm d x
+      | r = \int a \cos u \ F \left({\sqrt {a^2 - x^2} }\right) \ \mathrm d u
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \int a \cos u \ F \left({a \cos u}\right) \ \mathrm d u
+      | c = from $(1)$
+}}
+{{eqn | r = a \int \cos u \ F \left({a \cos u}\right) \ \mathrm d u
+      | c = [[Primitive of Constant Multiple of Function]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Function of Root of a squared plus x squared}
+Tags: Integral Substitutions
+
+\begin{theorem}
+:$\displaystyle \int F \left({\sqrt {a^2 + x^2}}\right) \ \mathrm d x = a \int \sec^2 u \ F \left({a \sec u}\right) \ \mathrm d u$
+where $x = a \tan u$.
+\end{theorem}
+
+\begin{proof}
+First note that:
+{{begin-eqn}}
+{{eqn | l = x 
+      | r = a \tan u
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \sqrt {a^2 + x^2}
+      | r = \sqrt {a^2 + \left({a \tan u}\right)^2}
+      | c =
+}}
+{{eqn | r = a \sqrt {1 + \tan^2 u}
+      | c = taking $a$ outside the square root
+}}
+{{eqn | r = a \sqrt {\sec^2 u}
+      | c = [[Difference of Squares of Secant and Tangent]]
+}}
+{{eqn | n = 1
+      | r = a \sec u
+      | c = 
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = x 
+      | r = a \tan u
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \frac{\mathrm d x} {\mathrm d u}
+      | r = a \sec^2 u
+      | c = [[Derivative of Tangent Function]]
+}}
+{{eqn | ll= \implies
+      | l = \int F \left({\sqrt {a^2 + x^2} }\right) \ \mathrm d x
+      | r = \int a \sec^2 u \ F \left({\sqrt {a^2 + x^2} }\right) \ \mathrm d u
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \int a \sec^2 u \ F \left({a \sec u}\right) \ \mathrm d u
+      | c = from $(1)$
+}}
+{{eqn | r = a \int \sec^2 u \ F \left({a \sec u}\right) \ \mathrm d u
+      | c = [[Primitive of Constant Multiple of Function]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Function of Root of x squared minus a squared}
+Tags: Integral Substitutions
+
+\begin{theorem}
+:$\displaystyle \int F \left({\sqrt {x^2 - a^2}}\right) \ \mathrm d x = a \int \sec u \tan u \ F \left({a \tan u}\right) \ \mathrm d u$
+where $x = a \sec u$.
+\end{theorem}
+
+\begin{proof}
+First note that:
+{{begin-eqn}}
+{{eqn | l = x 
+      | r = a \sec u
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \sqrt {x^2 - a^2}
+      | r = \sqrt {\left({a \sec u}\right)^2 - a^2}
+      | c =
+}}
+{{eqn | r = a \sqrt {\sec^2 u - 1}
+      | c = taking $a$ outside the square root
+}}
+{{eqn | r = a \sqrt {\tan^2 u}
+      | c = [[Difference of Squares of Secant and Tangent]]
+}}
+{{eqn | n = 1
+      | r = a \tan u
+      | c = 
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = x 
+      | r = a \sec u
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \frac{\mathrm d x} {\mathrm d u}
+      | r = a \sec u \tan u
+      | c = [[Derivative of Secant Function]]
+}}
+{{eqn | ll= \implies
+      | l = \int F \left({\sqrt {x^2 - a^2} }\right) \ \mathrm d x
+      | r = \int a \sec u \tan u \ F \left({\sqrt {x^2 - a^2} }\right) \ \mathrm d u
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \int a \sec u \tan u \ F \left({a \tan u}\right) \ \mathrm d u
+      | c = from $(1)$
+}}
+{{eqn | r = a \int \sec u \tan u \ F \left({a \tan u}\right) \ \mathrm d u
+      | c = [[Primitive of Constant Multiple of Function]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Function of Exponential Function}
+Tags: Integral Substitutions
+
+\begin{theorem}
+:$\displaystyle \int \map F {e^{a x} } \rd x = \frac 1 a \int \frac {\map F u} u \rd u$
+where $u = e^{a x}$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = u 
+      | r = e^{a x}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d u} {\d x}
+      | r = a e^{a x}
+      | c = [[Derivative of Exponential Function/Corollary 1|Derivative of Exponential Function: Corollary 1]]
+}}
+{{eqn | r = a u
+      | c = Definition of $u$
+}}
+{{eqn | ll= \leadsto
+      | l = \int \map F {e^{a x} } \rd x
+      | r = \int \frac {\map F u} {a u} \rd u
+      | c = [[Primitive of Composite Function]]
+}}
+{{eqn | r = \frac 1 a \int \frac {\map F u} u \rd u
+      | c = [[Primitive of Constant Multiple of Function]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Function of Natural Logarithm}
+Tags: Integral Substitutions
+
+\begin{theorem}
+:$\displaystyle \int F \left({\ln x}\right) \rd x = \int F \left({u}\right) e^u \rd u$
+where $u = \ln x$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = u 
+      | r = \ln x
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \frac {\d u} {\d x}
+      | r = \frac 1 x
+      | c = [[Derivative of Natural Logarithm Function]]
+}}
+{{eqn | ll= \implies
+      | l = \int F \left({\ln x}\right) \rd x
+      | r = \int F \left({u}\right) \ x \rd u
+      | c = [[Primitive of Composite Function]]
+}}
+{{eqn | r = \int F \left({u}\right) \ e^u \rd u
+      | c = {{Defof|Exponential Function/Real|subdef = Sum of Series|Exponential Function}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Function of Arcsine}
+Tags: Integral Substitutions, Arcsine Function, Primitives involving Inverse Sine Function
+
+\begin{theorem}
+:$\displaystyle \int F \left({\arcsin \frac x a}\right) \ \mathrm d x = a \int F \left({u}\right) \cos u \ \mathrm d u$
+where $u = \arcsin \dfrac x a$.
+\end{theorem}
+
+\begin{proof}
+First note that:
+{{begin-eqn}}
+{{eqn | l = u
+      | r = \arcsin \frac x a
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = x
+      | r = a \sin u
+      | c = Definition of [[Definition:Arcsine|Arcsine]]
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = u 
+      | r = \arcsin \frac x a
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \frac{\mathrm d u} {\mathrm d x}
+      | r = \frac 1 {\sqrt {a^2 - x^2} }
+      | c = [[Derivative of Arcsine Function/Corollary|Derivative of Arcsine Function: Corollary]]
+}}
+{{eqn | ll= \implies
+      | l = \int F \left({\arcsin \frac x a}\right) \ \mathrm d x
+      | r = \int F \left({u}\right) \ \sqrt {a^2 - x^2} \ \mathrm d u
+      | c = [[Primitive of Composite Function]]
+}}
+{{eqn | r = \int F \left({u}\right) \ \sqrt {a^2 - a^2 \sin^2 u} \ \mathrm d u
+      | c = Definition of $x$
+}}
+{{eqn | r = \int F \left({u}\right) \ a \sqrt {1 - \sin^2 u} \ \mathrm d u
+      | c = 
+}}
+{{eqn | r = \int F \left({u}\right) \ a \cos u \ \mathrm d u
+      | c = [[Sum of Squares of Sine and Cosine]]
+}}
+{{eqn | r = a \int F \left({u}\right) \ \cos u \ \mathrm d u
+      | c = [[Primitive of Constant Multiple of Function]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Function of Arccosine}
+Tags: Integral Substitutions, Arccosine Function, Primitives involving Inverse Cosine Function
+
+\begin{theorem}
+:$\displaystyle \int F \paren {\arccos \frac x a} \rd x = -a \int F \paren u \sin u \rd u$
+where $u = \arccos \dfrac x a$.
+\end{theorem}
+
+\begin{proof}
+First note that:
+{{begin-eqn}}
+{{eqn | l = u
+      | r = \arccos \frac x a
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = x
+      | r = a \cos u
+      | c = Definition of [[Definition:Arccosine|Arccosine]]
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = u 
+      | r = \arccos \frac x a
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \frac {\d u} {\d x}
+      | r = \frac {-1} {\sqrt {a^2 - x^2} }
+      | c = [[Derivative of Arccosine Function/Corollary|Derivative of Arccosine Function: Corollary]]
+}}
+{{eqn | ll= \implies
+      | l = \int F \paren {\arccos \frac x a} \rd x
+      | r = \int F \paren u \ \paren {-\sqrt {a^2 - x^2} } \rd u
+      | c = [[Primitive of Composite Function]]
+}}
+{{eqn | r = \int F \paren u \ \paren {-\sqrt {a^2 - a^2 \cos^2 u} } \rd u
+      | c = Definition of $x$
+}}
+{{eqn | r = \int F \paren u \ a \paren {-\sqrt {1 - \cos^2 u} } \rd u
+      | c = 
+}}
+{{eqn | r = \int F \paren u \ a \paren {-\sin u} \rd u
+      | c = [[Sum of Squares of Sine and Cosine]]
+}}
+{{eqn | r = -a \int F \paren u \ \sin u \rd u
+      | c = [[Primitive of Constant Multiple of Function]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Function of Arctangent}
+Tags: Integral Substitutions, Arctangent Function, Primitives involving Inverse Tangent Function
+
+\begin{theorem}
+:$\displaystyle \int F \left({\arctan \frac x a}\right) \ \mathrm d x = a \int F \left({u}\right) \sec^2 u \ \mathrm d u$
+where $u = \arctan \dfrac x a$.
+\end{theorem}
+
+\begin{proof}
+First note that:
+{{begin-eqn}}
+{{eqn | l = u
+      | r = \arctan \frac x a
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = x
+      | r = a \tan u
+      | c = Definition of [[Definition:Arctangent|Arctangent]]
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = u 
+      | r = \arctan \frac x a
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \frac{\mathrm d u} {\mathrm d x}
+      | r = \frac a {a^2 + x^2}
+      | c = [[Derivative of Arctangent Function/Corollary|Derivative of Arctangent Function: Corollary]]
+}}
+{{eqn | ll= \implies
+      | l = \int F \left({\arctan \frac x a}\right) \ \mathrm d x
+      | r = \int F \left({u}\right) \ \frac {a^2 + x^2} a \ \mathrm d u
+      | c = [[Primitive of Composite Function]]
+}}
+{{eqn | r = \int F \left({u}\right) \ \frac {a^2 + a^2 \tan^2 u} a \ \mathrm d u
+      | c = Definition of $x$
+}}
+{{eqn | r = \int F \left({u}\right) \ a^2 \frac {1 + \tan^2 u} a \ \mathrm d u
+      | c = 
+}}
+{{eqn | r = \int F \left({u}\right) \ a \sec^2 u \ \mathrm d u
+      | c = [[Difference of Squares of Secant and Tangent]]
+}}
+{{eqn | r = a \int F \left({u}\right) \ \sec^2 u \ \mathrm d u
+      | c = [[Primitive of Constant Multiple of Function]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Function of Arccotangent}
+Tags: Integral Substitutions, Arccotangent Function, Primitives involving Inverse Cotangent Function
+
+\begin{theorem}
+:$\displaystyle \int F \left({\operatorname{arccot} \frac x a}\right) \ \mathrm d x = -a \int F \left({u}\right) \csc^2 u \ \mathrm d u$
+where $u = \operatorname{arccot} \dfrac x a$.
+\end{theorem}
+
+\begin{proof}
+First note that:
+{{begin-eqn}}
+{{eqn | l = u
+      | r = \operatorname{arccot} \frac x a
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = x
+      | r = a \cot u
+      | c = Definition of [[Definition:Arccotangent|Arccotangent]]
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = u 
+      | r = \operatorname{arccot} \frac x a
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \frac{\mathrm d u} {\mathrm d x}
+      | r = \frac {-a} {a^2 + x^2}
+      | c = [[Derivative of Arccotangent Function/Corollary|Derivative of Arccotangent Function: Corollary]]
+}}
+{{eqn | ll= \implies
+      | l = \int F \left({\operatorname{arccot} \frac x a}\right) \ \mathrm d x
+      | r = \int F \left({u}\right) \ \frac {a^2 + x^2} {-a} \ \mathrm d u
+      | c = [[Primitive of Composite Function]]
+}}
+{{eqn | r = \int F \left({u}\right) \ \frac {a^2 + a^2 \cot^2 u} {-a} \ \mathrm d u
+      | c = Definition of $x$
+}}
+{{eqn | r = \int F \left({u}\right) \ a^2 \frac {1 + \cot^2 u} {-a} \ \mathrm d u
+      | c = 
+}}
+{{eqn | r = \int F \left({u}\right) \ \left({-a}\right) \csc^2 u \ \mathrm d u
+      | c = [[Difference of Squares of Cosecant and Cotangent]]
+}}
+{{eqn | r = -a \int F \left({u}\right) \ \csc^2 u \ \mathrm d u
+      | c = [[Primitive of Constant Multiple of Function]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Function of Arcsecant}
+Tags: Integral Substitutions, Arcsecant Function, Primitives involving Inverse Secant Function
+
+\begin{theorem}
+:$\displaystyle \int \map F {\arcsec \frac x a} \rd x = a \int \map F u \sec u \tan u \rd u$
+where $u = \arcsec \dfrac x a$.
+\end{theorem}
+
+\begin{proof}
+First note that:
+{{begin-eqn}}
+{{eqn | l = u
+      | r = \arcsec \frac x a
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x
+      | r = a \sec u
+      | c = {{Defof|Arcsecant}}
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = u 
+      | r = \arcsec \frac x a
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d u} {\d x}
+      | r = \frac a {\size x {\sqrt {x^2 - a^2} } }
+      | c = [[Derivative of Arcsecant Function/Corollary 1|Derivative of Arcsecant Function: Corollary 1]]
+}}
+{{eqn | ll= \leadsto
+      | l = \int \map F {\arcsec \frac x a} \rd x
+      | r = \int \map F u \, \frac {\size x {\sqrt {x^2 - a^2} } } a \rd u
+      | c = [[Primitive of Composite Function]]
+}}
+{{eqn | r = \int \map F u \, \frac {\size {a \sec u} {\sqrt {a^2 \sec^2 u - a^2} } } a \rd u
+      | c = Definition of $x$
+}}
+{{eqn | r = \int \map F u \, \size {\sec u} {\sqrt {a^2 \sec^2 u - a^2} } \rd u
+      | c = 
+}}
+{{eqn | r = \int \map F u \, a \sec u {\sqrt {\sec^2 u - 1} } \rd u
+      | c = $\sec u > 0$ in this domain
+}}
+{{eqn | r = \int \map F u \, a \sec u \tan u \rd u
+      | c = [[Difference of Squares of Secant and Tangent]]
+}}
+{{eqn | r = a \int \map F u \sec u \tan u \rd u
+      | c = [[Primitive of Constant Multiple of Function]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Function of Arccosecant}
+Tags: Integral Substitutions, Arccosecant Function, Primitives involving Inverse Cosecant Function
+
+\begin{theorem}
+:$\displaystyle \int F \left({\operatorname{arccsc} \frac x a}\right) \ \mathrm d x = -a \int F \left({u}\right) \left\vert{\csc u}\right\vert \cot u \ \mathrm d u$
+where $u = \operatorname{arccsc} \dfrac x a$.
+\end{theorem}
+
+\begin{proof}
+First note that:
+{{begin-eqn}}
+{{eqn | l = u
+      | r = \operatorname{arccsc} \frac x a
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = x
+      | r = a \csc u
+      | c = Definition of [[Definition:Arccosecant|Arccosecant]]
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = u 
+      | r = \operatorname{arccsc} \frac x a
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \frac{\mathrm d u} {\mathrm d x}
+      | r = \frac {-a} {\left\vert{x}\right\vert {\sqrt {x^2 - a^2} } }
+      | c = [[Derivative of Arccosecant Function/Corollary|Derivative of Arccosecant Function: Corollary]]
+}}
+{{eqn | ll= \implies
+      | l = \int F \left({\operatorname{arccsc} \frac x a}\right) \ \mathrm d x
+      | r = \int F \left({u}\right) \ \frac {\left\vert{x}\right\vert {\sqrt {x^2 - a^2} } } {-a} \ \mathrm d u
+      | c = [[Primitive of Composite Function]]
+}}
+{{eqn | r = \int F \left({u}\right) \ \frac {\left\vert{a \csc u}\right\vert {\sqrt {a^2 \csc^2 u - a^2} } } {-a} \ \mathrm d u
+      | c = Definition of $x$
+}}
+{{eqn | r = \int F \left({u}\right) \ \left\vert{\csc u}\right\vert \left({-\sqrt {a^2 \csc^2 u - a^2} }\right) \ \mathrm d u
+      | c = 
+}}
+{{eqn | r = \int F \left({u}\right) \ \left({-a}\right) \left\vert{\csc u}\right\vert {\sqrt {\csc^2 u - 1} } \ \mathrm d u
+      | c = 
+}}
+{{eqn | r = \int F \left({u}\right) \ \left({-a}\right) \left\vert{\csc u}\right\vert \cot u \ \mathrm d u
+      | c = [[Difference of Squares of Cosecant and Cotangent]]
+}}
+{{eqn | r = -a \int F \left({u}\right) \ \left\vert{\csc u}\right\vert \cot u \ \mathrm d u
+      | c = [[Primitive of Constant Multiple of Function]]
+}}
+{{end-eqn}}
+{{qed}}
+{{finish|Review the signs of the trig functions and their inverses rigorously -- I'm not sure about them.}}
+\end{proof}<|endoftext|>
+\section{Primitive of Reciprocal of a x + b}
+Tags: Primitives involving a x + b, Primitives involving Reciprocals
+
+\begin{theorem}
+:$\displaystyle \int \frac {\d x} {a x + b} = \frac 1 a \ln \size {a x + b} + C$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int \frac {\d x} {a x + b}
+      | r = \frac 1 a \int \frac {\map \d {a x + b} } {a x + b}
+      | c = [[Primitive of Function of a x + b|Primitive of Function of $a x + b$]]
+}}
+{{eqn | r = \frac 1 a \ln \size {a x + b} + C
+      | c = [[Primitive of Reciprocal]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $u = a x + b$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \int \frac {\mathrm d x} {\left({a x + b}\right)^2}
+      | r = \frac 1 a \int \frac {\mathrm d u} {u^2}
+      | c = [[Primitive of Function of a x + b|Primitive of Function of $a x + b$]]
+}}
+{{eqn | r = \frac 1 a \frac {-1} u + C
+      | c = [[Primitive of Power]]
+}}
+{{eqn | r = -\frac 1 {a \left({a x + b}\right)} + C
+      | c = substituting for $u$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+From [[Primitive of Power of a x + b|Primitive of Power of $a x + b$]]:
+:$\displaystyle \int \left({a x + b}\right)^n \ \mathrm d x = \frac {\left({a x + b}\right)^{n + 1} } {\left({n + 1}\right) a} + C$
+where $n \ne 1$.
+The result follows by setting $n = -2$.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $u = a x + b$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \int \frac {\mathrm d x} {\left({a x + b}\right)^3}
+      | r = \frac 1 a \int \frac {\mathrm d u} {u^3}
+      | c = [[Primitive of Function of a x + b|Primitive of Function of $a x + b$]]
+}}
+{{eqn | r = \frac 1 a \frac {-1} {2 u^2} + C
+      | c = [[Primitive of Power]]
+}}
+{{eqn | r = -\frac 1 {2 a \left({a x + b}\right)^2} + C
+      | c = substituting for $u$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+From [[Primitive of Power of a x + b|Primitive of Power of $a x + b$]]:
+:$\displaystyle \int \left({a x + b}\right)^n \ \mathrm d x = \frac {\left({a x + b}\right)^{n + 1} } {\left({n + 1}\right) a} + C$
+where $n \ne 1$.
+The result follows by setting $n = -3$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of x over a x + b}
+Tags: Primitive of x over a x + b
+
+\begin{theorem}
+:$\displaystyle \int \frac {x \rd x} {a x + b} = \frac x a - \frac b {a^2} \ln \size {a x + b} + C$
+\end{theorem}
+
+\begin{proof}
+Put $u = a x + b$.
+Then:
+{{begin-eqn}}
+{{eqn | l = x
+      | r = \frac {u - b} a
+      | c = 
+}}
+{{eqn | l = \frac {\mathrm d u} {\mathrm d x}
+      | r = \frac 1 a
+      | c = 
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = \int \frac {x \ \mathrm d x} {\left({a x + b}\right)^3}
+      | r = \int \frac 1 a \frac {u - b} {a u^3} \ \mathrm d u
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \frac 1 {a^2} \int \frac {\mathrm d u} {u^2} - \frac b {a^2} \int \frac {\mathrm d u} {u^3}
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \frac 1 {a^2} \frac {-1} {u} - \frac b {a^2} \frac {-1} {2 u^2} + C
+      | c = [[Primitive of Power]]
+}}
+{{eqn | r = \frac {-1} {a^2 \left({a x + b}\right)} + \frac b {2 a^2 \left({a x + b}\right)^2} + C
+      | c = substituting for $u$ and rearranging
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+From [[Primitive of x by Power of a x + b|Primitive of $x$ by Power of $a x + b$]]:
+:$\displaystyle \int x \left({a x + b}\right)^n \ \mathrm d x = \frac {\left({a x + b}\right)^{n + 2} } {\left({n + 2}\right) a^2} - \frac {b \left({a x + b}\right)^{n + 1} } {\left({n + 1}\right) a^2} + C$
+where $n \ne - 1$ and $n \ne - 2$.
+The result follows by setting $n = -3$.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Put $u = a x + b$
+Then:
+{{begin-eqn}}
+{{eqn | l = x
+      | r = \frac {u - b} a
+      | c = 
+}}
+{{eqn | l = \frac {\d x} {\d u}
+      | r = \frac 1 a
+      | c = 
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = \int \frac {x \rd x} {a x + b}
+      | r = \int \frac 1 a \frac {u - b} {a u} \rd u
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \frac 1 {a^2} \int \rd u - \frac b {a^2} \int \frac {\d u} u
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \frac u {a^2} + C - \frac b {a^2} \int \frac {\d u} u
+      | c = [[Primitive of Constant]]
+}}
+{{eqn | r = \frac u {a^2} - \frac b {a^2} \ln \size u + C
+      | c = [[Primitive of Reciprocal]]
+}}
+{{eqn | r = \frac {a x + b} {a^2} - \frac b {a^2} \ln \size {a x + b} + C
+      | c = substituting for $u$
+}}
+{{eqn | r = \frac x a + \frac b {a^2} - \frac b {a^2} \ln \size {a x + b} + C
+      | c = simplifying
+}}
+{{eqn | r = \frac x a - \frac b {a^2} \ln \size {a x + b} + C
+      | c = subsuming $\dfrac b {a^2}$ into the [[Definition:Arbitrary Constant (Calculus)|arbitrary constant]] $C$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+From [[Reduction Formula for Primitive of Power of x by Power of a x + b/Decrement of Power of x|Reduction Formula for Primitive of Power of $x$ by Power of $a x + b$: Decrement of Power of $x$]]:
+:$\displaystyle \int x^m \paren {a x + b}^n \rd x = \frac {x^m \paren {a x + b}^{n + 1} } {\paren {m + n + 1} a} - \frac {m b} {\paren {m + n + 1} a} \int x^{m - 1} \paren {a x + b}^n \rd x$
+Let $m = 1$ and $n = -1$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \int \frac {x \rd x} {a x + b}
+      | r = \int x^1 \paren {a x + b}^{-1} \rd x
+      | c = 
+}}
+{{eqn | r = \frac {x^1 \paren {a x + b}^0} {\paren 1 a} - \frac {1 b} {\paren 1 a} \int x^0 \paren {a x + b}^{-1} \rd x
+      | c = 
+}}
+{{eqn | r = \frac x a - \frac b a \int \frac {\d x} {a x + b}
+      | c = simplifying
+}}
+{{eqn | r = \frac x a - \frac b {a^2} \ln \size {a x + b} + C
+      | c = [[Primitive of Reciprocal of a x + b|Primitive of Reciprocal of $a x + b$]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int \frac {x \ \mathrm d x} {a x + b}
+      | r = \int \frac 1 a \frac {a x \ \mathrm d x} {a x + b}
+      | c = 
+}}
+{{eqn | r = \int \frac 1 a \frac {\left({a x + b - b}\right) \ \mathrm d x} {a x + b}
+      | c = 
+}}
+{{eqn | r = \frac 1 a \int \frac {\left({a x + b}\right) \ \mathrm d x} {a x + b} - \frac b a \int \frac {\mathrm d x} {a x + b}
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \frac 1 a \int \mathrm d x - \frac b a \int \frac {\mathrm d x} {a x + b}
+      | c = simplification
+}}
+{{eqn | r = \frac x a - \frac b a \int \frac {\mathrm d x} {a x + b}
+      | c = [[Primitive of Constant]]
+}}
+{{eqn | r = \frac x a - \frac b {a^2} \ln \left\vert{a x + b}\right\vert + C
+      | c = [[Primitive of Reciprocal of a x + b|Primitive of $\dfrac 1 {a x + b}$]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Put $u = a x + b$.
+Then:
+{{begin-eqn}}
+{{eqn | l = x
+      | r = \frac {u - b} a
+      | c = 
+}}
+{{eqn | l = \frac {\mathrm d u} {\mathrm d x}
+      | r = \frac 1 a
+      | c = 
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = \int \frac {x \ \mathrm d x} {\left({a x + b}\right)^2}
+      | r = \int \frac 1 a \frac {u - b} {a u^2} \ \mathrm d u
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \frac 1 {a^2} \int \frac {\mathrm d u} u - \frac b {a^2} \int \frac {\mathrm d u} {u^2}
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \frac 1 {a^2} \ln \left\vert{u}\right\vert + C - \frac b {a^2} \int \frac {\mathrm d u} u
+      | c = [[Primitive of Reciprocal]]
+}}
+{{eqn | r = \frac 1 {a^2} \ln \left\vert{u}\right\vert - \frac b {a^2} \frac {-1} u + C
+      | c = [[Primitive of Power]]
+}}
+{{eqn | r = \frac b {a^2 \left({a x + b}\right)} + \frac 1 {a^2} \ln \left\vert{a x + b}\right\vert + C
+      | c = substituting for $u$ and rearranging
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int \frac {x \ \mathrm d x} {\left({a x + b}\right)^2}
+      | r = \int \frac {a x \ \mathrm d x} {a \left({a x + b}\right)^2}
+      | c = multiplying [[Definition:Numerator|top]] and [[Definition:Denominator|bottom]] by $a$
+}}
+{{eqn | r = \int \frac {\left({a x + b - b}\right) \ \mathrm d x} {a \left({a x + b}\right)^2}
+      | c = adding and subtracting $b$
+}}
+{{eqn | r = \frac 1 a \int \frac {\left({a x + b}\right) \ \mathrm d x} {\left({a x + b}\right)^2} - \frac b a \int \frac {\mathrm d x} {\left({a x + b}\right)^2}
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \frac 1 a \int \frac {\mathrm d x} {a x + b} - \frac b a \int \frac {\mathrm d x} {\left({a x + b}\right)^2}
+      | c = simplification
+}}
+{{eqn | r = \frac 1 a \left({\frac 1 a \ln \left\vert{a x + b}\right\vert}\right) - \frac b a \int \frac {\mathrm d x} {\left({a x + b}\right)^2} + C
+      | c = [[Primitive of Reciprocal of a x + b|Primitive of Reciprocal of $\dfrac 1 {\left({a x + b}\right)}$]]
+}}
+{{eqn | r = \frac 1 a \left({\frac 1 a \ln \left\vert{a x + b}\right\vert}\right) - \frac b a \left({-\frac 1 {a \left({a x + b}\right)} }\right) + C
+      | c = [[Primitive of Reciprocal of a x + b squared|Primitive of Reciprocal of $\dfrac 1 {\left({a x + b}\right)^2}$]]
+}}
+{{eqn | r = \frac b {a^2 \left({a x + b}\right)} + \frac 1 {a^2} \ln \left\vert{a x + b}\right\vert + C
+      | c = simplification
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Laplace Transform of Exponential times Function}
+Tags: Laplace Transforms, Exponential Function
+
+\begin{theorem}
+:$\laptrans {e^{a t} \map f t} = \map F {s - a}$
+everywhere that $\laptrans f$ exists, for $\map \Re s > a$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \laptrans {e^{a t} \map f t}
+      | r = \int_0^{\to +\infty} e^{-s t} e^{a t} \map f t \rd t
+      | c = {{Defof|Laplace Transform}}
+}}
+{{eqn | r = \int_0^{\to +\infty} e^{-s t + a t} \map f t \rd t
+      | c = [[Exponent Combination Laws]]
+}}
+{{eqn | r = \int_0^{\to +\infty} e^{-\paren {s - a} t} \map f t \rd t
+}}
+{{eqn | r = \map F {s - a}
+      | c = {{Defof|Laplace Transform}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Laplace Transform of Derivative}
+Tags: Laplace Transforms, Derivatives, Laplace Transforms of Derivatives
+
+\begin{theorem}
+Let $f'$ be [[Definition:Piecewise Continuous Function with One-Sided Limits|piecewise continuous with one-sided limits]] on said [[Definition:Real Interval|intervals]].
+Then $\laptrans f$ exists for $\map \Re s > a$, and:
+:$\laptrans {\map {f'} t} = s \laptrans {\map f t} - \map f 0$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \laptrans {\map {f'} t}
+      | r = \int_0^{\mathop \to +\infty} e^{-s t} \map {f'} t \rd t
+      | c = {{Defof|Laplace Transform}}
+}}
+{{eqn | r = \lim_{A \mathop \to +\infty} \int_0^A e^{-s t} \map {f'} t \rd t
+      | c = {{Defof|Improper Integral on Closed Interval Unbounded Above}}
+}}
+{{end-eqn}}
+Consider:
+:$\displaystyle \int_0^A e^{-s t} \map {f'} t \rd t$
+[[Definition:By Hypothesis|By hypothesis]], $f'$ is [[Definition:Piecewise Continuous Function with One-Sided Limits|piecewise continuous with one-sided limits]].
+So by [[Piecewise Continuous Function with One-Sided Limits is Darboux Integrable]], this integral exists.
+This means that [[Integration by Parts|integration by parts]] can be invoked:
+:$\displaystyle \int h j\,' \rd t = h j - \int h' j \rd t$
+Here:
+{{begin-eqn}}
+{{eqn | l = h
+      | r = e^{-s t}
+}}
+{{eqn | ll = \leadsto
+      | l = h'
+      | r = -s e^{-s t}
+}}
+{{eqn | l = j\,'
+      | r = \map {f'} t
+}}
+{{eqn | ll = \leadsto
+      | l = j
+      | r = \map f t
+}}
+{{end-eqn}}
+So:
+{{begin-eqn}}
+{{eqn | l = \int_0^A e^{-s t} \map {f'} t \rd t
+      | r = \bigintlimits {e^{-s t} \map f t} {t \mathop = 0} {t \mathop = A} + s \int_0^A e^{-s t} \map f t \rd t
+}}
+{{end-eqn}}
+Now, take the limit as $t = A \to +\infty$:
+{{begin-eqn}}
+{{eqn | l = \laptrans {\map {f'} t}
+      | r = \lim_{A \mathop \to +\infty} e^{-s A} \map f A - \map f 0 + s \laptrans {\map f t}
+}}
+{{end-eqn}}
+Recall that $f$ is of [[Definition:Exponential Order|exponential order $a$]]:
+{{begin-eqn}}
+{{eqn | l = \size {\map f A}
+      | o = <
+      | r = K e^{a t}
+}}
+{{eqn | ll= \leadsto
+      | l = \size {\map f A} \size {e^{-s A} }
+      | o = <
+      | r = K e^{a t} \size {e^{-s A} }
+}}
+{{eqn | ll= \leadsto
+      | l = \size {e^{-s A} \map f A}
+      | o = <
+      | r = \size {K e^{a t} e^{-s A} }
+      | c = [[Modulus of Product]], [[Exponential Tends to Zero and Infinity]]
+}}
+{{eqn | r = \size {K e^{\paren {a - s} t} }
+      | c = [[Exponent Combination Laws]], definition of $A$
+}}
+{{eqn | r = \size {K e^{\paren {a - \map \Re s - i \map \Im s} t} }
+}}
+{{eqn | r = \size {K e^{\paren {a - \map \Re s} t} }
+      | c = [[Modulus of Exponential is Modulus of Real Part]]
+}}
+{{eqn | r = K e^{\paren {a - \map \Re s} t} 
+      | c = [[Exponential Tends to Zero and Infinity]]
+}}
+{{end-eqn}}
+This implies, from [[Complex Exponential Tends to Zero]] and the [[Squeeze Theorem/Functions|Squeeze Theorem]]:
+:$\displaystyle \lim_{A \mathop \to +\infty} e^{-s A} \map f A = 0$
+which produces:
+:$\laptrans {\map {f'} t} = s \laptrans {\map f t} - \map f 0$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of x squared over a x + b}
+Tags: Primitive of x squared over a x + b
+
+\begin{theorem}
+:$\displaystyle \int \frac {x^2 \rd x} {a x + b} = \frac {\paren {a x + b}^2} {2 a^3} - \frac {2 b \paren {a x + b} } {a^3} + \frac {b^2} {a^3} \ln \size {a x + b} + C$
+\end{theorem}
+
+\begin{proof}
+Put $u = a x + b$.
+Then:
+{{begin-eqn}}
+{{eqn | l = x
+      | r = \frac {u - b} a
+      | c = 
+}}
+{{eqn | l = \frac {\d u} {\d x}
+      | r = \frac 1 a
+      | c = 
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = \int \frac {x^2 \rd x} {a x + b}
+      | r = \int \frac 1 a \paren {\frac {u - b} a}^2 \frac {\d u} u
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \frac 1 {a^3} \int \frac {\paren {u - b}^2} u \rd u
+      | c = [[Primitive of Constant Multiple of Function]]
+}}
+{{eqn | r = \frac 1 {a^3} \int \paren {u - 2 b + \frac {b^2} u} \rd u
+      | c = multiplying out and simplifying
+}}
+{{eqn | r = \frac 1 {a^3} \paren {\int u \rd u - \int 2 b \rd u + \int \frac {b^2} u \rd u}
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \frac 1 {a^3} \paren {\frac {u^2} 2 + C - \int 2 b \rd u + \int \frac {b^2} u \rd u}
+      | c = [[Primitive of Power]]
+}}
+{{eqn | r = \frac 1 {a^3} \paren {\frac {u^2} 2 - 2 b u + C + \int \frac {b^2} u \rd u}
+      | c = [[Primitive of Constant]]
+}}
+{{eqn | r = \frac 1 {a^3} \paren {\frac {u^2} 2 - 2 b u + C + b^2 \int \frac {\d u} u}
+      | c = [[Primitive of Constant Multiple of Function]]
+}}
+{{eqn | r = \frac 1 {a^3} \paren {\frac {u^2} 2 - 2 b u + b^2 \ln \size u} + C
+      | c = [[Primitive of Reciprocal]] and subsuming [[Definition:Arbitrary Constant (Calculus)|arbitrary constant]] $C$
+}}
+{{eqn | r = \frac {\paren {a x + b}^2} {2 a^3} - \frac {2 b \paren {a x + b} } {a^3} + \frac {b^2} {a^3} \ln \size {a x + b} + C
+      | c = substituting for $u$ and simplifying
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+From [[Reduction Formula for Primitive of Power of x by Power of a x + b/Decrement of Power of x|Reduction Formula for Primitive of $x^m \paren {a x + b}^n$: Decrement of Power of $x$]]:
+:$\displaystyle \int x^m \paren {a x + b}^n \rd x = \frac {x^m \paren {a x + b}^{n + 1} } {\paren {m + n + 1} a} - \frac {m b} {\paren {m + n + 1} a} \int x^{m - 1} \paren {a x + b}^n \rd x$
+Let $m = 2$ and $n = -1$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \int \frac {x^2 \rd x} {a x + b}
+      | r = \int x^2 \paren {a x + b}^{-1} \rd x
+      | c = 
+}}
+{{eqn | r = \frac {x^2 \paren {a x + b}^0} {\paren 2 a} - \frac {2 b} {\paren 2 a} \int x^1 \paren {a x + b}^{-1} \rd x
+      | c = 
+}}
+{{eqn | r = \frac {x^2} {2 a} - \frac b a \int \frac {x \rd x} {a x + b}
+      | c = simplifying
+}}
+{{eqn | r = \frac {x^2} {2 a} - \frac b a \paren {\frac x a - \frac b {a^2} \ln \size {a x + b} } + C
+      | c = [[Primitive of x over a x + b|Primitive of $x$ over $a x + b$]]
+}}
+{{eqn | r = \frac {x^2} {2 a} - \frac {b x} {a^2} + \frac {b^2} {a^3} \ln \size {a x + b} + C
+      | c = simplifying
+}}
+{{eqn | r = \frac {a^2 x^2} {2 a^3} - \frac {2 a b x} {2 a^3} + \frac {b^2} {a^3} \ln \size {a x + b} + C
+      | c = 
+}}
+{{eqn | r = \frac {a^2 x^2 + 2 a b x} {2 a^3} - \frac {4 a b x} {2 a^3} + \frac {b^2} {a^3} \ln \size {a x + b} + C
+      | c = 
+}}
+{{eqn | r = \frac {a^2 x^2 + 2 a b x + b^2} {2 a^3} - \frac {4 a b x + b^2} {2 a^3} + \frac {b^2} {a^3} \ln \size {a x + b} + C
+      | c = 
+}}
+{{eqn | r = \frac {a^2 x^2 + 2 a b x + b^2} {2 a^3} - \frac {4 a b x + 4 b^2} {2 a^3} + \frac {b^2} {a^3} \ln \size {a x + b} + C
+      | c = subsuminging $\dfrac {3 b^2} {2 a^3}$ into the [[Definition:Arbitrary Constant (Calculus)|arbitrary constant]]
+}}
+{{eqn | r = \frac {\paren {a x + b}^2} {2 a^3} - \frac {2 a b x + 2 b^2} {a^3} + \frac {b^2} {a^3} \ln \size {a x + b} + C
+      | c = 
+}}
+{{eqn | r = \frac {\paren {a x + b}^2} {2 a^3} - \frac {2 b \paren {a x + b} } {a^3} + \frac {b^2} {a^3} \ln \size {a x + b} + C
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Put $u = a x + b$.
+Then:
+{{begin-eqn}}
+{{eqn | l = x
+      | r = \frac {u - b} a
+      | c = 
+}}
+{{eqn | l = \frac {\d u} {\d x}
+      | r = \frac 1 a
+      | c = 
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = \int \frac {x^2 \rd x} {\paren {a x + b}^2}
+      | r = \int \frac 1 a \paren {\frac {u - b} a}^2 \frac 1 {u^2} \rd u
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \int \frac 1 {a^3} \paren {1 - \frac {2 b} u + \frac {b^2} {u^2} } \rd u
+      | c = [[Square of Difference]], and simplification
+}}
+{{eqn | r = \frac 1 {a^3} \int \d u - \frac {2 b} {a^3} \int \frac {\d u} u + \frac {b^2} {a^3} \int \frac {\d u} {u^2}
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \frac u {a^3} - \frac {2 b} {a^3} \int \frac {\d u} u + \frac {b^2} {a^3} \int \frac {\d u} {u^2} + C
+      | c = [[Primitive of Constant]]
+}}
+{{eqn | r = \frac u {a^3} - \frac {2 b} {a^3} \ln \size u + \frac {b^2} {a^3} \int \frac {\d u} {u^2} + C
+      | c = [[Primitive of Reciprocal]]
+}}
+{{eqn | r = \frac u {a^3} - \frac {2 b} {a^3} \ln \size u + \frac {b^2} {a^3} \frac {-1} u + C
+      | c = [[Primitive of Power]]
+}}
+{{eqn | r = \frac {a x + b} {a^3} - \frac {b^2} {a^3 \paren {a x + b} } - \frac {2 b} {a^3} \ln \size {a x + b} + C
+      | c = substituting for $u$ and rearranging
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int \frac {x^2 \ \mathrm d x} {\left({a x + b}\right)^2}
+      | r = \int \frac {a x^2 \ \mathrm d x} {a \left({a x + b}\right)^2}
+      | c = multiplying [[Definition:Numerator|top]] and [[Definition:Denominator|bottom]] by $a$
+}}
+{{eqn | r = \int \frac {x \left({a x + b - b}\right) \ \mathrm d x} {a \left({a x + b}\right)^2}
+      | c = adding and subtracting $b x$
+}}
+{{eqn | r = \frac 1 a \int \frac {x \left({a x + b}\right) \ \mathrm d x} {\left({a x + b}\right)^2} - \frac b a \int \frac {x \ \mathrm d x} {\left({a x + b}\right)^2}
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \frac 1 a \int \frac {x \ \mathrm d x} {a x + b} - \frac b a \int \frac {x \ \mathrm d x} {\left({a x + b}\right)^2}
+      | c = simplification
+}}
+{{eqn | r = \frac 1 a \left({\frac x a - \frac b {a^2} \ln \left\vert{a x + b}\right\vert}\right) - \frac b a \int \frac {\mathrm d x} {\left({a x + b}\right)^2} + C
+      | c = [[Primitive of x over a x + b|Primitive of Reciprocal of $\dfrac x {\left({a x + b}\right)}$]]
+}}
+{{eqn | r = \frac 1 a \left({\frac x a - \frac b {a^2} \ln \left\vert{a x + b}\right\vert}\right) - \frac b a \left({\frac b {a^2 \left({a x + b}\right)} + \frac 1 {a^2} \left\vert{a x + b}\right\vert }\right) + C
+      | c = [[Primitive of x over a x + b squared|Primitive of Reciprocal of $\dfrac x {\left({a x + b}\right)^2}$]]
+}}
+{{eqn | r = \frac x {a^2} - \frac b {a^3} \ln \left\vert{a x + b}\right\vert - \frac {b^2} {a^3 \left({a x + b}\right)} - \frac b {a^3} \ln \left\vert{a x + b}\right\vert + C
+      | c = multiplying out
+}}
+{{eqn | r = \frac x {a^2} - \frac {b^2} {a^3 \left({a x + b}\right)} - \frac {2 b} {a^3} \ln \left\vert{a x + b}\right\vert + C
+      | c = gathering terms
+}}
+{{eqn | r = \frac {a x} {a^3} + \frac b {a^3} - \frac {b^2} {a^3 \left({a x + b}\right)} - \frac {2 b} {a^3} \ln \left\vert{a x + b}\right\vert + C
+      | c = where $\dfrac b {a^3}$ is subsumed into $C$
+}}
+{{eqn | r = \frac {a x + b} {a^3} - \frac {b^2} {a^3 \left({a x + b}\right)} - \frac {2 b} {a^3} \ln \left\vert{a x + b}\right\vert + C
+      | c = simplification
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of x cubed over a x + b}
+Tags: Primitives involving a x + b
+
+\begin{theorem}
+:$\displaystyle \int \frac {x^3 \rd x} {a x + b} = \frac {\paren {a x + b}^3} {3 a^4} - \frac {3 b \paren {a x + b}^2} {2 a^4} - \frac {3 b^2 \paren {a x + b} } {a^4} + \frac {b^3} {a^4} \ln \size {a x + b} + C$
+\end{theorem}
+
+\begin{proof}
+Put $u = a x + b$.
+Then:
+{{begin-eqn}}
+{{eqn | l = x
+      | r = \frac {u - b} a
+      | c = 
+}}
+{{eqn | l = \frac {\d u} {\d x}
+      | r = \frac 1 a
+      | c = 
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = \int \frac {x^3 \rd x} {a x + b}
+      | r = \int \frac 1 a \paren {\frac {u - b} a}^3 \frac {\d u} u
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \frac 1 {a^4} \int \frac {\paren {u - b}^3} u \rd u
+      | c = [[Primitive of Constant Multiple of Function]]
+}}
+{{eqn | r = \frac 1 {a^4} \int \paren {u^2 - 3 b u + 3 b^2 - \frac {b^3} u} \rd u
+      | c = multiplying out and simplifying
+}}
+{{eqn | r = \frac 1 {a^4} \paren {\int u^2 \rd u - \int 3 b u \rd u + \int 3 b^2 \rd u - \int \frac {b^3 \rd u} u}
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \frac 1 {a^4} \paren {\frac {u^3} 3 - 3 b \frac {u^2} 2 + \int 3 b^2 \rd u - \int \frac {b^3} u \rd u} + C
+      | c = [[Primitive of Power]] (two instances)
+}}
+{{eqn | r = \frac 1 {a^4} \paren {\frac {u^3} 3 - 3 b \frac {u^2} 2 + b^2 u - \int \frac {b^3} u \rd u} + C
+      | c = [[Primitive of Constant]] and subsuming [[Definition:Arbitrary Constant (Calculus)|arbitrary constant]] $C$
+}}
+{{eqn | r = \frac 1 {a^4} \paren {\frac {u^3} 3 - 3 b \frac {u^2} 2 + 3 b^2 u - b^3 \int \frac {\d u} u} + C
+      | c = [[Primitive of Constant Multiple of Function]]
+}}
+{{eqn | r = \frac 1 {a^4} \paren {\frac {u^3} 3 - 3 b \frac {u^2} 2 + 3 b^2 u - b^3 \ln \size u} + C
+      | c = [[Primitive of Reciprocal]]
+}}
+{{eqn | r = \frac {\paren {a x + b}^3} {3 a^4} - \frac {3 b \paren {a x + b}^2} {2 a^4} - \frac {3 b^2 \paren {a x + b} } {a^4} + \frac {b^3} {a^4} \ln \size {a x + b} + C
+      | c = substituting for $u$ and simplifying
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Reciprocal of x by a x + b}
+Tags: Primitive of Reciprocal of x by a x + b, Primitives involving Reciprocals
+
+\begin{theorem}
+:$\displaystyle \int \frac {\rd x} {x \paren {a x + b} } = \frac 1 b \ln \size {\frac x {a x + b} } + C$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int \frac {\d x} {x \paren {a x + b} }
+      | r = \int \paren {\dfrac 1 {b x} - \dfrac a {b \paren {a x + b} } } \rd x
+      | c = [[Primitive of Reciprocal of x by a x + b/Partial Fraction Expansion|Partial Fraction Expansion]]
+}}
+{{eqn | r = \frac 1 b \int \frac {\d x} x - \frac a b \int \frac {\d x} {a x + b}
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \frac 1 b \ln \size x - \frac a b \int \frac {\d x} {a x + b} + C
+      | c = [[Primitive of Reciprocal]]
+}}
+{{eqn | r = \frac 1 b \ln \size x - \frac 1 b \ln \size {a x + b} + C
+      | c = [[Primitive of Reciprocal of a x + b|Primitive of $\dfrac 1 {a x + b}$]]
+}}
+{{eqn | r = \frac 1 b \ln \size {\frac x {a x + b} } + C
+      | c = [[Difference of Logarithms]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int \frac {\d x} {x \paren {a x + b} }
+      | r = \int \frac {b \rd x} {b x \paren {a x + b} }
+      | c = multiplying [[Definition:Numerator|top]] and [[Definition:Denominator|bottom]] by $b$
+}}
+{{eqn | r = \int \frac {\paren {a x + b - a x} \rd x} {b x \paren {a x + b} }
+      | c = adding and subtracting $a x$
+}}
+{{eqn | r = \frac 1 b \int \frac {\paren {a x + b} \rd x} {x \paren {a x + b} } - \frac a b \int \frac {x \rd x} {x \paren {a x + b} }
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \frac 1 b \int \frac {\d x} x - \frac a b \int \frac {\d x} {a x + b}
+      | c = simplifying
+}}
+{{eqn | r = \frac 1 b \ln \size x - \frac a b \int \frac {\d x} {a x + b} + C
+      | c = [[Primitive of Reciprocal]]
+}}
+{{eqn | r = \frac 1 b \ln \size x - \frac 1 b \ln \size {a x + b} + C
+      | c = [[Primitive of Reciprocal of a x + b|Primitive of $\dfrac 1 {a x + b}$]]
+}}
+{{eqn | r = \frac 1 b \ln \size {\frac x {a x + b} } + C
+      | c = [[Difference of Logarithms]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int \frac {\mathrm d x} {x \left({a x + b}\right)^2}
+      | r = \int \left({\frac 1 {b^2 x} - \frac a {b^2 \left({a x + b}\right)} - \frac a {b \left({a x + b}\right)^2} }\right) \ \mathrm d x
+      | c = [[Primitive of Reciprocal of x by a x + b squared/Partial Fraction Expansion|Partial Fraction Expansion]]
+}}
+{{eqn | r = \frac 1 {b^2} \int \frac {\mathrm d x} x - \frac a {b^2} \int \frac {\mathrm d x} {a x + b} - \frac a b \int \frac {\mathrm d x} {\left({a x + b}\right)^2}
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \frac 1 {b^2} \ln \left\vert{x}\right\vert - \frac a {b^2} \int \frac {\mathrm d x} {a x + b} - \frac a b \int \frac {\mathrm d x} {\left({a x + b}\right)^2} + C
+      | c = [[Primitive of Reciprocal]]
+}}
+{{eqn | r = \frac 1 {b^2} \ln \left\vert{x}\right\vert - \frac a {b^2} \ln \left\vert{a x + b}\right\vert - \frac a b \int \frac {\mathrm d x} {\left({a x + b}\right)^2} + C
+      | c = [[Primitive of Reciprocal of a x + b|Primitive of $\dfrac 1 {a x + b}$]]
+}}
+{{eqn | r = \frac 1 {b^2} \ln \left\vert{x}\right\vert - \frac a {b^2} \ln \left\vert{a x + b}\right\vert - \frac a b \frac {-1} {a \left({a x + b}\right)} + C
+      | c = [[Primitive of Reciprocal of a x + b squared|Primitive of $\dfrac 1 {\left({a x + b}\right)^2}$]]
+}}
+{{eqn | r = \frac 1 {b \left({a x + b}\right)} + \frac 1 {b^2} \ln \left\vert{\frac x {a x + b} }\right\vert + C
+      | c = [[Difference of Logarithms]] and rearranging
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int \frac {\d x} {x \paren {a x + b}^2}
+      | r = \int \frac {b \rd x} {b x \paren {a x + b}^2}
+      | c = multiplying [[Definition:Numerator|top]] and [[Definition:Denominator|bottom]] by $b$
+}}
+{{eqn | r = \int \frac {\paren {a x + b - a x} \rd x} {b x \paren {a x + b}^2}
+      | c = adding and subtracting $a x$
+}}
+{{eqn | r = \frac 1 b \int \frac {\paren {a x + b} \rd x} {x \paren {a x + b}^2} - \frac a b \int \frac {x \rd x} {x \paren {a x + b}^2}
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \frac 1 b \int \frac {\d x} {x \paren {a x + b} } - \frac a b \int \frac {\d x} {\paren {a x + b}^2}
+      | c = simplifying
+}}
+{{eqn | r = \frac 1 b \paren {\frac 1 b \ln \size {\frac x {a x + b} } } - \frac a b \int \frac {\d x} {\paren {a x + b}^2} + C
+      | c = [[Primitive of Reciprocal of x by a x + b|Primitive of $\dfrac 1 {x \paren {a x + b} }$]]
+}}
+{{eqn | r = \frac 1 {b^2} \ln \size {\frac x {a x + b} } - \frac a b \paren {-\frac 1 {a \paren {a x + b} } } + C
+      | c = [[Primitive of Reciprocal of a x + b squared|Primitive of $\dfrac 1 {\paren {a x + b}^2}$]]
+}}
+{{eqn | r = \frac 1 {b \paren {a x + b} } + \frac 1 {b^2} \ln \size {\frac x {a x + b} } + C
+      | c = simplifying
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Reciprocal of x squared by a x + b}
+Tags: Primitive of Reciprocal of x squared by a x + b, Primitives involving Reciprocals
+
+\begin{theorem}
+:$\ds \int \frac {\d x} {x^2 \paren {a x + b} } = -\frac 1 {b x} + \frac a {b^2} \ln \size {\frac {a x + b} x} + C$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int \frac {\d x} {x^2 \paren {a x + b} }
+      | r = \int \paren {-\frac a {b^2 x} + \frac 1 {b x^2} + \frac {a^2} {b^2 \paren {a x + b} } } \rd x
+      | c = [[Primitive of Reciprocal of x squared by a x + b/Partial Fraction Expansion|Partial Fraction Expansion]]
+}}
+{{eqn | r = -\frac a {b^2} \int \frac {\d x} x + \frac 1 b \int \frac {\d x} {x^2} + \frac {a^2} {b^2} \int \frac {\d x} {a x + b}
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = -\frac a {b^2} \int \frac {\d x} x + \frac 1 b \frac {-1} x + \frac {a^2} {b^2} \int \frac {\d x} {a x + b} + C
+      | c = [[Primitive of Power]]
+}}
+{{eqn | r = -\frac a {b^2} \ln \size x - \frac 1 {b x} + \frac {a^2} {b^2} \int \frac {\d x} {a x + b} + C
+      | c = [[Primitive of Reciprocal]]
+}}
+{{eqn | r = -\frac a {b^2} \ln \size x - \frac 1 {b x} + \frac a {b^2} \ln \size {a x + b} + C
+      | c = [[Primitive of Reciprocal of a x + b]]
+}}
+{{eqn | r = -\frac 1 {b x} + \frac a {b^2} \ln \size {\frac {a x + b} x} + C
+      | c = [[Difference of Logarithms]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int \frac {\d x} {x^2 \paren {a x + b} }
+      | r = \int \frac {b \rd x} {b x^2 \paren {a x + b} }
+      | c = multiplying [[Definition:Numerator|top]] and [[Definition:Denominator|bottom]] by $b$
+}}
+{{eqn | r = \int \frac {\paren {a x + b - a x} \rd x} {b x^2 \paren {a x + b} }
+      | c = adding and subtracting $a x$
+}}
+{{eqn | r = \frac 1 b \int \frac {\paren {a x + b } \rd x} {x^2 \paren {a x + b} } - \frac a b \int \frac {x \rd x} {x^2 \paren {a x + b} }
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \frac 1 b \int \frac {\d x} {x^2} - \frac a b \int \frac {\d x} {x \paren {a x + b} }
+      | c = simplification
+}}
+{{eqn | r = -\frac 1 {b x} - \frac a b \int \frac {\d x} {x \paren {a x + b} } + C
+      | c = [[Primitive of Power]]
+}}
+{{eqn | r = -\frac 1 {b x} - \frac a b \paren {\frac 1 b \ln \size {\frac x {a x + b} } } + C
+      | c = [[Primitive of Reciprocal of x by a x + b|Primitive of $\dfrac 1 {x \paren {a x + b} }$]]
+}}
+{{eqn | r = -\frac 1 {b x} + \frac a {b^2} \ln \size {\frac {a x + b} x} + C
+      | c = [[Logarithm of Reciprocal]] and simplification
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Reciprocal of x cubed by a x + b}
+Tags: Primitive of Reciprocal of x cubed by a x + b, Primitives involving Reciprocals
+
+\begin{theorem}
+:$\displaystyle \int \frac {\mathrm d x} {x^3 \left({a x + b}\right)} = \frac {2 a x - b} {2 b^2 x^2} + \frac {a^2} {b^3} \ln \left\vert{\frac x {a x + b} }\right\vert + C$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int \frac {\mathrm d x} {x^3 \left({a x + b}\right)}
+      | r = \int \frac {b \ \mathrm d x} {b x^3 \left({a x + b}\right)}
+      | c = multiplying [[Definition:Numerator|top]] and [[Definition:Denominator|bottom]] by $b$
+}}
+{{eqn | r = \int \frac {\left({a x + b - a x}\right) \ \mathrm d x} {b x^3 \left({a x + b}\right)}
+      | c = adding and subtracting $a x$
+}}
+{{eqn | r = \frac 1 b \int \frac {\left({a x + b}\right) \ \mathrm d x} {x^3 \left({a x + b}\right)} - \frac a b \int \frac {x \ \mathrm d x} {x^3 \left({a x + b}\right)}
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \frac 1 b \int \frac {\mathrm d x} {x^3} - \frac a b \int \frac {\mathrm d x} {x^2 \left({a x + b}\right)}
+      | c = simplification
+}}
+{{eqn | r = -\frac 1 {2 b x^2} - \frac a b \int \frac {\mathrm d x} {x^2 \left({a x + b}\right)} + C
+      | c = [[Primitive of Power]]
+}}
+{{eqn | r = -\frac 1 {2 b x^2} - \frac a b \left({-\frac 1 {b x} + \frac a {b^2} \ln \left\vert{\frac {a x + b} x}\right\vert}\right) + C
+      | c = [[Primitive of Reciprocal of x squared by a x + b|Primitive of $\dfrac 1 {x^2 \left({a x + b}\right)}$]]
+}}
+{{eqn | r = -\frac 1 {2 b x^2} + \frac a {b^2 x} - \frac {a^2} {b^3} \ln \left\vert{\frac {a x + b} x}\right\vert + C
+      | c = multiplying out
+}}
+{{eqn | r = \frac a {b^2 x} - \frac 1 {2 b x^2} + \frac {a^2} {b^3} \ln \left\vert{\frac x {a x + b} }\right\vert + C
+      | c = [[Difference of Logarithms]]
+}}
+{{eqn | r = \frac {2 a x - b} {2 b^2 x^2} + \frac {a^2} {b^3} \ln \left\vert{\frac x {a x + b} }\right\vert + C
+      | c = rearranging
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int \frac {\d x} {x^3 \paren {a x + b} }
+      | r = \int \paren {\frac {a^2} {b^3 x} + \frac {-a} {b^2 x^2} + \frac 1 {b x^3} + \frac {-a^3} {b^3 \paren {a x + b} } } \rd x
+      | c = [[Primitive of Reciprocal of x cubed by a x + b/Partial Fraction Expansion|Partial Fraction Expansion]]
+}}
+{{eqn | r = \frac {a^2} {b^3} \int \frac {\d x} x + \frac {-a} {b^2} \int \frac {\d x} {x^2} + \frac 1 b \int \frac {\d x} {x^3} + \frac {-a^3} {b^3} \int \frac {\d x} {a x + b}
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \frac {a^2} {b^3} \int \frac {\d x} x + \frac {-a} {b^2} \frac {-1} x + \frac 1 b \frac {-1} {2 x^2} + \frac {-a^3} {b^3} \int \frac {\d x} {a x + b} + C
+      | c = [[Primitive of Power]]
+}}
+{{eqn | r = \frac {a^2} {b^3} \ln \size x + \frac a {b^2 x} - \frac 1 {2 b x^2} + \frac {-a^3} {b^3} \int \frac {\d x} {a x + b} + C
+      | c = [[Primitive of Reciprocal]]
+}}
+{{eqn | r = \frac {a^2} {b^3} \ln \size x + \frac a {b^2 x} - \frac 1 {2 b x^2} + \frac {-a^3} {b^3} \frac 1 a \ln \size {a x + b} + C
+      | c = [[Primitive of Reciprocal of a x + b]]
+}}
+{{eqn | r = \frac a {b^2 x} - \frac 1 {2 b x^2} + \frac {a^2} {b^3} \ln \size {\frac x {a x + b} } + C
+      | c = [[Difference of Logarithms]]
+}}
+{{eqn | r = \frac {2 a x - b} {2 b^2 x^2} + \frac {a^2} {b^3} \ln \size {\frac x {a x + b} } + C
+      | c = rearranging
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitives of Rational Functions involving a x + b}
+Tags: Primitives, Primitives of Rational Functions, Primitives involving a x + b
+
+\begin{theorem}
+This page gathers together the [[Definition:Primitive (Calculus)|primitives]] of some [[Definition:Rational Function|rational functions]] involving $a x + b$.
+\end{theorem}<|endoftext|>
+\section{Primitive of Reciprocal of a x + b squared}
+Tags: Primitive of Reciprocal of a x + b squared, Primitives involving a x + b, Primitives involving Reciprocals
+
+\begin{theorem}
+:$\displaystyle \int \frac {\d x} {\paren {a x + b}^2} = -\frac 1 {a \paren {a x + b} } + C$
+\end{theorem}
+
+\begin{proof}
+Let $u = a x + b$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \int \frac {\mathrm d x} {\left({a x + b}\right)^2}
+      | r = \frac 1 a \int \frac {\mathrm d u} {u^2}
+      | c = [[Primitive of Function of a x + b|Primitive of Function of $a x + b$]]
+}}
+{{eqn | r = \frac 1 a \frac {-1} u + C
+      | c = [[Primitive of Power]]
+}}
+{{eqn | r = -\frac 1 {a \left({a x + b}\right)} + C
+      | c = substituting for $u$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+From [[Primitive of Power of a x + b|Primitive of Power of $a x + b$]]:
+:$\displaystyle \int \left({a x + b}\right)^n \ \mathrm d x = \frac {\left({a x + b}\right)^{n + 1} } {\left({n + 1}\right) a} + C$
+where $n \ne 1$.
+The result follows by setting $n = -2$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of x over a x + b squared}
+Tags: Primitive of x over a x + b squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {x \ \mathrm d x} {\left({a x + b}\right)^2} = \frac b {a^2 \left({a x + b}\right)} + \frac 1 {a^2} \ln \left\vert{a x + b}\right\vert + C$
+\end{theorem}
+
+\begin{proof}
+Put $u = a x + b$.
+Then:
+{{begin-eqn}}
+{{eqn | l = x
+      | r = \frac {u - b} a
+      | c = 
+}}
+{{eqn | l = \frac {\mathrm d u} {\mathrm d x}
+      | r = \frac 1 a
+      | c = 
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = \int \frac {x \ \mathrm d x} {\left({a x + b}\right)^2}
+      | r = \int \frac 1 a \frac {u - b} {a u^2} \ \mathrm d u
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \frac 1 {a^2} \int \frac {\mathrm d u} u - \frac b {a^2} \int \frac {\mathrm d u} {u^2}
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \frac 1 {a^2} \ln \left\vert{u}\right\vert + C - \frac b {a^2} \int \frac {\mathrm d u} u
+      | c = [[Primitive of Reciprocal]]
+}}
+{{eqn | r = \frac 1 {a^2} \ln \left\vert{u}\right\vert - \frac b {a^2} \frac {-1} u + C
+      | c = [[Primitive of Power]]
+}}
+{{eqn | r = \frac b {a^2 \left({a x + b}\right)} + \frac 1 {a^2} \ln \left\vert{a x + b}\right\vert + C
+      | c = substituting for $u$ and rearranging
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int \frac {x \ \mathrm d x} {\left({a x + b}\right)^2}
+      | r = \int \frac {a x \ \mathrm d x} {a \left({a x + b}\right)^2}
+      | c = multiplying [[Definition:Numerator|top]] and [[Definition:Denominator|bottom]] by $a$
+}}
+{{eqn | r = \int \frac {\left({a x + b - b}\right) \ \mathrm d x} {a \left({a x + b}\right)^2}
+      | c = adding and subtracting $b$
+}}
+{{eqn | r = \frac 1 a \int \frac {\left({a x + b}\right) \ \mathrm d x} {\left({a x + b}\right)^2} - \frac b a \int \frac {\mathrm d x} {\left({a x + b}\right)^2}
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \frac 1 a \int \frac {\mathrm d x} {a x + b} - \frac b a \int \frac {\mathrm d x} {\left({a x + b}\right)^2}
+      | c = simplification
+}}
+{{eqn | r = \frac 1 a \left({\frac 1 a \ln \left\vert{a x + b}\right\vert}\right) - \frac b a \int \frac {\mathrm d x} {\left({a x + b}\right)^2} + C
+      | c = [[Primitive of Reciprocal of a x + b|Primitive of Reciprocal of $\dfrac 1 {\left({a x + b}\right)}$]]
+}}
+{{eqn | r = \frac 1 a \left({\frac 1 a \ln \left\vert{a x + b}\right\vert}\right) - \frac b a \left({-\frac 1 {a \left({a x + b}\right)} }\right) + C
+      | c = [[Primitive of Reciprocal of a x + b squared|Primitive of Reciprocal of $\dfrac 1 {\left({a x + b}\right)^2}$]]
+}}
+{{eqn | r = \frac b {a^2 \left({a x + b}\right)} + \frac 1 {a^2} \ln \left\vert{a x + b}\right\vert + C
+      | c = simplification
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of x squared over a x + b squared}
+Tags: Primitive of x squared over a x + b squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {x^2 \rd x} {\paren {a x + b}^2} = \frac {a x + b} {a^3} - \frac {b^2} {a^3 \paren {a x + b} } - \frac {2 b} {a^3} \ln \size {a x + b} + C$
+\end{theorem}
+
+\begin{proof}
+Put $u = a x + b$.
+Then:
+{{begin-eqn}}
+{{eqn | l = x
+      | r = \frac {u - b} a
+      | c = 
+}}
+{{eqn | l = \frac {\d u} {\d x}
+      | r = \frac 1 a
+      | c = 
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = \int \frac {x^2 \rd x} {\paren {a x + b}^2}
+      | r = \int \frac 1 a \paren {\frac {u - b} a}^2 \frac 1 {u^2} \rd u
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \int \frac 1 {a^3} \paren {1 - \frac {2 b} u + \frac {b^2} {u^2} } \rd u
+      | c = [[Square of Difference]], and simplification
+}}
+{{eqn | r = \frac 1 {a^3} \int \d u - \frac {2 b} {a^3} \int \frac {\d u} u + \frac {b^2} {a^3} \int \frac {\d u} {u^2}
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \frac u {a^3} - \frac {2 b} {a^3} \int \frac {\d u} u + \frac {b^2} {a^3} \int \frac {\d u} {u^2} + C
+      | c = [[Primitive of Constant]]
+}}
+{{eqn | r = \frac u {a^3} - \frac {2 b} {a^3} \ln \size u + \frac {b^2} {a^3} \int \frac {\d u} {u^2} + C
+      | c = [[Primitive of Reciprocal]]
+}}
+{{eqn | r = \frac u {a^3} - \frac {2 b} {a^3} \ln \size u + \frac {b^2} {a^3} \frac {-1} u + C
+      | c = [[Primitive of Power]]
+}}
+{{eqn | r = \frac {a x + b} {a^3} - \frac {b^2} {a^3 \paren {a x + b} } - \frac {2 b} {a^3} \ln \size {a x + b} + C
+      | c = substituting for $u$ and rearranging
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int \frac {x^2 \ \mathrm d x} {\left({a x + b}\right)^2}
+      | r = \int \frac {a x^2 \ \mathrm d x} {a \left({a x + b}\right)^2}
+      | c = multiplying [[Definition:Numerator|top]] and [[Definition:Denominator|bottom]] by $a$
+}}
+{{eqn | r = \int \frac {x \left({a x + b - b}\right) \ \mathrm d x} {a \left({a x + b}\right)^2}
+      | c = adding and subtracting $b x$
+}}
+{{eqn | r = \frac 1 a \int \frac {x \left({a x + b}\right) \ \mathrm d x} {\left({a x + b}\right)^2} - \frac b a \int \frac {x \ \mathrm d x} {\left({a x + b}\right)^2}
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \frac 1 a \int \frac {x \ \mathrm d x} {a x + b} - \frac b a \int \frac {x \ \mathrm d x} {\left({a x + b}\right)^2}
+      | c = simplification
+}}
+{{eqn | r = \frac 1 a \left({\frac x a - \frac b {a^2} \ln \left\vert{a x + b}\right\vert}\right) - \frac b a \int \frac {\mathrm d x} {\left({a x + b}\right)^2} + C
+      | c = [[Primitive of x over a x + b|Primitive of Reciprocal of $\dfrac x {\left({a x + b}\right)}$]]
+}}
+{{eqn | r = \frac 1 a \left({\frac x a - \frac b {a^2} \ln \left\vert{a x + b}\right\vert}\right) - \frac b a \left({\frac b {a^2 \left({a x + b}\right)} + \frac 1 {a^2} \left\vert{a x + b}\right\vert }\right) + C
+      | c = [[Primitive of x over a x + b squared|Primitive of Reciprocal of $\dfrac x {\left({a x + b}\right)^2}$]]
+}}
+{{eqn | r = \frac x {a^2} - \frac b {a^3} \ln \left\vert{a x + b}\right\vert - \frac {b^2} {a^3 \left({a x + b}\right)} - \frac b {a^3} \ln \left\vert{a x + b}\right\vert + C
+      | c = multiplying out
+}}
+{{eqn | r = \frac x {a^2} - \frac {b^2} {a^3 \left({a x + b}\right)} - \frac {2 b} {a^3} \ln \left\vert{a x + b}\right\vert + C
+      | c = gathering terms
+}}
+{{eqn | r = \frac {a x} {a^3} + \frac b {a^3} - \frac {b^2} {a^3 \left({a x + b}\right)} - \frac {2 b} {a^3} \ln \left\vert{a x + b}\right\vert + C
+      | c = where $\dfrac b {a^3}$ is subsumed into $C$
+}}
+{{eqn | r = \frac {a x + b} {a^3} - \frac {b^2} {a^3 \left({a x + b}\right)} - \frac {2 b} {a^3} \ln \left\vert{a x + b}\right\vert + C
+      | c = simplification
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of x cubed over a x + b squared}
+Tags: Primitives involving a x + b
+
+\begin{theorem}
+:$\displaystyle \int \frac {x^3 \ \mathrm d x} {\left({a x + b}\right)^2} = \frac {\left({a x + b}\right)^2} {2 a^4} - \frac {3 b \left({a x + b}\right)} {a^4} + \frac {b^3} {a^4 \left({a x + b}\right)} + \frac {3 b^2} {a^4} \ln \left\vert{a x + b}\right\vert + C$
+\end{theorem}
+
+\begin{proof}
+Put $u = a x + b$.
+Then:
+{{begin-eqn}}
+{{eqn | l = x
+      | r = \frac {u - b} a
+      | c = 
+}}
+{{eqn | l = \frac {\mathrm d u} {\mathrm d x}
+      | r = \frac 1 a
+      | c = 
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l =  \int \frac {x^3 \ \mathrm d x} {\left({a x + b}\right)^2}
+      | r = \int \frac 1 a \left({\frac {u - b} a}\right)^3 \frac 1 {u^2} \ \mathrm d u
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \int \frac 1 {a^4} \left({u - 3 b + \frac {3 b^2} u - \frac {b^3} {u^2} }\right) \ \mathrm d u
+      | c = multiplying out and simplification
+}}
+{{eqn | r = \frac 1 {a^4} \int u \ \mathrm d u - \frac {3 b} {a^4} \int \mathrm d u + \frac {3 b^2} {a^4} \int \frac {\mathrm d u} u - \frac {b^3} {a^4} \int \frac {\mathrm d u} {u^2}
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \frac {u^2} {2 a^4} - \frac {3 b} {a^4} \int \mathrm d u + \frac {3 b^2} {a^4} \int \frac {\mathrm d u} u - \frac {b^3} {a^4} \frac {-1} u + C
+      | c = [[Primitive of Power]]
+}}
+{{eqn | r = \frac {u^2} {2 a^4} - \frac {3 b u} {a^4} + \frac {3 b^2} {a^4} \int \frac {\mathrm d u} u + \frac {b^3} {a^4 u} + C
+      | c = [[Primitive of Constant]]
+}}
+{{eqn | r = \frac {u^2} {2 a^4} - \frac {3 b u} {a^4} + \frac {3 b^2} {a^4} \ln \left\vert{u}\right\vert + \frac {b^3} {a^4 u} + C
+      | c = [[Primitive of Reciprocal]]
+}}
+{{eqn | r = \frac {\left({a x + b}\right)^2} {2 a^4} -  \frac {3 b \left({a x + b}\right)} {a^4} + \frac {b^3} {a^4 \left({a x + b}\right)} + \frac {3 b^2} {a^4} \ln \left\vert{a x + b}\right\vert + C
+      | c = substituting for $u$ and rearranging
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Reciprocal of x by a x + b squared}
+Tags: Primitive of Reciprocal of x by a x + b squared, Primitives involving Reciprocals
+
+\begin{theorem}
+:$\displaystyle \int \frac {\d x} {x \paren {a x + b}^2} = \frac 1 {b \paren {a x + b} } + \frac 1 {b^2} \ln \size {\frac x {a x + b} } + C$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int \frac {\mathrm d x} {x \left({a x + b}\right)^2}
+      | r = \int \left({\frac 1 {b^2 x} - \frac a {b^2 \left({a x + b}\right)} - \frac a {b \left({a x + b}\right)^2} }\right) \ \mathrm d x
+      | c = [[Primitive of Reciprocal of x by a x + b squared/Partial Fraction Expansion|Partial Fraction Expansion]]
+}}
+{{eqn | r = \frac 1 {b^2} \int \frac {\mathrm d x} x - \frac a {b^2} \int \frac {\mathrm d x} {a x + b} - \frac a b \int \frac {\mathrm d x} {\left({a x + b}\right)^2}
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \frac 1 {b^2} \ln \left\vert{x}\right\vert - \frac a {b^2} \int \frac {\mathrm d x} {a x + b} - \frac a b \int \frac {\mathrm d x} {\left({a x + b}\right)^2} + C
+      | c = [[Primitive of Reciprocal]]
+}}
+{{eqn | r = \frac 1 {b^2} \ln \left\vert{x}\right\vert - \frac a {b^2} \ln \left\vert{a x + b}\right\vert - \frac a b \int \frac {\mathrm d x} {\left({a x + b}\right)^2} + C
+      | c = [[Primitive of Reciprocal of a x + b|Primitive of $\dfrac 1 {a x + b}$]]
+}}
+{{eqn | r = \frac 1 {b^2} \ln \left\vert{x}\right\vert - \frac a {b^2} \ln \left\vert{a x + b}\right\vert - \frac a b \frac {-1} {a \left({a x + b}\right)} + C
+      | c = [[Primitive of Reciprocal of a x + b squared|Primitive of $\dfrac 1 {\left({a x + b}\right)^2}$]]
+}}
+{{eqn | r = \frac 1 {b \left({a x + b}\right)} + \frac 1 {b^2} \ln \left\vert{\frac x {a x + b} }\right\vert + C
+      | c = [[Difference of Logarithms]] and rearranging
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int \frac {\d x} {x \paren {a x + b}^2}
+      | r = \int \frac {b \rd x} {b x \paren {a x + b}^2}
+      | c = multiplying [[Definition:Numerator|top]] and [[Definition:Denominator|bottom]] by $b$
+}}
+{{eqn | r = \int \frac {\paren {a x + b - a x} \rd x} {b x \paren {a x + b}^2}
+      | c = adding and subtracting $a x$
+}}
+{{eqn | r = \frac 1 b \int \frac {\paren {a x + b} \rd x} {x \paren {a x + b}^2} - \frac a b \int \frac {x \rd x} {x \paren {a x + b}^2}
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \frac 1 b \int \frac {\d x} {x \paren {a x + b} } - \frac a b \int \frac {\d x} {\paren {a x + b}^2}
+      | c = simplifying
+}}
+{{eqn | r = \frac 1 b \paren {\frac 1 b \ln \size {\frac x {a x + b} } } - \frac a b \int \frac {\d x} {\paren {a x + b}^2} + C
+      | c = [[Primitive of Reciprocal of x by a x + b|Primitive of $\dfrac 1 {x \paren {a x + b} }$]]
+}}
+{{eqn | r = \frac 1 {b^2} \ln \size {\frac x {a x + b} } - \frac a b \paren {-\frac 1 {a \paren {a x + b} } } + C
+      | c = [[Primitive of Reciprocal of a x + b squared|Primitive of $\dfrac 1 {\paren {a x + b}^2}$]]
+}}
+{{eqn | r = \frac 1 {b \paren {a x + b} } + \frac 1 {b^2} \ln \size {\frac x {a x + b} } + C
+      | c = simplifying
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Reciprocal of x squared by a x + b squared}
+Tags: Primitives involving a x + b, Primitives involving Reciprocals
+
+\begin{theorem}
+:$\displaystyle \int \frac {\mathrm d x} {x^2 \left({a x + b}\right)^2} = \frac {-a} {b^2 \left({a x + b}\right)} - \frac 1 {b^2 x} + \frac {2 a} {b^3} \ln \left\vert{\frac {a x + b} x}\right\vert + C$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int \frac {\mathrm d x} {x^2 \left({a x + b}\right)^2}
+      | r = \int \left({-\frac {2 a} {b^3 x} + \frac 1 {b^2 x^2} + \frac {2 a^2} {b^3 \left({a x + b}\right)} + \frac {a^2} {b^2 \left({a x + b}\right)^2} }\right) \ \mathrm d x
+      | c = [[Primitive of Reciprocal of x squared by a x + b squared/Partial Fraction Expansion|Partial Fraction Expansion]]
+}}
+{{eqn | r = -\frac {2 a} {b^3} \int \frac {\mathrm d x} x + \frac 1 {b^2} \int \frac {\mathrm d x} {x^2} + \frac {2 a^2} {b^3} \int \frac {\mathrm d x} {\left({a x + b}\right)} + \frac {a^2} {b^2} \int \frac {\mathrm d x} {\left({a x + b}\right)^2}
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = -\frac {2 a} {b^3} \int \frac {\mathrm d x} x + \frac 1 {b^2} \frac {-1} x + \frac {2 a^2} {b^3} \int \frac {\mathrm d x} {\left({a x + b}\right)} + \frac {a^2} {b^2} \int \frac {\mathrm d x} {\left({a x + b}\right)^2} + C
+      | c = [[Primitive of Power]]
+}}
+{{eqn | r = -\frac {2 a} {b^3} \ln \left\vert{x}\right\vert + \frac 1 {b^2} \frac {-1} x + \frac {2 a^2} {b^3} \int \frac {\mathrm d x} {\left({a x + b}\right)} + \frac {a^2} {b^2} \int \frac {\mathrm d x} {\left({a x + b}\right)^2} + C
+      | c = [[Primitive of Reciprocal|Primitive of $\dfrac 1 x$]]
+}}
+{{eqn | r = -\frac {2 a} {b^3} \ln \left\vert{x}\right\vert - \frac 1 {b^2 x} + \frac {2 a^2} {a b^3} \ln \left\vert{a x + b}\right\vert + \frac {a^2} {b^2} \int \frac {\mathrm d x} {\left({a x + b}\right)^2} + C
+      | c = [[Primitive of Reciprocal of a x + b|Primitive of $\dfrac 1 {a x + b}$]]
+}}
+{{eqn | r = -\frac {2 a} {b^3} \ln \left\vert{x}\right\vert - \frac 1 {b^2 x} + \frac {2 a} {b^3} \ln \left\vert{a x + b}\right\vert + \frac {a^2} {b^2} \frac {-1} {a \left({a x + b}\right)} + C
+      | c = [[Primitive of Reciprocal of a x + b squared|Primitive of $\dfrac 1 {\left({a x + b}\right)^2}$]]
+}}
+{{eqn | r = \frac {-a} {b^2 \left({a x + b}\right)} - \frac 1 {b^2 x} + \frac {2 a} {b^3} \ln \left\vert{\frac {a x + b} x}\right\vert + C
+      | c = [[Difference of Logarithms]] and rearranging
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Reciprocal of x cubed by a x + b squared}
+Tags: Primitives involving a x + b, Primitives involving Reciprocals
+
+\begin{theorem}
+:$\displaystyle \int \frac {\mathrm d x} {x^3 \left({a x + b}\right)^2} = - \frac {\left({a x + b}\right)^2} {2 b^4 x^2} + \frac {3 a \left({a x + b}\right)} {b^4 x} - \frac {a^3 x} {b^4 \left({a x + b}\right)} + \frac {3 a^2} {b^4} \ln \left\vert{\frac x {a x + b} }\right\vert + C$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int \frac {\mathrm d x} {x^3 \left({a x + b}\right)^2}
+      | r = \int \left({\frac {3 a^2} {b^4 x} + \frac {-2 a} {b^3 x^2} + \frac 1 {b^2 x^3} + \frac {-3 a^3} {b^4 \left({a x + b}\right)} + \frac {-a^3} {b^3 \left({a x + b}\right)^2} }\right) \ \mathrm d x
+      | c = [[Primitive of Reciprocal of x cubed by a x + b squared/Partial Fraction Expansion|Partial Fraction Expansion]]
+}}
+{{eqn | r = \frac {3 a^2} {b^4} \int \frac {\mathrm d x} x + \frac {-2 a} {b^3} \int \frac {\mathrm d x} {x^2} + \frac 1 {b^2} \int \frac {\mathrm d x} {x^3} + \frac {-3 a^3} {b^4} \int \frac {\mathrm d x} {a x + b} + \frac {-a^3} {b^3} \int \frac {\mathrm d x} {\left({a x + b}\right)^2}
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \frac {3 a^2} {b^4} \int \frac {\mathrm d x} x + \frac {-2 a} {b^3} \frac {-1} x + \frac 1 {b^2} \frac {-1} {2 x^2} + \frac {-3 a^3} {b^4} \int \frac {\mathrm d x} {a x + b} + \frac {-a^3} {b^3} \int \frac {\mathrm d x} {\left({a x + b}\right)^2} + C
+      | c = [[Primitive of Power]]
+}}
+{{eqn | r = \frac {3 a^2} {b^4} \ln \left\vert{x}\right\vert + \frac {2 a} {b^3 x} - \frac 1 {2 b^2 x^2} + \frac {-3 a^3} {b^4} \int \frac {\mathrm d x} {a x + b} + \frac {-a^3} {b^3} \int \frac {\mathrm d x} {\left({a x + b}\right)^2} + C
+      | c = [[Primitive of Reciprocal]]
+}}
+{{eqn | r = \frac {3 a^2} {b^4} \ln \left\vert{x}\right\vert + \frac {2 a} {b^3 x} - \frac 1 {2 b^2 x^2} + \frac {-3 a^3} {b^4} \frac 1 a \ln \left\vert{a x + b}\right\vert + \frac {-a^3} {b^3} \int \frac {\mathrm d x} {\left({a x + b}\right)^2} + C
+      | c = [[Primitive of Reciprocal of a x + b]]
+}}
+{{eqn | r = \frac {3 a^2} {b^4} \ln \left\vert{x}\right\vert + \frac {2 a} {b^3 x} - \frac 1 {2 b^2 x^2} + \frac {-3 a^2} {b^4} \ln \left\vert{a x + b}\right\vert + \frac {-a^3} {b^3} \frac {-1} {a \left({a x + b}\right)} + C
+      | c = [[Primitive of Reciprocal of a x + b squared]]
+}}
+{{eqn | r = \frac {2 a} {b^3 x} - \frac 1 {2 b^2 x^2} + \frac {a^2} {b^3 \left({a x + b}\right)} + \frac {3 a^2} {b^4} \ln \left\vert{\frac x {a x + b} }\right\vert + C
+      | c = [[Difference of Logarithms]]
+}}
+{{eqn | r = \frac {2 a} {b^3 x} - \frac 1 {2 b^2 x^2} + \frac {a^2} {b^3 \left({a x + b}\right)} + \frac {3 a^2} {b^4} \ln \left\vert{\frac x {a x + b} }\right\vert + \frac {3 a^2} {2 b^4} + C
+      | c = C is an arbitrary constant
+}}
+{{eqn | r = \frac {4 a b \left({a x + b}\right) x} {2 b^4 \left({a x + b}\right) x^2} - \frac {b^2 \left({a x + b}\right)} {2 b^4 \left({a x + b}\right) x^2} + \frac {2 a^2 b x^2} {2 b^4 \left({a x + b}\right) x^2} + \frac {3 a^2} {b^4} \ln \left\vert{\frac x {a x + b} }\right\vert + \frac {3 a^2 \left({a x + b}\right) x^2} {2 b^4 \left({a x + b}\right) x^2} + C
+      | c = Represent with common denominators
+}}
+{{eqn | r = \frac {4 a b \left({a x + b}\right) x - b^2 \left({a x + b}\right) + 2 a^2 b x^2 + 3 a^2 \left({a x + b}\right) x^2} {2 b^4 \left({a x + b}\right) x^2} + \frac {3 a^2} {b^4} \ln \left\vert{\frac x {a x + b} }\right\vert + C
+      | c = Combine fractions
+}}
+{{eqn | r = \frac {4 a^2 b x^2 + 4 a b^2 x - a b^2 x - b^3 + 2 a^2 b x^2 + 3 a^3 x^3 + 3 a^2 b x^2} {2 b^4 \left({a x + b}\right) x^2} + \frac {3 a^2} {b^4} \ln \left\vert{\frac x {a x + b} }\right\vert + C
+      | c = Expansion
+}}
+{{eqn | r = \frac {6 a \left({a x + b}\right)^2 x - \left({a x + b}\right)^3 - 2 a^3 x^3} {2 b^4 \left({a x + b}\right) x^2} + \frac {3 a^2} {b^4} \ln \left\vert{\frac x {a x + b} }\right\vert + C
+      | c = Factorisation
+}}
+{{eqn | r = \frac {6 a \left({a x + b}\right)^2 x} {2 b^4 \left({a x + b}\right) x^2} - \frac {\left({a x + b}\right)^3} {2 b^4 \left({a x + b}\right) x^2} - \frac {2 a^3 x^3} {2 b^4 \left({a x + b}\right) x^2} + \frac {3 a^2} {b^4} \ln \left\vert{\frac x {a x + b} }\right\vert + C
+      | c = Split into separate fractions
+}}
+{{eqn | r = \frac {3 a \left({a x + b}\right)} {b^4 x} - \frac {\left({a x + b}\right)^2} {2 b^4 x^2} - \frac {a^3 x} {b^4 \left({a x + b}\right)} + \frac {3 a^2} {b^4} \ln \left\vert{\frac x {a x + b} }\right\vert + C
+      | c = Simplify
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitives of Rational Functions involving a x + b squared}
+Tags: Primitives, Primitives of Rational Functions, Primitives involving a x + b
+
+\begin{theorem}
+This page gathers together the [[Definition:Primitive (Calculus)|primitives]] of some [[Definition:Rational Function|rational functions]] involving $\left({a x + b}\right)^2$.
+\end{theorem}<|endoftext|>
+\section{Primitive of Reciprocal of a x + b cubed}
+Tags: Primitive of Reciprocal of a x + b cubed, Primitives involving a x + b, Primitives involving Reciprocals
+
+\begin{theorem}
+:$\displaystyle \int \frac {\d x} {\paren {a x + b}^3} = -\frac 1 {2 a \paren {a x + b}^2} + C$
+\end{theorem}
+
+\begin{proof}
+Let $u = a x + b$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \int \frac {\mathrm d x} {\left({a x + b}\right)^3}
+      | r = \frac 1 a \int \frac {\mathrm d u} {u^3}
+      | c = [[Primitive of Function of a x + b|Primitive of Function of $a x + b$]]
+}}
+{{eqn | r = \frac 1 a \frac {-1} {2 u^2} + C
+      | c = [[Primitive of Power]]
+}}
+{{eqn | r = -\frac 1 {2 a \left({a x + b}\right)^2} + C
+      | c = substituting for $u$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+From [[Primitive of Power of a x + b|Primitive of Power of $a x + b$]]:
+:$\displaystyle \int \left({a x + b}\right)^n \ \mathrm d x = \frac {\left({a x + b}\right)^{n + 1} } {\left({n + 1}\right) a} + C$
+where $n \ne 1$.
+The result follows by setting $n = -3$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of x over a x + b cubed}
+Tags: Primitives involving a x + b
+
+\begin{theorem}
+:$\displaystyle \int \frac {x \ \mathrm d x} {\left({a x + b}\right)^3} = \frac {-1} {a^2 \left({a x + b}\right)} + \frac b {2 a^2 \left({a x + b}\right)^2} + C$
+\end{theorem}
+
+\begin{proof}
+Put $u = a x + b$.
+Then:
+{{begin-eqn}}
+{{eqn | l = x
+      | r = \frac {u - b} a
+      | c = 
+}}
+{{eqn | l = \frac {\mathrm d u} {\mathrm d x}
+      | r = \frac 1 a
+      | c = 
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = \int \frac {x \ \mathrm d x} {\left({a x + b}\right)^3}
+      | r = \int \frac 1 a \frac {u - b} {a u^3} \ \mathrm d u
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \frac 1 {a^2} \int \frac {\mathrm d u} {u^2} - \frac b {a^2} \int \frac {\mathrm d u} {u^3}
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \frac 1 {a^2} \frac {-1} {u} - \frac b {a^2} \frac {-1} {2 u^2} + C
+      | c = [[Primitive of Power]]
+}}
+{{eqn | r = \frac {-1} {a^2 \left({a x + b}\right)} + \frac b {2 a^2 \left({a x + b}\right)^2} + C
+      | c = substituting for $u$ and rearranging
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+From [[Primitive of x by Power of a x + b|Primitive of $x$ by Power of $a x + b$]]:
+:$\displaystyle \int x \left({a x + b}\right)^n \ \mathrm d x = \frac {\left({a x + b}\right)^{n + 2} } {\left({n + 2}\right) a^2} - \frac {b \left({a x + b}\right)^{n + 1} } {\left({n + 1}\right) a^2} + C$
+where $n \ne - 1$ and $n \ne - 2$.
+The result follows by setting $n = -3$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of x squared over a x + b cubed}
+Tags: Primitives involving a x + b
+
+\begin{theorem}
+:$\ds \int \frac {x^2 \rd x} {\paren {a x + b}^3} = \frac {2 b} {a^3 \paren {a x + b} } - \frac {b^2} {2 a^3 \paren {a x + b}^2} + \frac 1 {a^3} \ln \size {a x + b} + C$
+\end{theorem}
+
+\begin{proof}
+Put $u = a x + b$.
+Then:
+{{begin-eqn}}
+{{eqn | l = x
+      | r = \frac {u - b} a
+      | c = 
+}}
+{{eqn | l = \frac {\d u} {\d x}
+      | r = \frac 1 a
+      | c = 
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = \int \frac {x^2 \rd x} {\paren {a x + b}^3}
+      | r = \int \frac 1 a \paren {\frac {u - b} a}^2 \frac 1 {u^2} \rd u
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \int \frac 1 {a^3} \paren {\frac 1 u - \frac {2 b} {u^2} + \frac {b^2} {u^3} } \rd u
+      | c = [[Square of Difference]], and simplification
+}}
+{{eqn | r = \frac 1 {a^3} \int \frac {\d u} u - \frac {2 b} {a^3} \int \frac {\d u} {u^2} + \frac {b^2} {a^3} \int \frac {\d u} {u^3}
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \frac 1 {a^3} \ln \size u - \frac {2 b} {a^3} \int \frac {\d u} {u^2} + \frac {b^2} {a^3} \int \frac {\d u} {u^3} + C
+      | c = [[Primitive of Reciprocal]]
+}}
+{{eqn | r = \frac 1 {a^3} \ln \size u - \frac {2 b} {a^3} \frac {-1} u + \frac {b^2} {a^3} \frac {-1} {2 u^2} + C
+      | c = [[Primitive of Power]]
+}}
+{{eqn | r = \frac {2 b} {a^3 \paren {a x + b} } - \frac {b^2} {2 a^3 \paren {a x + b}^2} + \frac 1 {a^3} \ln \size {a x + b} + C
+      | c = substituting for $u$ and rearranging
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of x cubed over a x + b cubed}
+Tags: Primitives involving a x + b
+
+\begin{theorem}
+:$\displaystyle \int \frac {x^3 \rd x} {\paren {a x + b}^3} = \frac x {a^3} - \frac {3 b^2} {a^4 \paren {a x + b} } + \frac {b^3} {2 a^4 \paren {a x + b}^2} - \frac {3 b} {a^4} \ln \size {a x + b} + C$
+\end{theorem}
+
+\begin{proof}
+Put $u = a x + b$.
+Then:
+{{begin-eqn}}
+{{eqn | l = x
+      | r = \frac {u - b} a
+      | c = 
+}}
+{{eqn | l = \frac {\d u} {\d x}
+      | r = \frac 1 a
+      | c = 
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l =  \int \frac {x^3 \rd x} {\paren {a x + b}^3}
+      | r = \int \frac 1 a \paren {\frac {u - b} a}^3 \frac 1 {u^3} \rd u
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \int \frac 1 {a^4} \paren {1 - \frac {3 b} u + \frac {3 b^2} {u^2} - \frac {b^3} {u^3} } \rd u
+      | c = multiplying out and simplification
+}}
+{{eqn | r = \frac 1 {a^4} \int \d u - \frac {3 b} {a^4} \int \frac {\d u} u + \frac {3 b^2} {a^4} \int \frac {\d u} {u^2} - \frac {b^3} {a^4} \int \frac {\d u} {u^3}
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \frac u {a^4} - \frac {3 b} {a^4} \int \frac {\d u} u + \frac {3 b^2} {a^4} \int \frac {\d u} {u^2} - \frac {b^3} {a^4} \int \frac {\d u} {u^3} + C
+      | c = [[Primitive of Constant]]
+}}
+{{eqn | r = \frac u {a^4} - \frac {3 b} {a^4} \int \frac {\d u} u + \frac {3 b^2} {a^4} \frac {-1} u - \frac {b^3} {a^4} \frac {-1} {2 u^2} + C
+      | c = [[Primitive of Power]]
+}}
+{{eqn | r = \frac u {a^4} - \frac {3 b} {a^4} \ln \size u - \frac {3 b^2} {a^4 u} + \frac {b^3} {2 a^4 u^2} + C
+      | c = [[Primitive of Reciprocal]]
+}}
+{{eqn | r = \frac {a x + b} {a^4} - \frac {3 b^2} {a^4 \paren {a x + b} )} + \frac {b^3} {2 a^4 \paren {a x + b}^2} - \frac {3 b} {a^4} \ln \size {a x + b} + C
+      | c = substituting for $u$ and rearranging
+}}
+{{eqn | r = \frac x {a^3} + \frac b {a^4} - \frac {3 b^2} {a^4 \paren {a x + b} } + \frac {b^3} {2 a^4 \paren {a x + b}^2} - \frac {3 b} {a^4} \ln \size {a x + b} + C
+      | c = separating first term
+}}
+{{eqn | r = \frac x {a^3} - \frac {3 b^2} {a^4 \paren {a x + b} } + \frac {b^3} {2 a^4 \paren {a x + b}^2} - \frac {3 b} {a^4} \ln \size {a x + b} + C
+      | c = subsuming $\dfrac b {a^4}$ into the [[Definition:Arbitrary Constant (Calculus)|arbitrary constant]] $C$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Reciprocal of x by a x + b cubed}
+Tags: Primitives involving a x + b
+
+\begin{theorem}
+:$\displaystyle \int \frac {\mathrm d x} {x \left({a x + b}\right)^3} = \frac {a^2 x^2} {2 b^3 \left({a x + b}\right)^2} - \frac {2 a x} {b^3 \left({a x + b}\right)} + \frac 1 {b^3} \ln \left\vert{\frac x {a x + b} }\right\vert + C$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int \frac {\mathrm d x} {x \left({a x + b}\right)^3}
+      | r = \int \left({\frac 1 {b^3 x} - \frac a {b^3 \left({a x + b}\right)} - \frac a {b^2 \left({a x + b}\right)^2} - \frac a {b \left({a x + b}\right)^3} }\right) \ \mathrm d x
+      | c = [[Primitive of Reciprocal of x by a x + b cubed/Partial Fraction Expansion|Partial Fraction Expansion]]
+}}
+{{eqn | r = \frac 1 {b^3} \int \frac {\mathrm d x} x - \frac a {b^3} \int \frac {\mathrm d x} {a x + b} - \frac a {b^2} \int \frac {\mathrm d x} {\left({a x + b}\right)^2} - \frac a b \int \frac {\mathrm d x} {\left({a x + b}\right)^3}
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \frac 1 {b^3} \ln \left\vert{x}\right\vert - \frac a {b^3} \int \frac {\mathrm d x} {a x + b} - \frac a {b^2} \int \frac {\mathrm d x} {\left({a x + b}\right)^2} - \frac a b \int \frac {\mathrm d x} {\left({a x + b}\right)^3} + C
+      | c = [[Primitive of Reciprocal]]
+}}
+{{eqn | r = \frac 1 {b^3} \ln \left\vert{x}\right\vert - \frac 1 {b^3} \ln \left\vert{a x + b}\right\vert - \frac a {b^2} \int \frac {\mathrm d x} {\left({a x + b}\right)^2} - \frac a b \int \frac {\mathrm d x} {\left({a x + b}\right)^3} + C
+      | c = [[Primitive of Reciprocal of a x + b]]
+}}
+{{eqn | r = \frac 1 {b^3} \ln \left\vert{x}\right\vert - \frac 1 {b^3} \ln \left\vert{a x + b}\right\vert - \frac a {b^2} \frac {-1} {a \left({a x + b}\right)} - \frac a b \int \frac {\mathrm d x} {\left({a x + b}\right)^3} + C
+      | c = [[Primitive of Reciprocal of a x + b squared]]
+}}
+{{eqn | r = \frac 1 {b^3} \ln \left\vert{x}\right\vert - \frac 1 {b^3} \ln \left\vert{a x + b}\right\vert - \frac a {b^2} \frac {-1} {a \left({a x + b}\right)} - \frac a b \frac {-1} {2 a \left({a x + b}\right)^2} + C
+      | c = [[Primitive of Reciprocal of a x + b cubed]]
+}}
+{{eqn | r = \frac 1 {b^2 \left({a x + b}\right)} + \frac 1 {2 b \left({a x + b}\right)^2} + \frac 1 {b^3} \ln \left\vert{\frac  x {a x + b} }\right\vert + C
+      | c = [[Difference of Logarithms]] and rearranging
+}}
+{{eqn | r = \frac {b} {b^3 \left({a x + b}\right)} + \frac {b^2} {2 b^3 \left({a x + b}\right)^2} + \frac 1 {b^3} \ln \left\vert{\frac  x {a x + b} }\right\vert + C
+      | c = 
+}}
+{{eqn | r = \left({ \frac {a x + b} {b^3 \left({a x + b}\right)} - \frac {a x} {b^3 \left({a x + b}\right)} }\right) + \left({ \frac {a^2 x^2 + 2 a b x + b^2} {2 b^3 \left({a x + b}\right)^2} - \frac {a^2 x^2 + 2 a b x} {2 b^3 \left({a x + b}\right)^2} }\right) + \frac 1 {b^3} \ln \left\vert{\frac  x {a x + b} }\right\vert + C
+      | c = 
+}}
+{{eqn | r = \frac {a x + b} {b^3 \left({a x + b}\right)} - \frac {a x} {b^3 \left({a x + b}\right)} + \frac {\left({a x + b}\right)^2} {2 b^3 \left({a x + b}\right)^2} - \frac {a^2 x^2 + 2 a b x} {2 b^3 \left({a x + b}\right)^2} + \frac 1 {b^3} \ln \left\vert{\frac  x {a x + b} }\right\vert + C
+      | c = 
+}}
+{{eqn | r = \frac 1 {b^3} - \frac {a x} {b^3 \left({a x + b}\right)} + \frac 1 {2 b^3} - \frac {a^2 x^2 + 2 a b x} {2 b^3 \left({a x + b}\right)^2} + \frac 1 {b^3} \ln \left\vert{\frac  x {a x + b} }\right\vert + C
+      | c = 
+}}
+{{eqn | r = - \frac {a x} {b^3 \left({a x + b}\right)} - \frac {a^2 x^2 + 2 a b x} {2 b^3 \left({a x + b}\right)^2} + \frac 1 {b^3} \ln \left\vert{\frac  x {a x + b} }\right\vert + C + \frac 3 {2 b^3}
+      | c = 
+}}
+{{eqn | r = - \frac {a x} {b^3 \left({a x + b}\right)} - \frac {a^2 x^2} {2 b^3 \left({a x + b}\right)^2} - \frac {2 a b x} {2 b^3 \left({a x + b}\right)^2} + \frac 1 {b^3} \ln \left\vert{\frac  x {a x + b} }\right\vert + C + \frac 3 {2 b^3}
+      | c = 
+}}
+{{eqn | r = - \frac {a x} {b^3 \left({a x + b}\right)} - \frac {a^2 x^2} {2 b^3 \left({a x + b}\right)^2} - \frac {2 a x \left({a x + b}\right)} {2 b^3 \left({a x + b}\right)^2} + \frac {2 a^2 x^2} {2 b^3 \left({a x + b}\right)^2} + \frac 1 {b^3} \ln \left\vert{\frac  x {a x + b} }\right\vert + C + \frac 3 {2 b^3}
+      | c = 
+}}
+{{eqn | r = - \frac {a x} {b^3 \left({a x + b}\right)} - \frac {a^2 x^2} {2 b^3 \left({a x + b}\right)^2} - \frac {2 a x} {2 b^3 \left({a x + b}\right)} + \frac {2 a^2 x^2} {2 b^3 \left({a x + b}\right)^2} + \frac 1 {b^3} \ln \left\vert{\frac  x {a x + b} }\right\vert + C + \frac 3 {2 b^3}
+      | c = 
+}}
+{{eqn | r = - \frac {a x} {b^3 \left({a x + b}\right)} + \frac {a^2 x^2} {2 b^3 \left({a x + b}\right)^2} - \frac {a x} {b^3 \left({a x + b}\right)} + \frac 1 {b^3} \ln \left\vert{\frac  x {a x + b} }\right\vert + C + \frac 3 {2 b^3}
+      | c = 
+}}
+{{eqn | r = \frac {a^2 x^2} {2 b^3 \left({a x + b}\right)^2} - \frac {2 a x} {b^3 \left({a x + b}\right)} + \frac 1 {b^3} \ln \left\vert{\frac  x {a x + b} }\right\vert + C + \frac 3 {2 b^3}
+      | c = 
+}}
+{{eqn | r = \frac {a^2 x^2} {2 b^3 \left({a x + b}\right)^2} - \frac {2 a x} {b^3 \left({a x + b}\right)} + \frac 1 {b^3} \ln \left\vert{\frac  x {a x + b} }\right\vert + C
+      | c = $C + \frac 3 {2 b^3}$ is constant
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Reciprocal of x squared by a x + b cubed}
+Tags: Primitives involving a x + b, Primitives involving Reciprocals
+
+\begin{theorem}
+:$\displaystyle \int \frac {\mathrm d x} {x^2 \left({a x + b}\right)^3} = \frac {-a} {2 b^2 \left({a x + b}\right)^2} - \frac {2 a} {b^3 \left({a x + b}\right)} - \frac 1 {b^3 x} + \frac {3 a} {b^4} \ln \left\vert{\frac {a x + b} x}\right\vert + C$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int \frac {\mathrm d x} {x^2 \left({a x + b}\right)^3}
+      | r = \int \left({\frac {-3 a} {b^4 x} + \frac 1 {b^3 x^2} + \frac {3 a^2} {b^4 \left({a x + b}\right)} + \frac {2 a^2} {b^3 \left({a x + b}\right)^2} + \frac {a^2} {b^2 \left({a x + b}\right)^3} }\right) \ \mathrm d x
+      | c = [[Primitive of Reciprocal of x squared by a x + b cubed/Partial Fraction Expansion|Partial Fraction Expansion]]
+}}
+{{eqn | r = \frac {-3 a} {b^4} \int \frac {\mathrm d x} x + \frac 1 {b^3} \int \frac {\mathrm d x} {x^2} + \frac {3 a^2} {b^4} \int \frac {\mathrm d x} {\left({a x + b}\right)} + \frac {2 a^2} {b^3} \int \frac {\mathrm d x} {\left({a x + b}\right)^2} + \frac {a^2} {b^2} \int \frac {\mathrm d x} {\left({a x + b}\right)^3}
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \frac {-3 a} {b^4} \ln \left\vert{x}\right\vert + \frac 1 {b^3} \int \frac {\mathrm d x} {x^2} + \frac {3 a^2} {b^4} \int \frac {\mathrm d x} {\left({a x + b}\right)} + \frac {2 a^2} {b^3} \int \frac {\mathrm d x} {\left({a x + b}\right)^2} + \frac {a^2} {b^2} \int \frac {\mathrm d x} {\left({a x + b}\right)^3} + C
+      | c = [[Primitive of Reciprocal]]
+}}
+{{eqn | r = \frac {-3 a} {b^4} \ln \left\vert{x}\right\vert + \frac 1 {b^3} \int \frac {\mathrm d x} {x^2} + \frac {3 a^2} {b^4} \ln \left\vert{a x + b}\right\vert + \frac {2 a^2} {b^3} \int \frac {\mathrm d x} {\left({a x + b}\right)^2} + \frac {a^2} {b^2} \int \frac {\mathrm d x} {\left({a x + b}\right)^3} + C
+      | c = [[Primitive of Reciprocal of a x + b]]
+}}
+{{eqn | r = \frac {-3 a} {b^4} \ln \left\vert{x}\right\vert + \frac 1 {b^3} \int \frac {\mathrm d x} {x^2} + \frac {3 a^2} {b^4} \ln \left\vert{a x + b}\right\vert + \frac {2 a^2} {b^3} \frac {-1} {a \left({a x + b}\right)} + \frac {a^2} {b^2} \int \frac {\mathrm d x} {\left({a x + b}\right)^3} + C
+      | c = [[Primitive of Reciprocal of a x + b squared]]
+}}
+{{eqn | r = \frac {-3 a} {b^4} \ln \left\vert{x}\right\vert + \frac 1 {b^3} \int \frac {\mathrm d x} {x^2} + \frac {3 a^2} {b^4} \ln \left\vert{a x + b}\right\vert - \frac {2 a} {b^3 \left({a x + b}\right)} + \frac {a^2} {b^2} \frac {-1} {2 a \left({a x + b}\right)^2} + C
+      | c = [[Primitive of Reciprocal of a x + b cubed]]
+}}
+{{eqn | r = \frac {-3 a} {b^4} \ln \left\vert{x}\right\vert + \frac 1 {b^3} \frac {-1} x + \frac {3 a^2} {b^4} \ln \left\vert{a x + b}\right\vert - \frac {2 a} {b^3 \left({a x + b}\right)} - \frac a {2 b^2 \left({a x + b}\right)^2} + C
+      | c = [[Primitive of Power]]
+}}
+{{eqn | r = \frac {-a} {2 b^2 \left({a x + b}\right)^2} - \frac {2 a} {b^3 \left({a x + b}\right)} - \frac 1 {b^3 x} + \frac {3 a} {b^4} \ln \left\vert{\frac {a x + b} x}\right\vert + C
+      | c = [[Difference of Logarithms]] and rearranging
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Reciprocal of x cubed by a x + b cubed}
+Tags: Primitives involving a x + b
+
+\begin{theorem}
+:$\displaystyle \int \frac {\d x} {x^3 \paren {a x + b}^3} = \frac {a^4 x^2} {2 b^5 \paren {a x + b}^2} - \frac {4 a^3 x} {b^5 \paren {a x + b} } - \frac {\paren {a x + b}^2} {2 b^5 x^2} + \frac {4 a} {b^4 x} + \frac {6 a^2} {b^5} \ln \size {\frac x {a x + b} } + C$
+\end{theorem}
+
+\begin{proof}
+A [[Primitive of Reciprocal of x cubed by a x + b cubed/Partial Fraction Expansion|partial fraction expansion]] of the [[Definition:Integrand|integrand]] gives:
+:$\dfrac 1 {x^3 \paren {a x + b}^3} = \dfrac {6 a^2} {b^5 x} - \dfrac {3 a} {b^4 x^2} + \dfrac 1 {b^3 x^3} - \dfrac {6 a^3} {b^5 \paren {a x + b} } - \dfrac {3 a^3} {b^4 \paren {a x + b}^2} - \dfrac {a^3} {b^3 \paren {a x + b}^3}$
+From [[Linear Combination of Integrals]] it follows that the expression can be rendered as:
+:$\displaystyle \int \frac {\d x} {x^3 \paren {a x + b}^3} = \frac {6 a^2} {b^5} \int \frac {\d x} x - \frac {3 a} {b^4} \int \frac {\d x} {x^2} + \frac 1 {b^3} \int \frac {\d x} {x^3} - \frac {6 a^3} {b^5} \int \frac {\d x} {\paren {a x + b} } - \frac {3 a^3} {b^4} \int \frac {\d x} {\paren {a x + b}^2} - \frac {a^3} {b^3} \int \frac {\d x} {\paren {a x + b}^3}$
+This can be integrated termwise as follows (ignoring for now the [[Definition:Arbitrary Constant (Calculus)|arbitrary constant]]):
+{{begin-eqn}}
+{{eqn | l = \frac {6 a^2} {b^5} \int \frac {\d x} x 
+      | r = \frac {6 a^2} {b^5} \ln \size x
+      | c = [[Primitive of Reciprocal]]
+}}
+{{eqn | l = -\frac {3 a} {b^4} \int \frac {\d x} {x^2}
+      | r = \frac {3 a} {b^4 x}
+      | c = [[Primitive of Power]]
+}}
+{{eqn | l = \frac 1 {b^3} \int \frac {\d x} {x^3}
+      | r = -\frac 1 {2 b^3 x^2}
+      | c = [[Primitive of Power]]
+}}
+{{eqn | l = -\frac {6 a^3} {b^5} \int \frac {\d x} {\paren {a x + b} }
+      | r = -\frac {6 a^2} {b^5} \ln \size {a x + b}
+      | c = [[Primitive of Reciprocal of a x + b|Primitive of $\dfrac 1 {a x + b}$]]
+}}
+{{eqn | l = - \frac {3 a^3} {b^4} \int \frac {\d x} {\paren {a x + b}^2}
+      | r = \frac {3 a^3} {b^4 a \paren {a x + b} }
+      | c = [[Primitive of Reciprocal of a x + b squared|Primitive of $\dfrac 1 {\paren {a x + b}^2}$]]
+}}
+{{eqn | l = -\frac {a^3} {b^3} \int \frac {\d x} {\paren {a x + b}^3}
+      | r = \frac {a^2} {2 b^3 \paren {a x + b}^2}
+      | c = [[Primitive of Reciprocal of a x + b cubed|Primitive of $\dfrac 1 {\paren {a x + b}^3}$]]
+}}
+{{end-eqn}}
+The terms with the [[Definition:Natural Logarithm|logarithm]] can be combined thus:
+{{begin-eqn}}
+{{eqn | l = \frac {6 a^2} {b^5} \ln \size x - \frac {6 a^2} {b^5} \ln \size {a x + b}
+      | r = \frac {6 a^2} {b^5} \ln \size {\frac x {a x + b} }
+      | c = [[Difference of Logarithms]]
+}}
+{{end-eqn}}
+and so the solution to the [[Definition:Primitive (Calculus)|integral]] can be expressed as:
+{{begin-eqn}}
+{{eqn | l = \int \frac {\d x} {x^3 \paren {a x + b}^3}
+      | r = \frac {3 a} {b^4 x} - \frac 1 {2 b^3 x^2} + \frac {3 a^2} {b^4 \paren {a x + b} } + \frac {a^2} {2 b^3 \paren {a x + b}^2} + \frac {6 a^2} {b^5} \ln \size {\frac x {a x + b} } + C
+}}
+{{end-eqn}}
+While this would usually be considered as an acceptable form to leave such an expression, there is some way to go to obtain the result requested.
+So, placing the non-logarithmic terms over a common denominator and rearranging, with a view to our destination:
+{{begin-eqn}}
+{{eqn | r = \frac {3 a} {b^4 x} - \frac 1 {2 b^3 x^2} + \frac {3 a^2} {b^4 \paren {a x + b} } + \frac {a^2} {2 b^3 \paren {a x + b}^2}
+      | o = 
+}}
+{{eqn | r = \frac {6 a x \paren {a x + b}^2} {2 b^4 x^2 \\paren {a x + b}^2} - \frac {b \paren {a x + b}^2} {2 b^4 x^2 \paren {a x + b}^2} + \frac {6 a^2 x^2 \paren {a x + b} } {2 b^4 x^2 \paren {a x + b}^2} + \frac {a^2 b x^2} {2 b^4 x^2 \paren {a x + b}^2}
+}}
+{{eqn | r = \frac {12 a^3 x^3 + 18 a^2 b x^2 + 4 a b^2 x - b^3} {2 b^4 \paren {a x + b}^2 x^2}
+}}
+{{eqn | r = \frac {a^4 x^4 + 16 a^3 b x^3 + 24 a^2 b^2 x^2 + 8 a b^3 x} {2 b^5 \paren {a x + b}^2 x^2} - \frac {a^4 x^4 + 4 a^3 b x^3 + 6 a^2 b^2 x^2 + 4 a b^3 x + b^4} {2 b^5 \paren {a x + b}^2 x^2}
+      | c = 
+}}
+{{eqn | r = \frac {a^4 x^4 + 16 a^3 b x^3 + 24 a^2 b^2 x^2 + 8 a b^3 x} {2 b^5 \paren {a x + b}^2 x^2} - \frac {\paren {a x + b}^4} {2 b^5 \paren {a x + b}^2 x^2}
+      | c = 
+}}
+{{eqn | r = \frac {a^4 x^4 + 16 a^3 b x^3 + 24 a^2 b^2 x^2 + 8 a b^3 x} {2 b^5 \paren {a x + b}^2 x^2} - \frac {\paren {a x + b}^2} {2 b^5 x^2}
+      | c = 
+}}
+{{eqn | r = \frac {a^4 x^2} {2 b^5 \paren {a x + b}^2} + \frac {16 a^3 b x^3 + 24 a^2 b^2 x^2 + 8 a b^3 x} {2 b^5 \paren {a x + b}^2 x^2} - \frac {\paren {a x + b}^2} {2 b^5 x^2}
+      | c = 
+}}
+{{eqn | r = \frac {a^4 x^2} {2 b^5 \paren {a x + b}^2} + \frac {8 a^3 b x^3 + 8 a^2 b^2 x^2} {2 b^5 \paren {a x + b}^2 x^2} + \frac {8 a^3 b x^3 + 16 a^2 b^2 x^2 + 8 a b^3 x} {2 b^5 \paren {a x + b}^2 x^2} - \frac {\paren {a x + b}^2} {2 b^5 x^2}
+      | c = 
+}}
+{{eqn | r = \frac {a^4 x^2} {2 b^5 \paren {a x + b}^2} + \frac {8 a^2 b x^2 \paren {a x + b} } {2 b^5 \paren {a x + b}^2 x^2} + \frac {8 a b x \paren {a x + b}^2} {2 b^5 \paren {a x + b}^2 x^2} - \frac {\paren {a x + b}^2} {2 b^5 x^2}
+      | c = 
+}}
+{{eqn | r = \frac {a^4 x^2} {2 b^5 \paren {a x + b}^2} + \frac {4 a^2} {b^4 \paren {a x + b} } + \frac {4 a} {b^4 x} - \frac {\paren {a x + b}^2} {2 b^5 x^2}
+      | c = 
+}}
+{{eqn | r = \frac {a^4 x^2} {2 b^5 \paren {a x + b}^2} + \frac {4 a^2 b} {b^5 \paren {a x + b} } + \frac {4 a} {b^4 x} - \frac {\paren {a x + b}^2} {2 b^5 x^2}
+      | c = 
+}}
+{{eqn | r = \frac {a^4 x^2} {2 b^5 \paren {a x + b}^2} + \frac {4 a^2 \paren {a x + b} } {b^5 \paren {a x + b} } - \frac {4 a^3 x} {b^5 \paren {a x + b} } + \frac {4 a} {b^4 x} - \frac {\paren {a x + b}^2} {2 b^5 x^2}
+      | c = 
+}}
+{{eqn | r = \frac {a^4 x^2} {2 b^5 \paren {a x + b}^2} + \frac {4 a^2} {b^5} - \frac {4 a^3 x} {b^5 \paren {a x + b} } + \frac {4 a} {b^4 x} - \frac {\paren {a x + b}^2} {2 b^5 x^2}
+      | c = 
+}}
+{{end-eqn}}
+Thus the full solution to the [[Definition:Primitive (Calculus)|integral]] can be assembled as follows, where $\dfrac {4 a^2} {b^5}$ is subsumed into the [[Definition:Arbitrary Constant (Calculus)|arbitrary constant]]:
+{{begin-eqn}}
+{{eqn | l = \int \frac {\d x} {x^3 \paren {a x + b}^3}
+      | r = \frac {a^4 x^2} {2 b^5 \paren {a x + b}^2} - \frac {4 a^3 x} {b^5 \paren {a x + b} } - \frac {\paren {a x + b}^2} {2 b^5 x^2} + \frac {4 a} {b^4 x} + \frac {6 a^2} {b^5} \ln \size {\frac x {a x + b} } + C
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Laplace Transform of Second Derivative}
+Tags: Laplace Transforms of Derivatives
+
+\begin{theorem}
+Let $f: \R \to \R$ or $\R \to \C$ be a [[Definition:Continuous Mapping|continuous]] [[Definition:Function|function]] on any interval of the form $0 \le t \le a$.
+Let $f$ be [[Definition:Second Derivative|twice differentiable]].
+Let $f'$ be [[Definition:Continuous Mapping|continuous]] and $f''$ [[Definition:Piecewise Continuous Function with One-Sided Limits|piecewise continuous with one-sided limits]] on said intervals.
+Let $f$ and $f'$ be of [[Definition:Exponential Order|exponential order]].
+Let $\laptrans f$ denote the [[Definition:Laplace Transform|Laplace transform]] of $f$.
+Then $\laptrans {f''}$ exists for $\map \Re s > a$, and:
+:$\laptrans {\map {f''} t} = s^2 \laptrans {\map f t} - s \, \map f 0 - \map {f'} 0$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \laptrans {\map {f''} t}
+      | r = s \laptrans {\map {f'} t} - \map {f'} 0
+      | c = [[Laplace Transform of Derivative]]
+}}
+{{eqn | r = s \paren {s \laptrans {\map f t} - \map f 0} - \map {f'} 0
+      | c = [[Laplace Transform of Derivative]]
+}}
+{{eqn | r = s^2 \laptrans {\map f t} - s \, \map f 0 - \map {f'} 0
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Euler's Number is Transcendental}
+Tags: Euler's Number, Transcendental Number Theory, Euler's Number is Transcendental
+
+\begin{theorem}
+[[Definition:Euler's Number|Euler's Number]] $e$ is [[Definition:Transcendental|transcendental]].
+\end{theorem}
+
+\begin{proof}
+{{AimForCont}} there exist [[Definition:Integer|integers]] $a_0, \ldots, a_n$ with $a_0 \ne 0$ such that:
+:$(1): \quad a_n e^n + a_{n - 1} e^{n - 1} + \cdots + a_0 = 0$
+Define $M$, $M_1, \ldots, M_n$ and $\epsilon_1, \ldots, \epsilon_n$ as follows:
+{{begin-eqn}}
+{{eqn | l = M
+      | r = \int_0^\infty \frac {x^{p - 1} \sqbrk {\paren {x - 1} \cdots \paren {x - n} }^p e^{-x} } {\paren {p - 1}!} \rd x
+}}
+{{eqn | l = M_k
+      | r = e^k \int_k^\infty \frac {x^{p - 1} \sqbrk {\paren {x - 1} \cdots \paren {x - n} }^p e^{-x} } {\paren {p - 1}!} \rd x
+}}
+{{eqn | l = \epsilon_k
+      | r = e^k \int_0^k \frac {x^{p - 1} \sqbrk {\paren {x - 1} \cdots \paren {x - n} }^p e^{-x} } {\paren {p - 1}!} \rd x
+}}
+{{end-eqn}}
+where $p$ is a [[Definition:Prime Number|prime number]] with $p > n$ and $p > \size {a_0}$.
+The expression $\sqbrk {\paren {x - 1} \cdots \paren {x - n} }^p$ is a [[Definition:Polynomial (Abstract Algebra)|polynomial]] of [[Definition:Degree of Polynomial/Field|degree]] $n p$ with integer coefficients.
+Hence:
+{{begin-eqn}}
+{{eqn | l = M
+      | r = \sum_{\alpha \mathop = 0}^{n p} \frac {C_\alpha} {\paren {p - 1}!} \int_0^\infty x^{p - 1 + \alpha} e^{-x} \rd x
+      | c = 
+}}
+{{eqn | r = \sum_{\alpha \mathop = 0}^{n p} C_\alpha \dfrac {\paren {p - 1 + \alpha}!} {\paren {p - 1}!}
+      | c = [[Gamma Function Extends Factorial]]
+}}
+{{end-eqn}}
+where $C_\alpha$ are integers and $C_0 = \pm \paren {n!}^p$.
+For $\alpha = 0$ we have:
+:$C_0 \dfrac {\paren {p - 1}!} {\paren {p - 1}!} = \pm \paren {n!}^p$
+Since $p > n$, it follows from [[Prime iff Coprime to all Smaller Positive Integers]] and [[Euclid's Lemma]] that this term is not divisible by $p$.
+For $\alpha \ge 1$ we have:
+:$C_\alpha \dfrac {\paren {p - 1 + \alpha}!} {\paren {p - 1}!} = C_\alpha \paren {p - 1 + \alpha} \paren {p - 2 + \alpha} \cdots p$
+which is clearly divisible by $p$.
+It follows from [[Common Divisor Divides Difference]] that $M$ is an integer not [[Definition:Divisor of Integer|divisible]] by $p$.
+We also have:
+{{begin-eqn}}
+{{eqn | l = M_k
+      | r = \int_k^\infty \frac {x^{p - 1} \sqbrk {\paren {x - 1} \cdots \paren {x - n} }^p e^{-\paren {x - k} } } {\paren {p - 1}!} \rd x
+      | c = 
+}}
+{{eqn | r = \int_0^\infty \frac {\paren {x + k}^{p - 1} \sqbrk {\paren {x + k - 1} \cdots \paren {x + k - n} }^p e^{- x} } {\paren {p - 1}!} \rd x
+      | c = [[Integration by Substitution|substituting]] $x - k \mapsto x$
+}}
+{{end-eqn}}
+The expression $\sqbrk {\paren {x + k - 1} \cdots \paren {x + k - n} }$ is divisible by $x$.
+So $\paren {x + k}^{p - 1} \sqbrk {\paren {x + k - 1} \cdots \paren {x + k - n} }^p$ is a polynomial of degree at least $p$ with integer coefficients.
+Hence:
+{{begin-eqn}}
+{{eqn | l = M_k
+      | r = \sum_{\alpha \mathop = 1}^{n p} \frac {D_{\alpha} } {\paren {p - 1}!} \int_0^\infty x^{p - 1 + \alpha} e^{-x} \rd x
+      | c = 
+}}
+{{eqn | r = \sum_{\alpha \mathop = 1}^{n p} D_{\alpha} \frac {\paren {p - 1 + \alpha}!} {\paren {p - 1}!}
+      | c = [[Gamma Function Extends Factorial]]
+}}
+{{end-eqn}}
+where $D_\alpha$ are integers.
+Since this sum begins with $\alpha = 1$, each term is divisible by $p$. 
+Thus each $M_k$ is an integer divisible by $p$.
+By the above definitions we have:
+:$e^k = \dfrac {M_k + \epsilon_k} M$
+Substituting this into $(1)$ and multiplying by $M$ we obtain:
+:$\paren {a_0 M + a_1 M_1 + \cdots + a_n M_n} + \paren {a_1 \epsilon_1 + \cdots + a_n \epsilon_n} = 0$
+Since $p > \size {a_0}$, it follows from [[Prime iff Coprime to all Smaller Positive Integers]] that $p$ does not divide $a_0$.
+So by [[Euclid's Lemma]], $a_0 M$ is not divisible by $p$.
+Since each $M_k$ is divisible by $p$, it follows from [[Common Divisor Divides Difference]] that $a_0 M + a_1 M_1 + \cdots + a_n M_n$ is not divisible by $p$.
+Therefore $a_0 M + a_1 M_1 + \cdots + a_n M_n$ is a non-zero integer.
+We also have:
+{{begin-eqn}}
+{{eqn | l = \size {\epsilon_k}
+      | r = e^k \size {\int_0^k \frac {x^{p - 1} \sqbrk {\paren {x - 1} \cdots \paren {x - n} }^p e^{-x} } {\paren {p - 1}!} \rd x}
+      | c = 
+}}
+{{eqn | o = \le
+      | r = e^k \int_0^k \frac {\size {x^{p - 1} \sqbrk {\paren {x - 1} \cdots \paren {x - n} }^p} e^{-x} } {\paren {p - 1}!} \rd x
+      | c = [[Absolute Value of Definite Integral]]
+}}
+{{eqn | o = \le
+      | r = e^n \int_0^n \frac {n^{p - 1} \size {\sqbrk {\paren {x - 1} \cdots \paren {x - n} }^p} e^{-x} } {\paren {p - 1}!} \rd x
+      | c = [[Relative Sizes of Definite Integrals]]
+}}
+{{end-eqn}}
+Let $A$ be the maximum value of $\size {\paren {x - 1} \cdots \paren {x - n} }$ for $x$ in the [[Definition:Closed Interval|interval]] $\closedint 0 n$.
+Then:
+{{begin-eqn}}
+{{eqn | o = \le
+      | l = \size {\epsilon_k}
+      | r = \frac {e^n n^{p - 1} A^p} {\paren {p - 1}!} \int_0^n e^{-x} \rd x
+      | c = 
+}}
+{{eqn | o = \le
+      | r = \frac {e^n n^{p - 1} A^p} {\paren {p - 1}!} n
+      | c = as $e^{- x} \le 1$ for $x \ge 0$ 
+}}
+{{eqn | o = \le
+      | r = \frac {e^n \paren {n A}^p} {p!}
+      | c = 
+}}
+{{end-eqn}}
+By [[Power over Factorial]]:
+:$\displaystyle \lim_{p \mathop \to \infty} \frac {e^n \paren {n A}^p} {p!} = 0$
+So $\size {\epsilon_k}$, and therefore $\size {a_1 \epsilon_1 + \cdots + a_n \epsilon_n}$, can be made arbitrarily small by choosing $p$ sufficiently large. 
+It follows that $\paren {a_0 M + a_1 M_1 + \cdots + a_n M_n} + \paren {a_1 \epsilon_1 + \cdots + a_n \epsilon_n}$ is non-zero.
+This is a [[Proof by Contradiction|contradiction]], so $e$ must be [[Definition:Transcendental|transcendental]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+{{ProofWanted}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Laplace Transform of Higher Order Derivatives}
+Tags: Laplace Transforms of Derivatives
+
+\begin{theorem}
+{{begin-eqn}}
+{{eqn | l = \laptrans {\map {f^{\paren n} } t}
+      | r = s^n \laptrans {\map f t} - \sum_{j \mathop = 1}^n s^{j - 1} \map {f^{\paren {n - j} } } 0
+}}
+{{eqn | r = s^n \map F s - s^{n - 1} \, \map f 0 - s^{n - 2} \, \map {f'} 0 - s^{n - 3} \, \map {f''} 0 - \ldots - s \, \map {f^{\paren {n - 2} } } 0 - \map {f^{\paren {n - 1} } } 0
+}}
+{{end-eqn}}
+\end{theorem}
+
+\begin{proof}
+The proof proceeds by [[Principle of Mathematical Induction|induction]] on $n$, the [[Definition:Order of Derivative|order of the derivative]] of $f$.
+=== Basis for the Induction ===
+[[Definition:By Hypothesis|By hypothesis]] $f$ is of [[Definition:Exponential Order|exponential order]].
+Thus [[Laplace Transform of Derivative]] can be invoked:
+:$\laptrans {\map {f'} t} = s \laptrans {\map f t} - \map f 0$
+This is the [[Definition:Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Fix $n - 1 \in \N$ with $n \ge 2$.
+Assume:
+:$\displaystyle \laptrans {\map {f^{\paren {n - 1} } } t} = s^{n - 1} \laptrans {\map f t}- \sum_{j \mathop = 1}^{n - 1} s^{j - 1} \map {f^{\paren {n - 1 - j} } } 0$
+This is our [[Definition:Induction Hypothesis|induction hypothesis]].
+=== Induction Step ===
+This is our [[Definition:Induction Step|induction step]]:
+[[Definition:By Hypothesis|By hypothesis]] $f^\paren {n - 1}$ is of [[Definition:Exponential Order|exponential order]].
+Thus [[Laplace Transform of Derivative]] can be invoked:
+{{begin-eqn}}
+{{eqn | l = \laptrans {\map {f^{\paren n} } t}
+      | r = s \laptrans {\map {f^{\paren {n - 1} } } t} - \map f 0
+      | c = [[Laplace Transform of Derivative]]
+}}
+{{eqn | r = s \paren {s^{n - 1} \laptrans {\map f t} - \sum_{j \mathop = 1}^{n - 1} s^{j - 1} \map {f^{\paren {n - 1 - j} } } 0} - \map f 0
+      | c = [[Laplace Transform of Higher Order Derivatives#Induction Hypothesis|Induction Hypothesis]]
+}}
+{{eqn | r = s^n \laptrans {\map f t} - \sum_{j \mathop = 1}^{n - 1} s^{j - 1 + 1} \map {f^{\paren {n - j} } } 0 - s^0 \map {f^{\paren 0} } 0
+      | c = {{Defof|Zeroth Derivative}}
+}}
+{{eqn | r = s^n \laptrans {\map f t} - \sum_{j \mathop = 0}^{n - 1} s^j \map {f^{\paren {n - j} } } 0
+}}
+{{eqn | r = s^n \laptrans {\map f t} - \sum_{j \mathop = 1}^n s^{j - 1} \map {f^{\paren {n - j} } } 0
+      | c = replace $j$ with $j - 1$
+}}
+{{end-eqn}}
+The result follows by the [[Principle of Mathematical Induction]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Laplace Transform of Hyperbolic Cosine}
+Tags: Hyperbolic Cosine Function, Laplace Transforms, Laplace Transform of Hyperbolic Cosine
+
+\begin{theorem}
+Let $\cosh t$ be the [[Definition:Hyperbolic Cosine|hyperbolic cosine]], where $t$ is [[Definition:Real Number|real]].
+Let $\laptrans f$ denote the [[Definition:Laplace Transform|Laplace transform]] of the [[Definition:Real Function|real function]] $f$.
+Then:
+:$\laptrans {\cosh a t} = \dfrac s {s^2 - a^2}$
+where $a \in \R_{>0}$ is [[Definition:Constant|constant]], and $\map \Re s > a$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \laptrans {\cosh at}
+      | r = \laptrans {\frac {e^{at} + e^{-at} } 2}
+      | c = {{Defof|Hyperbolic Cosine}}
+}}
+{{eqn | r = \frac 1 2 \paren {\laptrans {e^{at} } + \laptrans {e^{-at} } }
+      | c = [[Linear Combination of Laplace Transforms]]
+}}
+{{eqn | r = \frac 1 2 \paren {\frac 1 {s - a} + \frac 1 {s + a} }
+      | c = [[Laplace Transform of Exponential]]
+}}
+{{eqn | r = \frac 1 2 \paren {\frac {s + a + s - a} {\paren {s - a} \paren {s + a} } }
+}}
+{{eqn | r = \frac s {s^2 - a^2}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Laplace Transform of Hyperbolic Sine}
+Tags: Hyperbolic Sine Function, Laplace Transforms, Laplace Transform of Hyperbolic Sine
+
+\begin{theorem}
+Let $\sinh t$ be the [[Definition:Hyperbolic Sine|hyperbolic sine]], where $t$ is [[Definition:Real Number|real]].
+Let $\laptrans f$ denote the [[Definition:Laplace Transform|Laplace transform]] of the [[Definition:Real Function|real function]] $f$.
+Then:
+:$\laptrans {\sinh a t} = \dfrac a {s^2 - a^2}$
+where $a \in \R_{>0}$ is [[Definition:Constant|constant]], and $\map \Re s > a$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \laptrans {\sinh a t}
+      | r = \laptrans {\frac {e^{a t} - e^{-a t} } 2}
+      | c = {{Defof|Hyperbolic Sine}}
+}}
+{{eqn | r = \frac 1 2 \paren {\laptrans {e^{at} } - \laptrans {e^{-a t} } }
+      | c = [[Linear Combination of Laplace Transforms]]
+}}
+{{eqn | r = \frac 1 2 \paren {\frac 1 {s-a} - \frac 1 {s + a} }
+      | c = [[Laplace Transform of Exponential]]
+}}
+{{eqn | r = \frac 1 2 \paren {\frac {s + a - s + a} {\paren {s - a} \paren {s + a} } }
+}}
+{{eqn | r = \frac a {s^2 - a^2}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \laptrans {\sinh {a t} }
+      | r = \int_0^{\to +\infty} e^{-s t} \sinh {a t} \rd t
+      | c = {{Defof|Laplace Transform}}
+}}
+{{eqn | r = \intlimits {\frac {e^{-s t} \paren {-s \sinh a t - a \cosh a t} } {\paren {-s}^2 - a^2} } {t \mathop = 0} {t \mathop \to +\infty}
+      | c = [[Primitive of Exponential of a x by Hyperbolic Sine of b x|Primitive of $e^{a x} \sinh b x$]]
+}}
+{{eqn | r = 0 - \frac {-s \, \map \sinh {0 \times a} - a \, \map \cosh {0 \times a} } {s^2 - a^2}
+      | c = [[Exponential Tends to Zero and Infinity|Exponential Tends to Zero]], [[Exponential of Zero]]
+}}
+{{eqn | r = \frac {s \sinh 0 + a \cosh 0} {s^2 - a^2}
+      | c = simplifying
+}}
+{{eqn | r = \frac a {s^2 - a^2}
+      | c = [[Hyperbolic Sine of Zero is Zero]], [[Hyperbolic Cosine of Zero is One]]
+}}
+{{end-eqn}}
+{{Qed}}
+\end{proof}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \laptrans {\sinh a t}
+      | r = \laptrans {\frac {e^{a t} - e^{-a t} } 2}
+      | c = {{Defof|Hyperbolic Sine}}
+}}
+{{eqn | r = \int_0^{\to +\infty} e^{-s t} \paren {\frac {e^{a t} - e^{-a t} } 2} \rd t
+      | c = {{Defof|Laplace Transform}}
+}}
+{{eqn | r = \dfrac 1 2 \int_0^{\to +\infty} e^{-s t} e^{a t} \rd t - \dfrac 1 2 \int_0^{\to +\infty} e^{-s t} e^{-a t} \rd t
+      | c = [[Linear Combination of Laplace Transforms]]
+}}
+{{eqn | r = \dfrac 1 2 \laptrans {e^{a t} } - \dfrac 1 2 \laptrans {e^{-a t} }
+      | c = {{Defof|Laplace Transform}}
+}}
+{{eqn | r = \frac 1 2 \paren {\frac 1 {s-a} - \frac 1 {s + a} }
+      | c = [[Laplace Transform of Exponential]]
+}}
+{{eqn | r = \frac 1 2 \paren {\frac {s + a - s + a} {\paren {s - a} \paren {s + a} } }
+}}
+{{eqn | r = \frac a {s^2 - a^2}
+}}
+{{end-eqn}}
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Primitives of Rational Functions involving a x + b cubed}
+Tags: Primitives, Primitives of Rational Functions, Primitives involving a x + b
+
+\begin{theorem}
+This page gathers together the [[Definition:Primitive (Calculus)|primitives]] of some [[Definition:Rational Function|rational functions]] involving $\left({a x + b}\right)^3$.
+\end{theorem}<|endoftext|>
+\section{Primitive of Power of a x + b}
+Tags: Primitives involving a x + b, Primitive of Power of a x + b
+
+\begin{theorem}
+:$\displaystyle \int \paren {a x + b}^n \rd x = \frac {\paren {a x + b}^{n + 1} } {\paren {n + 1} a} + C$
+\end{theorem}
+
+\begin{proof}
+Let $u = a x + b$.
+Then:
+:$\dfrac {\d u} {\d x} = a$
+Then:
+{{begin-eqn}}
+{{eqn | l = \int \paren {a x + b}^n \rd x
+      | r = \int \dfrac {u^n} a \rd u
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \dfrac 1 a \dfrac {u^{n + 1} } {n + 1}
+      | c = [[Primitive of Power]]
+}}
+{{eqn | r = \frac {\paren {a x + b}^{n + 1} } {\paren {n + 1} a} + C
+      | c = substituting back for $u$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of x by Power of a x + b}
+Tags: Primitives involving a x + b
+
+\begin{theorem}
+:$\displaystyle \int x \left({a x + b}\right)^n \ \mathrm d x = \frac {\left({a x + b}\right)^{n + 2} } {\left({n + 2}\right) a^2} - \frac {b \left({a x + b}\right)^{n + 1} } {\left({n + 1}\right) a^2} + C$
+\end{theorem}
+
+\begin{proof}
+Let $u = a x + b$.
+Then:
+{{begin-eqn}}
+{{eqn | l = x
+      | r = \frac {u - b} a
+      | c = 
+}}
+{{eqn | l = \frac {\mathrm d u} {\mathrm d x}
+      | r = \frac 1 a
+      | c = 
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = \int x \left({a x + b}\right)^n \ \mathrm d x
+      | r = \frac 1 a \int \frac {u - b} a u^n \ \mathrm d u
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \frac 1 {a^2} \int u^{n+1} \ \mathrm d u - \frac b {a^2} \int u^n \ \mathrm d u
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \frac 1 {a^2} \frac {u^{n+2} } {n + 2} - \frac b {a^2} \frac {u^{n+1} } {n + 1} + C
+      | c = [[Primitive of Power]]
+}}
+{{eqn | r =  \frac {\left({a x + b}\right)^{n + 2} } {\left({n + 2}\right) a^2} - \frac {b \left({a x + b}\right)^{n + 1} } {\left({n + 1}\right) a^2} + C
+      | c = substituting for $u$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of x squared by Power of a x + b}
+Tags: Primitives involving a x + b
+
+\begin{theorem}
+:$\displaystyle \int x^2 \left({a x + b}\right)^n \ \mathrm d x = \frac {\left({a x + b}\right)^{n + 3} } {\left({n + 3}\right) a^3} - \frac {2 b \left({a x + b}\right)^{n + 2} } {\left({n + 2}\right) a^3} + \frac {b^2 \left({a x + b}\right)^{n + 1} } {\left({n + 1}\right) a^3} + C$
+\end{theorem}
+
+\begin{proof}
+Let $u = a x + b$.
+Then:
+{{begin-eqn}}
+{{eqn | l = x
+      | r = \frac {u - b} a
+      | c = 
+}}
+{{eqn | l = \frac {\mathrm d u} {\mathrm d x}
+      | r = \frac 1 a
+      | c = 
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = \int x \left({a x + b}\right)^n \ \mathrm d x
+      | r = \int \frac 1 a \left({\frac {u - b} a}\right)^2 u^n \ \mathrm d u
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \int \frac 1 {a^3} \left({u^{n + 2} - 2 b u^{n + 1} + b^2 u^n}\right) \ \mathrm d u
+      | c = [[Square of Difference]] and multiplying out
+}}
+{{eqn | r = \frac 1 {a^3}  \int u^{n+3} \ \mathrm d u - \frac {2 b} {a^3} \int u^{n + 1} \ \mathrm d u + \frac {b^2} {a^3} \int u^n \ \mathrm d u
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \frac 1 {a^3} \frac {u^{n+3} } {n + 3} - \frac {2 b} {a^3} \frac {u^{n+2} } {n + 2} + \frac {b^2} {a^3} \frac {u^{n+1} } {n + 1} + C
+      | c = [[Primitive of Power]]
+}}
+{{eqn | r = \frac {\left({a x + b}\right)^{n + 3} } {\left({n + 3}\right) a^3} - \frac {2 b \left({a x + b}\right)^{n + 2} } {\left({n + 2}\right) a^3} + \frac {b^2 \left({a x + b}\right)^{n + 1} } {\left({n + 1}\right) a^3} + C
+      | c = substituting for $u$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Reduction Formula for Primitive of Power of x by Power of a x + b/Decrement of Power of a x + b}
+Tags: Primitive of Power of x by Power of a x + b
+
+\begin{theorem}
+:$\displaystyle \int x^m \left({a x + b}\right)^n \rd x = \frac {x^{m+1} \left({a x + b}\right)^n} {m + n + 1} + \frac {n b} {m + n + 1} \int x^m \left({a x + b}\right)^{n - 1} \rd x$
+\end{theorem}<|endoftext|>
+\section{Primitive of x over a x + b cubed/Proof 1}
+Tags: Primitives involving a x + b
+
+\begin{theorem}
+:$\displaystyle \int \frac {x \ \mathrm d x} {\left({a x + b}\right)^3} = \frac {-1} {a^2 \left({a x + b}\right)} + \frac b {2 a^2 \left({a x + b}\right)^2} + C$
+\end{theorem}
+
+\begin{proof}
+Put $u = a x + b$.
+Then:
+{{begin-eqn}}
+{{eqn | l = x
+      | r = \frac {u - b} a
+      | c = 
+}}
+{{eqn | l = \frac {\mathrm d u} {\mathrm d x}
+      | r = \frac 1 a
+      | c = 
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = \int \frac {x \ \mathrm d x} {\left({a x + b}\right)^3}
+      | r = \int \frac 1 a \frac {u - b} {a u^3} \ \mathrm d u
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \frac 1 {a^2} \int \frac {\mathrm d u} {u^2} - \frac b {a^2} \int \frac {\mathrm d u} {u^3}
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \frac 1 {a^2} \frac {-1} {u} - \frac b {a^2} \frac {-1} {2 u^2} + C
+      | c = [[Primitive of Power]]
+}}
+{{eqn | r = \frac {-1} {a^2 \left({a x + b}\right)} + \frac b {2 a^2 \left({a x + b}\right)^2} + C
+      | c = substituting for $u$ and rearranging
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of x over a x + b cubed/Proof 2}
+Tags: Primitives involving a x + b
+
+\begin{theorem}
+:$\displaystyle \int \frac {x \ \mathrm d x} {\left({a x + b}\right)^3} = \frac {-1} {a^2 \left({a x + b}\right)} + \frac b {2 a^2 \left({a x + b}\right)^2} + C$
+\end{theorem}
+
+\begin{proof}
+From [[Primitive of x by Power of a x + b|Primitive of $x$ by Power of $a x + b$]]:
+:$\displaystyle \int x \left({a x + b}\right)^n \ \mathrm d x = \frac {\left({a x + b}\right)^{n + 2} } {\left({n + 2}\right) a^2} - \frac {b \left({a x + b}\right)^{n + 1} } {\left({n + 1}\right) a^2} + C$
+where $n \ne - 1$ and $n \ne - 2$.
+The result follows by setting $n = -3$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Reduction Formula for Primitive of Power of x by Power of a x + b}
+Tags: Primitives involving a x + b, Primitive of Power of x by Power of a x + b
+
+\begin{theorem}
+=== [[Reduction Formula for Primitive of Power of x by Power of a x + b/Decrement of Power of a x + b|Decrement of Power of $a x + b$]] ===
+{{:Reduction Formula for Primitive of Power of x by Power of a x + b/Decrement of Power of a x + b}}
+=== [[Reduction Formula for Primitive of Power of x by Power of a x + b/Decrement of Power of x|Decrement of Power of $x$]] ===
+{{:Reduction Formula for Primitive of Power of x by Power of a x + b/Decrement of Power of x}}
+=== [[Reduction Formula for Primitive of Power of x by Power of a x + b/Increment of Power of a x + b|Increment of Power of $a x + b$]] ===
+{{:Reduction Formula for Primitive of Power of x by Power of a x + b/Increment of Power of a x + b}}
+=== [[Reduction Formula for Primitive of Power of x by Power of a x + b/Increment of Power of x|Increment of Power of $x$]] ===
+{{:Reduction Formula for Primitive of Power of x by Power of a x + b/Increment of Power of x}}
+\end{theorem}<|endoftext|>
+\section{Distance from Subset to Element}
+Tags: Distance Function
+
+\begin{theorem}
+Let $\struct {M, d}$ be a [[Definition:Metric Space|metric space]].
+Let $S \subseteq M$ be a [[Definition:Subset|subset]] of $M$.
+Let $s \in S$.
+Then:
+:$\map d {s, S} = 0$
+where $\map d {s, S}$ denotes the [[Definition:Distance between Element and Subset of Metric Space|distance between $s$ and $S$]].
+\end{theorem}
+
+\begin{proof}
+By [[Distance between Element and Subset is Nonnegative]]:
+:$\map d {s, S} \ge 0$
+Also, because:
+:$\map d {s, S} = 0$
+and $s \in S$, it follows that:
+:$\map d {s, S} \le 0$
+Hence the result.
+{{qed}}
+[[Category:Distance Function]]
+d5vrsa7b3u83yiv2wxrnm1vrttq59n7
+\end{proof}<|endoftext|>
+\section{Reduction Formula for Primitive of Power of x by Power of a x + b/Decrement of Power of x}
+Tags: Primitive of Power of x by Power of a x + b
+
+\begin{theorem}
+:$\displaystyle \int x^m \left({a x + b}\right)^n \rd x = \frac {x^m \left({a x + b}\right)^{n + 1} } {\left({m + n + 1}\right) a} - \frac {m b} {\left({m + n + 1}\right) a} \int x^{m - 1} \left({a x + b}\right)^n \rd x$
+\end{theorem}<|endoftext|>
+\section{Distance from Subset to Supremum}
+Tags: Real Analysis
+
+\begin{theorem}
+Let $S \subseteq \R$ be a [[Definition:Subset|subset]] of the [[Definition:Real Numbers|real numbers]].
+Suppose that the [[Definition:Supremum of Subset of Real Numbers|supremum]] $\sup S$ of $S$ exists.
+Then:
+:$\map d {\sup S, S} = 0$
+where $\map d {\sup S, S}$ is the [[Definition:Distance between Element and Subset of Real Numbers|distance between $\sup S$ and $S$]].
+\end{theorem}
+
+\begin{proof}
+By [[Distance between Element and Subset is Nonnegative]]:
+:$\map d {\sup S, S} \ge 0$
+By definition of [[Definition:Supremum of Subset of Real Numbers|supremum]]:
+:$\forall \epsilon > 0: \exists s \in S: \map d {\sup S, s} < \epsilon$
+meaning that, by nature of the [[Definition:Infimum of Subset of Real Numbers|infimum]] and the [[Definition:Distance between Element and Subset of Real Numbers|definition of $\map d {\sup S, S}$]]:
+:$\forall \epsilon > 0: \map d {\sup S, S} < \epsilon$
+Together, these two observations lead to the conclusion that:
+:$\map d {\sup S, S} = 0$
+as desired.
+{{qed}}
+[[Category:Real Analysis]]
+9dx5tipm6gxmv12gx5qz1mcvrn3l8a6
+\end{proof}<|endoftext|>
+\section{Distance between Element and Subset is Nonnegative}
+Tags: Distance Function
+
+\begin{theorem}
+Let $\struct {M, d}$ be a [[Definition:Metric Space|metric space]].
+Let $x \in M$ and $S \subseteq M$.
+Then:
+:$\map d {x, S} \ge 0$
+where $\map d {x, S}$ is the [[Definition:Distance between Element and Subset of Metric Space|distance between $x$ and $S$]].
+\end{theorem}
+
+\begin{proof}
+By definition of the [[Definition:Distance between Element and Subset of Metric Space|distance between $x$ and $S$]]:
+:$\map d {x, S} = \displaystyle \inf_{s \mathop \in S} \map d {x, s}$
+From the [[Definition:Metric Space Axioms|metric space axioms]]:
+:$\forall s \in M: \map d {x, s} \ge 0$
+Hence by the nature of the [[Definition:Infimum of Subset of Real Numbers|infimum]]:
+:$\map d {x, S} \ge 0$
+as desired.
+{{qed}}
+[[Category:Distance Function]]
+k6h3621h7lo04cx0n3c2y9totl55ob7
+\end{proof}<|endoftext|>
+\section{Distance from Subset to Infimum}
+Tags: Real Analysis
+
+\begin{theorem}
+Let $S \subseteq \R$ be a [[Definition:Subset|subset]] of the [[Definition:Real Numbers|real numbers]].
+Suppose that the [[Definition:Infimum of Subset of Real Numbers|infimum]] $\inf S$ of $S$ exists.
+Then:
+:$\map d {\inf S, S} = 0$
+where $\map d {\inf S, S}$ is the [[Definition:Distance between Element and Subset of Real Numbers|distance between $\inf S$ and $S$]].
+\end{theorem}
+
+\begin{proof}
+By [[Distance between Element and Subset is Nonnegative]]:
+:$\map d {\inf S, S} \ge 0$
+By definition of [[Definition:Infimum of Subset of Real Numbers|infimum]]:
+:$\forall \epsilon > 0: \exists s \in S: \map d {\inf S, s} < \epsilon$
+meaning that, by nature of the [[Definition:Infimum of Subset of Real Numbers|infimum]] and the [[Definition:Distance between Element and Subset of Real Numbers|definition of $\map d {\inf S, S}$]]:
+:$\forall \epsilon > 0: \map d {\inf S, S} < \epsilon$
+Together, these two observations lead to the conclusion that:
+:$\map d {\inf S, S} = 0$
+as desired.
+{{qed}}
+[[Category:Real Analysis]]
+ge783khyr89w1rtnt8t2t2aw56bn4vm
+\end{proof}<|endoftext|>
+\section{Reduction Formula for Primitive of Power of x by Power of a x + b/Increment of Power of a x + b}
+Tags: Primitive of Power of x by Power of a x + b
+
+\begin{theorem}
+:$\displaystyle \int x^m \paren {a x + b}^n \rd x = \frac {-x^{m + 1} \paren {a x + b}^{n + 1} } {\paren {n + 1} b} + \frac {m + n + 2} {\paren {n + 1} b} \int x^m \paren {a x + b}^{n + 1} \rd x$
+\end{theorem}<|endoftext|>
+\section{Reduction Formula for Primitive of Power of x by Power of a x + b/Increment of Power of x}
+Tags: Primitive of Power of x by Power of a x + b
+
+\begin{theorem}
+:$\displaystyle \int x^m \left({a x + b}\right)^n \rd x = \frac {x^{m+1} \left({a x + b}\right)^{n + 1} } {\left({m + 1}\right) b} - \frac {\left({m + n + 2}\right) a} {\left({m + 1}\right) b} \int x^{m + 1} \left({a x + b}\right)^n \rd x$
+\end{theorem}<|endoftext|>
+\section{Primitive of Reciprocal of Root of a x + b}
+Tags: Primitives involving Root of a x + b, Primitives involving Reciprocals
+
+\begin{theorem}
+:$\displaystyle \int \frac {\d x} {\sqrt{a x + b} } = \frac {2 \sqrt {a x + b} } a + C$
+\end{theorem}
+
+\begin{proof}
+Put $u = \sqrt{a x + b}$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \int \frac {\d x} {\sqrt {a x + b} }
+      | r = \frac 2 a \int u \frac {\d u} u
+      | c = [[Primitive of Function of Root of a x + b|Primitive of Function of $\sqrt {a x + b}$]]
+}}
+{{eqn | r = \frac 2 a \int \rd u
+      | c = 
+}}
+{{eqn | r = \frac 2 a u + C
+      | c = [[Primitive of Constant]]
+}}
+{{eqn | r = \frac {2 \sqrt {a x + b} } a + C
+      | c = substituting for $u$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of x over Root of a x + b}
+Tags: Primitives involving Root of a x + b
+
+\begin{theorem}
+:$\displaystyle \int \frac {x \ \mathrm d x} {\sqrt{a x + b} } = \frac {2 \left({a x - 2 b}\right) \sqrt{a x + b} } {3 a^2}$
+\end{theorem}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = u
+      | r = \sqrt{a x + b}
+      | c = 
+}}
+{{eqn | l = x
+      | r = \frac {u^2 - b} a
+      | c = 
+}}
+{{end-eqn}}
+Thus:
+{{begin-eqn}}
+{{eqn | l = F \left({\sqrt{a x + b} }\right)
+      | r = \frac x {\sqrt{a x + b} }
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = F \left({u}\right)
+      | r = \left({\frac {u^2 - b} a}\right) \frac 1 u
+      | c = 
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = \int \frac {x \ \mathrm d x} {\sqrt{a x + b} }
+      | r = \frac 2 a \int u \left({\frac {u^2 - b} a}\right) \frac 1 u \ \mathrm d u
+      | c = [[Primitive of Function of Root of a x + b|Primitive of Function of $\sqrt{a x + b}$]]
+}}
+{{eqn | r = \frac 2 {a^2} \int \left({u^2 - b}\right) \ \mathrm d u
+      | c = [[Primitive of Constant Multiple of Function]]
+}}
+{{eqn | r = \frac 2 {a^2} \left({\frac {u^3} 3 - b u}\right) + C
+      | c = [[Primitive of Power]] and [[Primitive of Constant]]
+}}
+{{eqn | r = \frac 2 {a^2} \left({\frac {\left({a x + b}\right) \sqrt{a x + b} } 3 - b \sqrt{a x + b} }\right) + C
+      | c = substituting for $u$
+}}
+{{eqn | r = \frac 2 {3 a^2} \left({\left({a x + b}\right) - 3 b}\right) \sqrt{a x + b} + C
+      | c = extracting common factors
+}}
+{{eqn | r = \frac {2 \left({a x - 2 b}\right) \sqrt{a x + b} } {3 a^2} + C
+      | c = simplifying
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of x squared over Root of a x + b}
+Tags: Primitives involving Root of a x + b
+
+\begin{theorem}
+:$\displaystyle \int \frac {x^2 \ \mathrm d x} {\sqrt{a x + b} } = \frac {2 \left({3 a^2 x^2 - 4 a b x + 8 b^2}\right) \sqrt{a x + b} } {15 a^3}$
+\end{theorem}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = u
+      | r = \sqrt{a x + b}
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = x
+      | r = \frac {u^2 - b} a
+      | c = 
+}}
+{{end-eqn}}
+Thus:
+{{begin-eqn}}
+{{eqn | l = F \left({\sqrt{a x + b} }\right)
+      | r = \frac {x^2} {\sqrt{a x + b} }
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = F \left({u}\right)
+      | r = \left({\frac {u^2 - b} a}\right)^2 \frac 1 u
+      | c = 
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = \int \frac {x^2 \ \mathrm d x} {\sqrt{a x + b} }
+      | r = \frac 2 a \int u \left({\frac {u^2 - b} a}\right)^2 \frac 1 u \ \mathrm d u
+      | c = [[Primitive of Function of Root of a x + b|Primitive of Function of $\sqrt{a x + b}$]]
+}}
+{{eqn | r = \frac 2 {a^3} \int \left({u^4 - 2 b u^2 + b^2}\right) \ \mathrm d u
+      | c = [[Primitive of Constant Multiple of Function]]
+}}
+{{eqn | r = \frac 2 {a^3} \left({\frac {u^5} 5 - \frac {2 b u^3} 3 + b^2 u}\right) + C
+      | c = [[Primitive of Power]] and [[Primitive of Constant]]
+}}
+{{eqn | r = \frac 2 {a^3} \left({\frac {\left({a x + b}\right)^2} 5 - \frac {2 b \left({a x + b}\right)} 3 + b^2}\right)\sqrt{a x + b} + C
+      | c = substituting for $u$ and extracting common factors
+}}
+{{eqn | r = \frac 2 {15 a^2} \left({3 a^2 x^2 + 6 a b x + 3 b^2 - 10 a b x - 10 b^2 + 15 b^2}\right) \sqrt{a x + b} + C
+      | c = multiplying out and combining fractions
+}}
+{{eqn | r = \frac {2 \left({3 a^2 x^2 - 4 a b x + 8 b^2}\right) \sqrt{a x + b} } {15 a^3} + C
+      | c = simplifying
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Reciprocal of x squared minus a squared/Inverse Hyperbolic Cotangent Form}
+Tags: Primitive of Reciprocal of x squared minus a squared
+
+\begin{theorem}
+:$\ds \int \frac {\d x} {x^2 - a^2} = -\frac 1 a \coth^{-1} {\frac x a} + C$
+where $\size x > a$.
+\end{theorem}<|endoftext|>
+\section{Primitive of Reciprocal of x by Root of a x + b}
+Tags: Primitives involving Root of a x + b
+
+\begin{theorem}
+:$\displaystyle \int \frac {\d x} {x \sqrt{a x + b} } = \begin {cases}
+\dfrac 1 {\sqrt b} \map \ln {\dfrac {\sqrt {a x + b} - \sqrt b} {\sqrt {a x + b} + \sqrt b} } + C & : b > 0 \\
+\dfrac 2 {\sqrt {-b} } \arctan \sqrt {\dfrac {a x + b} {-b} } + C & : b < 0 \end {cases}$
+\end{theorem}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = u
+      | r = \sqrt {a x + b}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x
+      | r = \frac {u^2 - b} a
+      | c = 
+}}
+{{end-eqn}}
+Thus:
+{{begin-eqn}}
+{{eqn | l = \map F {\sqrt {a x + b} }
+      | r = \frac 1 {x \sqrt {a x + b} }
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \map F u
+      | r = \paren {\frac a {u^2 - b} } \frac 1 u
+      | c = 
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = \int \frac {\d x} {x \sqrt {a x + b} }
+      | r = \frac 2 a \int u \paren {\frac a {u^2 - b} } \frac 1 u \rd u
+      | c = [[Primitive of Function of Root of a x + b|Primitive of Function of $\sqrt {a x + b}$]]
+}}
+{{eqn | r = 2 \int \frac {\d u} {u^2 - b}
+      | c = [[Primitive of Constant Multiple of Function]]
+}}
+{{end-eqn}}
+Let $b > 0$.
+Let $d = \sqrt b$.
+Then:
+{{begin-eqn}}
+{{eqn | l = 2 \int \frac {\d u} {u^2 - b}
+      | r = 2 \int \frac {\d u} {u^2 - d^2}
+      | c = 
+}}
+{{eqn | r = 2 \paren {\frac 1 {2 d} } \map \ln {\frac {u - d} {u + d} } + C
+      | c = [[Primitive of Reciprocal of x squared minus a squared/Logarithm Form|Primitive of Reciprocal of $x^2 - a^2$: Logarithm Form]]
+}}
+{{eqn | r = \dfrac 1 {\sqrt b} \map \ln {\dfrac {\sqrt {a x + b} - \sqrt b} {\sqrt {a x + b} + \sqrt b} } + C
+      | c = substituting for $u$ and $d$
+}}
+{{end-eqn}}
+{{qed|lemma}}
+{{questionable|In the link to the above primitive evaluation, the ln is of the absolute value of the argument. If $u - d$ can possibly be negative, this needs to hold over to this page If $x$ is negative, then this could indeed be the case. Hence the source work may also need to be scrutinised to see whether it has glossed over this detail.}}
+Let $b < 0$.
+Let $d = \sqrt {-b}$.
+Then:
+{{begin-eqn}}
+{{eqn | l = 2 \int \frac {\d u} {u^2 - b}
+      | r = 2 \int \frac {\d u} {u^2 - \paren {-d^2} }
+      | c = 
+}}
+{{eqn | r = 2 \int \frac {\d u} {u^2 + d^2}
+      | c = 
+}}
+{{eqn | r = 2 \paren {\frac 1 d} \arctan {\frac u d} + C
+      | c = [[Primitive of Reciprocal of x squared plus a squared/Arctangent Form|Primitive of Reciprocal of $x^2 + a^2$]]
+}}
+{{eqn | r = \frac 2 {\sqrt {-b} } \arctan \sqrt {\dfrac {a x + b} {-b} } + C
+      | c = substituting for $u$ and $d$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Reciprocal of x squared by Root of a x + b}
+Tags: Primitives involving Root of a x + b, Primitives involving Reciprocals
+
+\begin{theorem}
+:$\displaystyle \int \frac {\d x} {x^2 \sqrt{a x + b} } = -\frac {\sqrt{a x + b} } {b x} - \frac a {2 b} \int \frac {\mathrm d x} {x \sqrt{a x + b} }$
+\end{theorem}
+
+\begin{proof}
+From [[Reduction Formula for Primitive of Power of x by Power of a x + b/Increment of Power of x|Reduction Formula for Primitive of Power of $x$ by Power of $a x + b$: Increment of Power of $x$]]:
+:$\displaystyle \int x^m \left({a x + b}\right)^n \rd x = \frac {x^{m+1} \left({a x + b}\right)^{n + 1} } {\left({m + 1}\right) b} - \frac {\left({m + n + 2}\right) a} {\left({m + 1}\right) b} \int x^{m + 1} \left({a x + b}\right)^n \rd x$
+Setting $m = -2$ and $n = -\dfrac 1 2$:
+{{begin-eqn}}
+{{eqn | l = \int \frac {\d x} {x^2 \sqrt{a x + b} }
+      | r = \int x^{-2} \left({a x + b}\right)^{-1/2} \rd x
+      | c = 
+}}
+{{eqn | r = \frac {x^{-1} \left({a x + b}\right)^{1/2} } {\left({-1}\right) b} - \frac {\left({-1/2}\right) a} {\left({-1}\right) b} \int x^{-1} \left({a x + b}\right)^{-1/2} \rd x
+      | c = 
+}}
+{{eqn | r = -\frac {\sqrt{a x + b} } {b x} - \frac a {2 b} \int \frac {\d x} {x \sqrt{a x + b} }
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Root of a x + b}
+Tags: Primitives involving Root of a x + b
+
+\begin{theorem}
+:$\displaystyle \int \sqrt{a x + b} \ \mathrm d x = \frac {2 \sqrt{\left({a x + b}\right)^3} } {3 a}$
+\end{theorem}
+
+\begin{proof}
+Let $u = \sqrt{a x + b}$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \int \sqrt{a x + b} \ \mathrm d x
+      | r = \frac 2 a \int u^2 \ \mathrm d u
+      | c = [[Primitive of Function of Root of a x + b|Primitive of Function of $\sqrt{a x + b}$]]
+}}
+{{eqn | r = \frac 2 a \frac {u^3} 3
+      | c = [[Primitive of Power]]
+}}
+{{eqn | r = \frac {2 \sqrt{\left({a x + b}\right)^3} } {3 a}
+      | c = substituting for $u$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+From [[Reduction Formula for Primitive of Power of x by Power of a x + b/Decrement of Power of a x + b|Reduction Formula for Primitive of Power of $x$ by Power of $a x + b$: Decrement of Power of $a x + b$]]:
+:$\displaystyle \int x^m \paren {a x + b}^n \rd x = \frac {x^{m + 1} \paren {a x + b}^n} {m + n + 1} + \frac {n b} {m + n + 1} \int x^m \paren {a x + b}^{n - 1} \rd x$
+Putting $m = -1$ and $n = \dfrac 1 2$:
+{{begin-eqn}}
+{{eqn | l = \int \frac {\sqrt {a x + b} } x \rd x
+      | r = \int x^{-1} \paren {a x + b}^{1/2} \rd x
+      | c = 
+}}
+{{eqn | r = \frac {x^0 \paren {a x + b}^{1/2} } {\frac 1 2} + \frac {\frac 1 2 b} {\frac 1 2} \int x^{-1} \paren {a x + b}^{- 1/2} \rd x
+      | c = 
+}}
+{{eqn | r = 2 \sqrt {a x + b} + b \int \frac {\d x} {x \sqrt {a x + b} }
+      | c = simplifying
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = v
+      | r = \sqrt x
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \frac {\d v} {\d x}
+      | r = \frac 1 {2 \sqrt x}
+      | c = [[Power Rule for Derivatives]]
+}}
+{{eqn | l = u
+      | r = \frac {2 \sqrt {a x + b} } {\sqrt x}
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \frac {\d u} {\d x}
+      | r = \frac {\frac {\sqrt x \cdot 2 a} {2 \sqrt{a x + b} } - \frac {2 \sqrt {a x + b} } {2 \sqrt x} } x
+      | c = [[Quotient Rule for Derivatives]] etc.
+}}
+{{eqn | r = \frac {-b} {x^{3/2} \sqrt {a x + b} }
+      | c = simplifying
+}}
+{{end-eqn}}
+From [[Integration by Parts]]:
+: $\displaystyle \int u \frac {\d v} {\d x} \rd x = u v - \int v \ \frac {\d u} {\d x} \rd x$
+from which:
+{{begin-eqn}}
+{{eqn | l = \int \frac {\sqrt{a x + b} } x \rd x
+      | r = \int \frac {2 \sqrt{a x + b} }{\sqrt x} \frac 1 {2 \sqrt x} \rd x
+      | c = 
+}}
+{{eqn | r = \frac {2 \sqrt{a x + b} } {\sqrt x} \sqrt x - \int {\sqrt x} \frac {-b} {x^{3/2} \sqrt {a x + b} } \rd x
+      | c = 
+}}
+{{eqn | r = 2 \sqrt {a x + b} + b \int \frac {\d x} {x \sqrt {a x + b} }
+      | c = simplification
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Reciprocal of a squared minus x squared/Inverse Hyperbolic Tangent Form}
+Tags: Primitive of Reciprocal of a squared minus x squared
+
+\begin{theorem}
+:$\ds \int \frac {\d x} {a^2 - x^2} = \frac 1 a \tanh^{-1} \frac x a + C$
+where $\size x < a$.
+\end{theorem}<|endoftext|>
+\section{Primitive of Reciprocal of Root of x squared plus a squared/Inverse Hyperbolic Sine Form}
+Tags: Primitive of Reciprocal of Root of x squared plus a squared
+
+\begin{theorem}
+:$\ds \int \frac {\d x} {\sqrt {x^2 + a^2} } = \sinh^{-1} {\frac x a} + C$
+\end{theorem}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = u
+      | r = \sinh^{-1} {\frac x a}
+      | c = 
+}}
+{{eqn | l = x
+      | r = a \sinh u
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d x} {\d u}
+      | r = a \cosh u
+      | c = [[Derivative of Hyperbolic Sine]]
+}}
+{{eqn | ll= \leadsto
+      | l = \int \frac {\d x} {\sqrt {x^2 + a^2} }
+      | r = \int \frac {a \cosh u} {\sqrt {a^2 \sinh^2 u + a^2} } \rd u
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \frac a a \int \frac {\cosh u} {\sqrt {\sinh^2 u + 1} } \rd u
+      | c = [[Primitive of Constant Multiple of Function]]
+}}
+{{eqn | r = \int \frac {\cosh u} {\cosh u} \rd u
+      | c = [[Difference of Squares of Hyperbolic Cosine and Sine]]
+}}
+{{eqn | r = \int 1 \rd u
+}}
+{{eqn | r = u + C
+      | c = [[Integral of Constant]]
+}}
+{{eqn | r = \sinh^{-1} {\frac x a} + C
+      | c = Definition of $u$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Reciprocal of Root of x squared minus a squared/Inverse Hyperbolic Cosine Form}
+Tags: Expressions whose Primitives are Inverse Hyperbolic Functions, Inverse Hyperbolic Cosine, Primitive of Reciprocal of Root of x squared minus a squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {\d x} {\sqrt {x^2 - a^2} } = \dfrac {\size x} x \cosh^{-1} {\size {\frac x a} } + C$
+for $x > a$.
+\end{theorem}
+
+\begin{proof}
+When $x = a$ we have that $\sqrt {x^2 - a^2} = 0$ and then $\dfrac 1 {\sqrt {x^2 - a^2} }$ is not defined.
+When $\size x < a$ we have that $x^2 - a^2 < 0$ and then $\sqrt {x^2 - a^2}$ is not defined.
+Hence the [[Definition:Domain of Real Function|domain]] needs to be restricted to $\size x > a$, or that is: $\size {\dfrac x a} > 1$.
+Let $x > a$.
+Let:
+{{begin-eqn}}
+{{eqn | l = u
+      | r = \map {\cosh^{-1} } {\frac x a}
+      | c = 
+}}
+{{eqn | l = x
+      | r = a \cosh u
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d x} {\d u}
+      | r = a \sinh u
+      | c = [[Derivative of Hyperbolic Cosine Function]]
+}}
+{{eqn | ll= \leadsto
+      | l = \int \frac {\d x} {\sqrt {x^2 - a^2} }
+      | r = \int \frac {a \sinh u} {\sqrt {a^2 \cosh^2 u - a^2} } \rd u
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \frac a a \int \frac {\sinh u} {\sqrt {\cosh^2 u - 1} } \rd u
+      | c = [[Primitive of Constant Multiple of Function]]
+}}
+{{eqn | r = \int \frac {\sinh u} {\sinh u} \rd u
+      | c = [[Difference of Squares of Hyperbolic Cosine and Sine]]
+}}
+{{eqn | r = \int 1 \rd u
+}}
+{{eqn | r = u + C
+      | c = [[Integral of Constant]]
+}}
+{{eqn | r = \map {\cosh^{-1} } {\frac x a} + C
+      | c = Definition of $u$
+}}
+{{eqn | r = \cosh^{-1} \size {\frac x a} + C
+      | c = as $\dfrac x a > 0$
+}}
+{{eqn | r = \dfrac {\size x} x \cosh^{-1} \size {\frac x a} + C
+      | c = as $x > 0$, so $\dfrac {\size x} x = 1$
+}}
+{{end-eqn}}
+Let $x < -a$.
+Let $z = -x$.
+Hence:
+:$\dfrac {\d z} {\d x} = -1$
+Then:
+{{begin-eqn}}
+{{eqn | l = \int \frac {\d x} {\sqrt {x^2 - a^2} }
+      | r = \int \frac {-\d z} {\sqrt {\paren {-z}^2 - a^2} }
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = -\int \frac {\d z} {\sqrt {z^2 - a^2} }
+      | c = [[Primitive of Constant Multiple of Function]]
+}}
+{{eqn | r = -\map {\cosh^{-1} } {\frac z a} + C
+      | c = from above
+}}
+{{eqn | r = -\map {\cosh^{-1} } {\frac {-x} a} + C
+      | c = substituting back
+}}
+{{eqn | r = -\cosh^{-1} {\size {\frac x a} } + C
+      | c = as $\dfrac x a < 0$
+}}
+{{eqn | r = \dfrac {\size x} x \cosh^{-1} {\size {\frac x a} } + C
+      | c = as $x < 0$, so $\dfrac {\size x} x = -1$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Reciprocal of x by Root of x squared plus a squared/Inverse Hyperbolic Cosecant Form}
+Tags: Primitive of Reciprocal of x by Root of x squared plus a squared, Expressions whose Primitives are Inverse Hyperbolic Functions, Inverse Hyperbolic Cosecant
+
+\begin{theorem}
+:$\displaystyle \int \frac {\d x} {x \sqrt {x^2 + a^2} } = -\frac 1 a \csch^{-1} {\frac x a} + C$
+\end{theorem}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = u
+      | r = \csch^{-1} {\frac x a}
+      | c = 
+}}
+{{eqn | l = x
+      | r = a \csch u
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d x} {\d u}
+      | r = -a \csch u \coth u
+      | c = [[Derivative of Hyperbolic Cosecant]]
+}}
+{{eqn | ll = \leadsto
+      | l = \int \frac {\d x} {x \sqrt {x^2 + a^2} }
+      | r = \int \frac {-a \csch u \coth u} {a \csch u \sqrt {a^2 \csch^2 u + a^2} } \rd u
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = -\frac a {a^2} \int \frac {\csch u \coth u} {\csch u \sqrt {\csch^2 u - 1} } \rd u
+      | c = [[Primitive of Constant Multiple of Function]]
+}}
+{{eqn | r = -\frac 1 a \int \frac {\csch u \coth u} {\csch u \coth u} \rd u
+      | c = [[Difference of Squares of Hyperbolic Cotangent and Cosecant]]
+}}
+{{eqn | r = -\frac 1 a \int 1 \rd u
+}}
+{{eqn | r = -\frac 1 a u + C
+      | c = [[Integral of Constant]]
+}}
+{{eqn | r = \frac 1 a \csch^{-1} {\frac x a} + C
+      | c = Definition of $u$
+}}
+{{end-eqn}}
+{{qed}}
+{{finish|There may be a complication with the sign of the argument which needs to be resolved}}
+\end{proof}<|endoftext|>
+\section{Primitive of Reciprocal of x by Root of a squared minus x squared/Inverse Hyperbolic Secant Form}
+Tags: Expressions whose Primitives are Inverse Hyperbolic Functions, Inverse Hyperbolic Secant, Primitive of Reciprocal of x by Root of a squared minus x squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {\d x} {x \sqrt {a^2 - x^2} } = -\frac 1 a \sech^{-1} {\frac x a} + C$
+\end{theorem}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = u
+      | r = \sech^{-1} {\frac x a}
+      | c = 
+}}
+{{eqn | l = x
+      | r = a \sech u
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d x} {\d u}
+      | r = -a \sech u \tanh u
+      | c = [[Derivative of Hyperbolic Secant]]
+}}
+{{eqn | ll= \leadsto
+      | l = \int \frac {\d x} {x \sqrt {a^2 - x^2} }
+      | r = \int \frac {-a \sech u \tanh u} {a \sech u \sqrt {a^2 - a^2 \sech^2 u} } \rd u
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = -\frac a {a^2} \int \frac {\sech u \tanh u} {\sech u \sqrt {1 - \sech^2 u} } \rd u
+      | c = [[Primitive of Constant Multiple of Function]]
+}}
+{{eqn | r = -\frac 1 a \int \frac {\sech u \tanh u} {\sech u \tanh u} \rd u
+      | c = [[Sum of Squares of Hyperbolic Secant and Tangent]]
+}}
+{{eqn | r = -\frac 1 a \int 1 \rd u
+}}
+{{eqn | r = -\frac 1 a u + C
+      | c = [[Integral of Constant]]
+}}
+{{eqn | r = -\frac 1 a \sech^{-1} {\frac x a} + C
+      | c = Definition of $u$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of x by Root of a x + b}
+Tags: Primitives involving Root of a x + b
+
+\begin{theorem}
+:$\displaystyle \int x \sqrt{a x + b} \ \mathrm d x = \frac {2 \left({3 a x - 2 b}\right) } {15 a^2} \sqrt{\left({a x + b}\right)^3}$
+\end{theorem}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = u
+      | r = \sqrt{a x + b}
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = x
+      | r = \frac {u^2 - b} a
+      | c = 
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = \int x \sqrt{a x + b} \ \mathrm d x
+      | r = \frac 2 a \int \frac {u^2 - b} a u^2 \ \mathrm d u
+      | c = [[Primitive of Function of Root of a x + b|Primitive of Function of $\sqrt{a x + b}$]]
+}}
+{{eqn | r = \frac 2 {a^2} \int u^4 \ \mathrm d u - \frac {2 b} {a^2} \int u^2 \ \mathrm d u
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \frac 2 {a^2} \frac {u^5} 5 - \frac {2 b} {a^2} \frac {u^3} 3 + C
+      | c = [[Primitive of Power]]
+}}
+{{eqn | r = \frac {6 u^5 - 10 b u^3} {15 a^2} + C
+      | c = common [[Definition:Denominator|denominator]]
+}}
+{{eqn | r = \frac {\left({6 u^2 - 10 b}\right) u^3} {15 a^2} + C
+      | c = extracting $u^3$ as a factor
+}}
+{{eqn | r = \frac {6 \left({a x + b}\right) - 10 b} {15 a^2} \sqrt{\left({a x + b}\right)^3} + C
+      | c = substituting for $u$
+}}
+{{eqn | r = \frac {2 \left({3 a x - 2 b}\right)  } {15 a^2} \sqrt{\left({a x + b}\right)^3} + C
+      | c = simplifying
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Laplace Transform of Heaviside Step Function times Function}
+Tags: Laplace Transforms, Heaviside Step Function
+
+\begin{theorem}
+Let $\map f t: \R \to \R$ or $\R \to \C$ be a [[Definition:Function|function]] of [[Definition:Exponential Order|exponential order $a$]] for some [[Definition:Constant|constant]] $a \in \R$.
+Let $f$ be [[Definition:Piecewise Continuous Function with One-Sided Limits|piecewise continuous with one-sided limits]] on any [[Definition:Closed Real Interval|closed interval]] of the form $\closedint 0 b$ where $b > 0$.
+Let $\map {u_c} t$ be the [[Definition:Heaviside Step Function|Heaviside step function]].
+Let $\laptrans {\map f t} = \map F s$ denote the [[Definition:Laplace Transform|Laplace transform]] of $f$.
+Then:
+:$\laptrans {\map {u_c} t \, \map f {t - c} } = e^{-s c} \map F s$
+for $\map \Re s > a$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \laptrans {\map {u_c} t \, \map f {t - c} }
+      | r = \int_0^{\to +\infty} \map {u_c} t e^{-s t} \map f {t - c} \rd t
+      | c = {{Defof|Laplace Transform}}
+}}
+{{eqn | r = \int_0^{\to c^-} \map {u_c} t e^{-s t} \map f {t - c} \rd t + \int_{\to c^+}^{\to +\infty} \map {u_c} t e^{-s t} \map f {t - c} \rd t
+      | c = [[Sum of Complex Integrals on Adjacent Intervals]]
+}}
+{{eqn | r = \int_0^{\to c^-} 0 \times e^{-s t} \map f {t - c} \rd t + \int_{\to c^+}^{\to +\infty} 1 \times e^{-s t} \map f {t - c} \rd t
+      | c = {{Defof|Heaviside Step Function}}
+}}
+{{eqn | r = \int_{\to c^+}^{\to +\infty} e^{-s t} \map f {t - c} \rd t
+}}
+{{end-eqn}}
+Let $u = t - c$.
+Then $\dfrac {\rd u} {\rd t} = 1$.
+Then $u \to 0^-$ as $t \to c^+$.
+Also, $u \to +\infty$ as $t \to +\infty$.
+So:
+{{begin-eqn}}
+{{eqn | l = \int_{\to c^+}^{\to +\infty} e^{-s t} \map f {t - c} \rd t
+      | r = \int_{\to c^+}^{\to +\infty} e^{-s \paren {u + c} } \map f u \frac {\rd u} {\rd t} \rd t
+}}
+{{eqn | r = \int_{\to 0^-}^{\to +\infty} e^{-s \paren {u + c} } \map f u \rd u
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r =  \int_0^{\to 0^+} e^{-s \paren {u + c} } \map f u \rd u + \int_{\to 0^-}^{\to +\infty} e^{-s \paren {u + c} } \map f u \rd u
+      | c = [[Integral on Zero Interval]]
+}}
+{{eqn | r = \int_0^{\to +\infty} e^{-s \paren {u + c} } \map f u \rd u
+      | c = [[Sum of Complex Integrals on Adjacent Intervals]]
+}}
+{{eqn | r = \int_0^{\to +\infty} e^{-s u - s c} \map f u \rd u
+}}
+{{eqn | r = e^{-s c} \int_0^{\to +\infty} e^{-s u} \map f u \rd u
+      | c = [[Exponent Combination Laws]]
+}}
+{{eqn | r = e^{-s c} \int_0^{\to +\infty} e^{-s t} \map f t \rd t
+      | c = renaming the variable of integration
+}}
+{{eqn | r = e^{-s c} \laptrans {\map f t}
+      | c = {{Defof|Laplace Transform}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of x squared by Root of a x + b}
+Tags: Primitives involving Root of a x + b
+
+\begin{theorem}
+:$\displaystyle \int x^2 \sqrt{a x + b} \ \mathrm d x = \frac {2 \left({15 a^2 x^2 - 12 a b x + 8 b^2}\right) } {105 a^3} \sqrt{\left({a x + b}\right)^3} + C$
+\end{theorem}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = u
+      | r = \sqrt{a x + b}
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = x
+      | r = \frac {u^2 - b} a
+      | c = 
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = \int x \sqrt{a x + b} \ \mathrm d x
+      | r = \frac 2 a \int \left({\frac {u^2 - b} a}\right)^2 u^2 \ \mathrm d u
+      | c = [[Primitive of Function of Root of a x + b|Primitive of Function of $\sqrt{a x + b}$]]
+}}
+{{eqn | r = \frac 2 {a^3} \left({\int u^6 \ \mathrm d u - 2 b \int u^4 \ \mathrm d u + b^2 \int u^2 \ \mathrm d u}\right)
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \frac 2 {a^3} \left({\frac {u^7} 7 - 2 b \frac {u^5} 5 + b^2 \frac {u^3} 3}\right) + C
+      | c = [[Primitive of Power]]
+}}
+{{eqn | r = \frac 2 {a^3} \left({\frac {15 u^4 - 42 b u^2 + 35 b^2} {105} }\right) u^3 + C
+      | c = common [[Definition:Denominator|denominator]] and $u^3$ as a factor
+}}
+{{eqn | r = \frac 2 {a^3} \left({\frac {15 \left({a x + b}\right)^2 - 42 b \left({a x + b}\right) + 35 b^2} {105} }\right) \sqrt{\left({a x + b}\right)^3} + C
+      | c = substituting for $u$
+}}
+{{eqn | r = \frac {2 \left({15 a^2 x^2 + 30 a b x + 15 b^2 - 42 a b x - 42 b^2 + 35 b^2}\right)} {105 a^3} \sqrt{\left({a x + b}\right)^3} + C
+      | c = multiplying out
+}}
+{{eqn | r = \frac {2 \left({15 a^2 x^2 - 12 a b x + 8 b^2}\right)} {105 a^3} \sqrt{\left({a x + b}\right)^3} + C
+      | c = simplifying
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Root of a x + b over x}
+Tags: Primitives involving Root of a x + b, Primitive of Root of a x + b over x
+
+\begin{theorem}
+:$\displaystyle \int \frac {\sqrt {a x + b} } x \rd x = 2 \sqrt {a x + b} + b \int \frac {\d x} {x \sqrt{a x + b} }$
+\end{theorem}
+
+\begin{proof}
+From [[Reduction Formula for Primitive of Power of x by Power of a x + b/Decrement of Power of a x + b|Reduction Formula for Primitive of Power of $x$ by Power of $a x + b$: Decrement of Power of $a x + b$]]:
+:$\displaystyle \int x^m \paren {a x + b}^n \rd x = \frac {x^{m + 1} \paren {a x + b}^n} {m + n + 1} + \frac {n b} {m + n + 1} \int x^m \paren {a x + b}^{n - 1} \rd x$
+Putting $m = -1$ and $n = \dfrac 1 2$:
+{{begin-eqn}}
+{{eqn | l = \int \frac {\sqrt {a x + b} } x \rd x
+      | r = \int x^{-1} \paren {a x + b}^{1/2} \rd x
+      | c = 
+}}
+{{eqn | r = \frac {x^0 \paren {a x + b}^{1/2} } {\frac 1 2} + \frac {\frac 1 2 b} {\frac 1 2} \int x^{-1} \paren {a x + b}^{- 1/2} \rd x
+      | c = 
+}}
+{{eqn | r = 2 \sqrt {a x + b} + b \int \frac {\d x} {x \sqrt {a x + b} }
+      | c = simplifying
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = v
+      | r = \sqrt x
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \frac {\d v} {\d x}
+      | r = \frac 1 {2 \sqrt x}
+      | c = [[Power Rule for Derivatives]]
+}}
+{{eqn | l = u
+      | r = \frac {2 \sqrt {a x + b} } {\sqrt x}
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \frac {\d u} {\d x}
+      | r = \frac {\frac {\sqrt x \cdot 2 a} {2 \sqrt{a x + b} } - \frac {2 \sqrt {a x + b} } {2 \sqrt x} } x
+      | c = [[Quotient Rule for Derivatives]] etc.
+}}
+{{eqn | r = \frac {-b} {x^{3/2} \sqrt {a x + b} }
+      | c = simplifying
+}}
+{{end-eqn}}
+From [[Integration by Parts]]:
+: $\displaystyle \int u \frac {\d v} {\d x} \rd x = u v - \int v \ \frac {\d u} {\d x} \rd x$
+from which:
+{{begin-eqn}}
+{{eqn | l = \int \frac {\sqrt{a x + b} } x \rd x
+      | r = \int \frac {2 \sqrt{a x + b} }{\sqrt x} \frac 1 {2 \sqrt x} \rd x
+      | c = 
+}}
+{{eqn | r = \frac {2 \sqrt{a x + b} } {\sqrt x} \sqrt x - \int {\sqrt x} \frac {-b} {x^{3/2} \sqrt {a x + b} } \rd x
+      | c = 
+}}
+{{eqn | r = 2 \sqrt {a x + b} + b \int \frac {\d x} {x \sqrt {a x + b} }
+      | c = simplification
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Root of a x + b over x squared}
+Tags: Primitives involving Root of a x + b
+
+\begin{theorem}
+:$\displaystyle \int \frac {\sqrt {a x + b} } {x^2} \rd x = -\frac {\sqrt {a x + b} } x + \frac a 2 \int \frac {\d x} {x \sqrt {a x + b} }$
+\end{theorem}
+
+\begin{proof}
+From [[Reduction Formula for Primitive of Power of x by Power of a x + b/Decrement of Power of a x + b|Reduction Formula for Primitive of Power of $x$ by Power of $a x + b$: Decrement of Power of $a x + b$]]:
+:$\displaystyle \int x^m \paren {a x + b}^n \rd x = \frac {x^{m + 1} \paren {a x + b}^n} {m + n + 1} + \frac {n b} {m + n + 1} \int x^m \paren {a x + b}^{n - 1} \rd x$
+Putting $m = -2$ and $n = \dfrac 1 2$:
+{{begin-eqn}}
+{{eqn | l = \int \frac {\sqrt {a x + b} } {x^2} \rd x
+      | r = \int x^{-2} \paren {a x + b}^{1/2} \rd x
+      | c = 
+}}
+{{eqn | r = \frac {x^{-1} \paren {a x + b}^{1/2} } {-2 + \frac 1 2 + 1} + \frac {\frac 1 2 b} {-2 + \frac 1 2 + 1} \int x^{-2} \paren {a x + b}^{-1/2} \rd x
+      | c = 
+}}
+{{eqn | r = \frac {-2 \sqrt {a x + b} } x - b \int \frac {\d x} {x^2 \sqrt {a x + b} }
+      | c = simplifying
+}}
+{{eqn | r = \frac {-2 \sqrt {a x + b} } x - b \paren {-\frac {\sqrt {a x + b} } {b x} - \frac a {2 b} \int \frac {\d x} {x \sqrt {a x + b} } }
+      | c = [[Primitive of Reciprocal of x squared by Root of a x + b]]
+}}
+{{eqn | r = \frac {-2 \sqrt {a x + b} } x + \frac {\sqrt {a x + b} } x + \frac a 2 \int \frac {\d x} {x \sqrt {a x + b} }
+      | c = multiplying out
+}}
+{{eqn | r = -\frac {\sqrt {a x + b} } x + \frac a 2 \int \frac {\d x} {x \sqrt {a x + b} }
+      | c = simplifying
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Power of x over Root of a x + b}
+Tags: Primitives involving Root of a x + b
+
+\begin{theorem}
+:$\ds \int \frac {x^m} {\sqrt{a x + b} } \rd x = \frac {2 x^m \sqrt{a x + b} } {\paren {2 m + 1} a} - \frac {2 m b} {\paren {2 m + 1} a} \int \frac {x^{m - 1} } {\sqrt{a x + b} } \rd x$
+\end{theorem}
+
+\begin{proof}
+From [[Reduction Formula for Primitive of Power of x by Power of a x + b/Decrement of Power of x|Reduction Formula for Primitive of Power of $x$ by Power of $a x + b$: Decrement of Power of $x$]]:
+:$\ds \int x^m \paren {a x + b}^n \rd x = \frac {x^m \paren {a x + b}^{n + 1} } {\paren {m + n + 1} a} - \frac {m b} {\paren {m + n + 1} a} \int x^{m - 1} \paren {a x + b}^n \rd x$
+Putting $n = -\dfrac 1 2$:
+{{begin-eqn}}
+{{eqn | l = \int \frac {x^m} {\sqrt {a x + b} } \rd x
+      | r = \int x^m \paren {a x + b}^{-1/2} \rd x
+      | c = 
+}}
+{{eqn | r = \frac {x^m \paren {a x + b}^{1/2} } {\paren {m - \frac 1 2 + 1} a} - \frac {m b} {\paren {m - \frac 1 2 + 1} a} \int x^{m - 1} \paren {a x + b}^{-1/2} \rd x
+      | c = 
+}}
+{{eqn | r = \frac {x^m \sqrt {a x + b} } {\paren {m + \frac 1 2} a} - \frac {m b} {\paren {m + \frac 1 2} a} \int \frac {x^{m - 1} } {\sqrt {a x + b} } \rd x
+      | c = simplifying
+}}
+{{eqn | r = \frac {2 x^m \sqrt {a x + b} } {\paren {2 m + 1} a} - \frac {2 m b} {\paren {2 m + 1} a} \int \frac {x^{m - 1} } {\sqrt {a x + b} } \rd x
+      | c = multiplying [[Definition:Numerator|top]] and [[Definition:Denominator|bottom]] by $2$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Reciprocal of Power of x by Root of a x + b}
+Tags: Primitives involving Root of a x + b, Primitives involving Reciprocals
+
+\begin{theorem}
+:$\displaystyle \int \frac {\d x} {x^m \sqrt {a x + b} } = -\frac {\sqrt {a x + b} } {\paren {m - 1} b x^{m - 1} } - \frac {\paren {2 m - 3} a} {\paren {2 m - 2} b} \int \frac {\d x} {x^{m - 1} \sqrt {a x + b} }$
+\end{theorem}
+
+\begin{proof}
+From [[Reduction Formula for Primitive of Power of x by Power of a x + b/Increment of Power of x|Reduction Formula for Primitive of Power of $x$ by Power of $a x + b$: Increment of Power of $x$]]:
+:$\displaystyle \int x^m \paren {a x + b}^n \rd x = \frac {x^{m + 1} \paren {a x + b}^{n + 1} } {\paren {m + 1} b} - \frac {\paren {m + n + 2} a} {\paren {m + 1} b} \int x^{m + 1} \paren {a x + b}^n \rd x$
+Putting $n := -\dfrac 1 2$ and $m := -m$:
+{{begin-eqn}}
+{{eqn | l = \int \frac {\d x} {x^m \sqrt{a x + b} }
+      | r = \int x^{-m} \paren {a x + b}^{-1/2} \rd x
+      | c = 
+}}
+{{eqn | r = \frac {x^{-m+1} \paren {a x + b}^{1/2} } {\paren {-m + 1} b} - \frac {\paren {-m - \frac 1 2 + 2} a} {\paren {-m + 1} b} \int x^{-m + 1} \paren {a x + b}^{-1/2} \rd x
+      | c = 
+}}
+{{eqn | r = -\frac {\sqrt {a x + b} } {\paren {m - 1} b x^{m - 1} } - \frac {\paren {m - \frac 3 2} a} {\paren {m - 1} b} \int \frac {\d x} {x^{m - 1} \sqrt {a x + b} }
+      | c = simplifying
+}}
+{{eqn | r = -\frac {\sqrt {a x + b} } {\paren {m - 1} b x^{m - 1} } - \frac {\paren {2 m - 3} a} {\paren {2 m - 2} b} \int \frac {\d x} {x^{m - 1} \sqrt {a x + b} }
+      | c = multiplying [[Definition:Numerator|top]] and [[Definition:Denominator|bottom]] by $2$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Root of a x + b over Power of x}
+Tags: Primitives involving Root of a x + b
+
+\begin{theorem}
+=== [[Primitive of Root of a x + b over Power of x/Formulation 1|Formulation 1]] ===
+{{:Primitive of Root of a x + b over Power of x/Formulation 1}}
+=== [[Primitive of Root of a x + b over Power of x/Formulation 2|Formulation 2]] ===
+{{:Primitive of Root of a x + b over Power of x/Formulation 2}}
+\end{theorem}<|endoftext|>
+\section{Primitive of Power of Root of a x + b}
+Tags: Primitives involving Root of a x + b
+
+\begin{theorem}
+:$\displaystyle \int \left({\sqrt{a x + b} }\right)^m \ \mathrm d x = \frac {2 \left({\sqrt{a x + b} }\right)^{m + 2} } {a \left({m + 2}\right)} + C$
+\end{theorem}
+
+\begin{proof}
+Let $u = \sqrt{a x + b}$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \int \left({\sqrt{a x + b} }\right)^m \ \mathrm d x
+      | r = \frac 2 a \int u^{m + 1} \ \mathrm d x
+      | c = [[Primitive of Function of a x + b|Primitive of Function of $a x + b$]]
+}}
+{{eqn | r = \frac 2 a \frac {u^{m + 2} } {m + 2} + C
+      | c = [[Primitive of Power]]
+}}
+{{eqn | r = \frac {2 \left({\sqrt{a x + b} }\right)^{m + 2} } {a \left({m + 2}\right)} + C
+      | c = substituting for $u$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Power of Root of a x + b over x}
+Tags: Primitives involving Root of a x + b
+
+\begin{theorem}
+:$\displaystyle \int \frac {\left({\sqrt{a x + b} }\right)^m} x \ \mathrm d x = \frac {2 \left({\sqrt{a x + b} }\right)^m } m + b \int \frac {\left({\sqrt{a x + b} }\right)^{m - 2} } x \ \mathrm d x$
+\end{theorem}
+
+\begin{proof}
+From [[Reduction Formula for Primitive of Power of x by Power of a x + b/Decrement of Power of a x + b|Reduction Formula for Primitive of Power of $x$ by Power of $a x + b$: Decrement of Power of $a x + b$]]:
+:$\displaystyle \int x^m \left({a x + b}\right)^n \ \mathrm d x = \frac {x^{m+1} \left({a x + b}\right)^n} {m + n + 1} + \frac {n b} {m + n + 1} \int x^m \left({a x + b}\right)^{n - 1} \ \mathrm d x$
+Putting $n := \dfrac m 2$ and $m := -1$:
+{{begin-eqn}}
+{{eqn | l = \int \frac {\left({\sqrt{a x + b} }\right)^m} x \ \mathrm d x
+      | r = \int x^{-1} \left({a x + b}\right)^{m/2} \ \mathrm d x
+      | c = 
+}}
+{{eqn | r = \frac {x^0 \left({a x + b}\right)^{m/2} } {-1 + \frac m 2 + 1} + \frac {\left({\frac m 2}\right) b} {-1 + \frac m 2 + 1} \int x^{-1} \left({a x + b}\right)^{m/2 - 1} \ \mathrm d x
+      | c = 
+}}
+{{eqn | r = \frac {2 \left({a x + b}\right)^{m/2} } m + b \int x^{-1} \left({a x + b}\right)^{\left({m - 2}\right)/2} \ \mathrm d x
+      | c = simplifying
+}}
+{{eqn | r = \frac {2 \left({\sqrt{a x + b} }\right)^m } m + b \int \frac {\left({\sqrt{a x + b} }\right)^{m - 2} } x \ \mathrm d x
+      | c =
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Reciprocal of x by Power of Root of a x + b}
+Tags: Primitives involving Root of a x + b, Primitives involving Reciprocals
+
+\begin{theorem}
+:$\displaystyle \int \frac {\mathrm d x} {x \left({\sqrt{a x + b} }\right)^m} = \frac 2 {\left({m - 2}\right) b \left({\sqrt{a x + b} }\right)^{m - 2} }  + \frac 1 b \int \frac {\mathrm d x} {x \left({\sqrt{a x + b} }\right)^{m - 2} }$
+\end{theorem}
+
+\begin{proof}
+From [[Reduction Formula for Primitive of Power of x by Power of a x + b/Increment of Power of a x + b|Reduction Formula for Primitive of Power of $x$ by Power of $a x + b$: Increment of Power of $a x + b$]]:
+:$\displaystyle \int x^m \left({a x + b}\right)^n \ \mathrm d x = \frac {-x^{m+1} \left({a x + b}\right)^{n + 1} } {\left({n + 1}\right) b} + \frac {m + n + 2} {\left({n + 1}\right) b} \int x^m \left({a x + b}\right)^{n + 1} \ \mathrm d x$
+Putting $n := -\dfrac m 2$ and $m := -1$:
+{{begin-eqn}}
+{{eqn | l = \int \frac {\mathrm d x} {x \left({\sqrt{a x + b} }\right)^m}
+      | r = \int x^{-1} \left({a x + b}\right)^{-m/2}
+      | c = 
+}}
+{{eqn | r = \frac {-x^0 \left({a x + b}\right)^{-m/2 + 1} } {\left({- \frac m 2 + 1}\right) b} + \frac {-1 - \frac m 2 + 2} {\left({- \frac m 2 + 1}\right) b} \int x^{-1} \left({a x + b}\right)^{- m/2 + 1} \ \mathrm d x
+      | c = 
+}}
+{{eqn | r = \frac 2 {\left({m - 2}\right) b \left({a x + b}\right)^{\left({m - 2}\right)/2} } - \frac 1 b \int \frac {\mathrm d x} {x \left({a x + b}\right)^{\left({m - 2}\right) / 2} }
+      | c = simplifying
+}}
+{{eqn | r = \frac 2 {\left({m - 2}\right) b \left({\sqrt{a x + b} }\right)^{m - 2} }  + \frac 1 b \int \frac {\mathrm d x} {x \left({\sqrt{a x + b} }\right)^{m - 2} }
+      | c =
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Reciprocal of a x + b by p x + q}
+Tags: Primitives involving a x + b and p x + q
+
+\begin{theorem}
+:$\displaystyle \int \frac {\mathrm d x} {\left({a x + b}\right) \left({p x + q}\right)} = \frac 1 {b p - a q} \ln \left\vert{\frac {p x + q} {a x + b} }\right\vert + C$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int \frac {\mathrm d x} {\left({a x + b}\right) \left({p x + q}\right)}
+      | r = \int \left({\frac {-a} {\left({b p - a q}\right) \left({a x + b}\right)} + \frac p {\left({b p - a q}\right) \left({p x + q}\right)} }\right)\ \mathrm d x
+      | c = [[Primitive of Reciprocal of a x + b by p x + q/Partial Fraction Expansion|Partial Fraction Expansion]]
+}}
+{{eqn | r = \frac 1 {b p - a q} \left({ {-a} \int \frac 1 {a x + b} \ \mathrm d x + p \int \frac 1 {p x + q} \ \mathrm d x}\right)
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \frac 1 {b p - a q} \left({\frac {-a} a \ln \left\vert{a x + b}\right\vert + \frac p p \ln \left\vert{p x + q}\right\vert}\right) + C
+      | c = [[Primitive of Reciprocal of a x + b|Primitive of $\dfrac 1 {a x + b}$]]
+}}
+{{eqn | r = \frac 1 {b p - a q} \left({\ln \left\vert{p x + q}\right\vert - \ln \left\vert{a x + b}\right\vert}\right) + C
+      | c = simplification
+}}
+{{eqn | r = \frac 1 {b p - a q} \ln \left\vert{\frac {p x + q} {a x + b} }\right\vert + C
+      | c = [[Difference of Logarithms]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of x over a x + b by p x + q}
+Tags: Primitives involving a x + b and p x + q
+
+\begin{theorem}
+:$\displaystyle \int \frac {x \ \mathrm d x} {\left({a x + b}\right) \left({p x + q}\right)} = \frac 1 {b p - a q} \left({\frac b a \ln \left\vert{a x + b}\right\vert - \frac q p \ln \left\vert{p x + q}\right\vert}\right) + C$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int \frac {x \ \mathrm d x} {\left({a x + b}\right) \left({p x + q}\right)}
+      | r = \int \left({\frac b {\left({b p - a q}\right) \left({a x + b}\right)} - \frac q {\left({b p - a q}\right) \left({p x + q}\right)} }\right)\ \mathrm d x
+      | c = [[Primitive of x over a x + b by p x + q/Partial Fraction Expansion|Partial Fraction Expansion]]
+}}
+{{eqn | r = \frac 1 {b p - a q} \left({b \int \frac 1 {a x + b} \ \mathrm d x - q \int \frac 1 {p x + q} \ \mathrm d x}\right)
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \frac 1 {b p - a q} \left({\frac b a \ln \left\vert{a x + b}\right\vert - \frac q p \ln \left\vert{p x + q}\right\vert}\right) + C
+      | c = [[Primitive of Reciprocal of a x + b|Primitive of $\dfrac 1 {a x + b}$]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Reciprocal of a x + b squared by p x + q}
+Tags: Primitives involving a x + b and p x + q
+
+\begin{theorem}
+:$\displaystyle \int \frac {\mathrm d x} {\left({a x + b}\right)^2 \left({p x + q}\right)} = \frac 1 {b p - a q} \left({\frac 1 {a x + b} + \frac p {b p - a q} \ln \left\vert{\frac {p x + q} {a x + b} }\right\vert}\right) + C$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | o = 
+      | r = \int \frac {\mathrm d x} {\left({a x + b}\right)^2 \left({p x + q}\right)}
+      | c = 
+}}
+{{eqn | r = \int \left({\frac 1 {b p - a q} \left({\frac {-a p} {\left({b p - a q}\right) \left({a x + b}\right)} + \frac {-a} {\left({a x + b}\right)^2} + \frac {p^2} {\left({b p - a q}\right) \left({p x + q}\right)} }\right)}\right) \ \mathrm d x
+      | c = [[Primitive of Reciprocal of a x + b squared by p x + q/Partial Fraction Expansion|Partial Fraction Expansion]]
+}}
+{{eqn | r = \frac 1 {b p - a q} \left({\frac {-a p} {b p - a q} \int \frac {\mathrm d x} {a x + b} - a \int \frac {\mathrm d x} {\left({a x + b}\right)^2} + \frac {p^2} {b p - a q} \int \frac {\mathrm d x} {p x + q} }\right)
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \frac 1 {b p - a q} \left({\frac {-a p} {b p - a q} \frac 1 a \ln \left\vert{a x + b}\right\vert - a \int \frac {\mathrm d x} {\left({a x + b}\right)^2} + \frac {p^2} {b p - a q} \frac 1 p \ln \left\vert{p x + q}\right\vert }\right) + C
+      | c = [[Primitive of Reciprocal of a x + b|Primitive of Reciprocal of $a x + b$]]
+}}
+{{eqn | r = \frac 1 {b p - a q} \left({\frac {-p} {b p - a q} \ln \left\vert{a x + b}\right\vert - a \frac {-1} {a \left({a x + b}\right)} + \frac p {b p - a q} \ln \left\vert{p x + q}\right\vert }\right) + C
+      | c = [[Primitive of Reciprocal of a x + b squared|Primitive of Reciprocal of $\left({a x + b}\right)^2$]]
+}}
+{{eqn | r = \frac 1 {b p - a q} \left({\frac 1 {a x + b} + \frac p {b p - a q} \ln \left\vert{\frac {p x + q} {a x + b} }\right\vert}\right) + C
+      | c = [[Difference of Logarithms]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of x over a x + b squared by p x + q}
+Tags: Primitives involving a x + b and p x + q
+
+\begin{theorem}
+:$\displaystyle \int \frac {x \rd x} {\paren {a x + b}^2 \paren {p x + q} } = \frac 1 {b p - a q} \paren {\frac q {b p - a q} \ln \size {\frac {a x + b} {p x + q} } - \frac b {a \paren {a x + b} } } + C$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | o = 
+      | r = \int \frac {x \rd x} {\paren {a x + b}^2 \paren {p x + q} }
+      | c = 
+}}
+{{eqn | r = \int \paren {\frac 1 {b p - a q} \paren {\frac {a q} {\paren {b p - a q} \paren {a x + b} } + \frac b {\paren {a x + b}^2} - \frac {p q} {\paren {b p - a q} \paren {p x + q} } } } \rd x
+      | c = [[Primitive of x over a x + b squared by p x + q/Partial Fraction Expansion|Partial Fraction Expansion]]
+}}
+{{eqn | r = \frac 1 {b p - a q} \paren {\frac {a q} {b p - a q} \int \frac {\d x} {a x + b} + b \int \frac {\d x} {\paren {a x + b}^2} - \frac {p q} {b p - a q} \int \frac {\d x} {p x + q} }
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \frac 1 {b p - a q} \paren {\frac {a q} {b p - a q} \frac 1 a \ln \size {a x + b} + b \int \frac {\d x} {\paren {a x + b}^2} + \frac {p q} {b p - a q} \frac 1 p \ln \size {p x + q} } + C
+      | c = [[Primitive of Reciprocal of a x + b|Primitive of $\dfrac 1 {a x + b}$]]
+}}
+{{eqn | r = \frac 1 {b p - a q} \paren {\frac q {b p - a q} \ln \size {a x + b} + b \frac {-1} {a \paren {a x + b} } - \frac q {b p - a q} \ln \size {p x + q} } + C
+      | c = [[Primitive of Reciprocal of a x + b squared|Primitive of $\dfrac 1 {\left({a x + b}\right)^2}$]]
+}}
+{{eqn | r = \frac 1 {b p - a q} \paren {\frac q {b p - a q} \ln \size {\frac {a x + b} {p x + q} } - \frac b {a \paren {a x + b} } } + C
+      | c = [[Difference of Logarithms]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of x squared over a x + b squared by p x + q}
+Tags: Primitives involving a x + b and p x + q
+
+\begin{theorem}
+:$\displaystyle \int \frac {x^2 \ \mathrm d x} {\left({a x + b}\right)^2 \left({p x + q}\right)} = \frac {b^2} {\left({b p - a q}\right) a^2 \left({a x + b}\right)} + \frac 1 {\left({b p - a q}\right)^2} \left({\frac {q^2} p \ln \left\vert{p x + q}\right\vert + \frac {b \left({b p - 2 a q}\right)} {a^2} \ln \left\vert{a x + b}\right\vert}\right) + C$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | o = 
+      | r = \int \frac {x^2 \ \mathrm d x} {\left({a x + b}\right)^2 \left({p x + q}\right)}
+      | c = 
+}}
+{{eqn | r = \int \left({\frac {-b^2} {a \left({b p - a q}\right) \left({a x + b}\right)^2} + \frac {q^2} {\left({b p - a q}\right)^2 \left({p x + q}\right)} + \frac {b \left({b p - 2 a q}\right)} {a \left({b p - a q}\right)^2 \left({a x + b}\right)} }\right) \ \mathrm d x
+      | c = [[Primitive of x squared over a x + b squared by p x + q/Partial Fraction Expansion|Partial Fraction Expansion]]
+}}
+{{eqn | r = \frac {-b^2} {a \left({b p - a q}\right)} \int \frac {\mathrm d x} {\left({a x + b}\right)^2} + \frac {q^2} {\left({b p - a q}\right)^2} \int \frac {\mathrm d x} {p x + q} + \frac {b \left({b p - 2 a q}\right)} {a \left({b p - a q}\right)^2} \int \frac {\mathrm d x} {a x + b}
+      | c = [[Linear Combination of Integrals]]
+}}
+{{end-eqn}}
+We have:
+{{begin-eqn}}
+{{eqn | l = \int \frac {\mathrm d x} {\left({a x + b}\right)^2}
+      | r = \frac {-1} {a \left({a x + b}\right)}
+      | c = [[Primitive of Reciprocal of a x + b squared|Primitive of Reciprocal of $\left({a x + b}\right)^2$]]
+}}
+{{eqn | l = \int \frac {\mathrm d x} {p x + q}
+      | r = \frac 1 p \ln \left \vert {p x + q} \right \vert
+      | c = [[Primitive of Reciprocal of a x + b|Primitive of Reciprocal of $a x + b$]]
+}}
+{{eqn | l = \int \frac {\mathrm d x} {a x + b}
+      | r = \frac 1 a \ln \left \vert {a x + b} \right \vert
+      | c = [[Primitive of Reciprocal of a x + b|Primitive of Reciprocal of $a x + b$]]
+}}
+{{end-eqn}}
+Therefore:
+{{begin-eqn}}
+{{eqn | o = 
+      | r = \int \frac {x^2 \ \mathrm d x} {\left({a x + b}\right)^2 \left({p x + q}\right)}
+      | c = 
+}}
+{{eqn | r = \frac {-b^2} {a \left({b p - a q}\right)} \frac {-1} {a \left({a x + b}\right)} + \frac {q^2} {\left({b p - a q}\right)^2} \left({\frac 1 p \ln \left\vert{p x + q}\right\vert}\right) + \frac {b \left({b p - 2 a q}\right)} {a \left({b p - a q}\right)^2} \left({\frac 1 a \ln \left\vert{a x + b}\right\vert}\right) + C
+      | c = Substitution
+}}
+{{eqn | r = \frac {b^2} {\left({b p - a q}\right) a^2 \left({a x + b}\right)} + \frac 1 {\left({b p - a q}\right)^2} \left({\frac {q^2} p \ln \left\vert{p x + q}\right\vert + \frac {b \left({b p - 2 a q}\right)} {a^2} \ln \left\vert{a x + b}\right\vert}\right) + C
+      | c = simplifying
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Friedrichs' Inequality}
+Tags: 
+
+\begin{theorem}
+Let $G \subset \R^n$ be bounded domain.
+Then for any $u \in \map {W^{2, 1}_0} G$:
+:$\norm u_{\map {L^2} G} \le \map {\operatorname {diam} } G \norm {\nabla u}_{\map {L^2} G}$
+where:
+:$\map {\operatorname {diam} } G := \sup \limits_{x, y \mathop \in G} \size {x - y}$
+\end{theorem}
+
+\begin{proof}
+=== Smooth functions with compact support ===
+{{explain|In several places in the below, this construct was seen: $\vert \nabla u (x_1, \dots, x_{m - 1}, t\vert^2$ where it appears that the close of the round bracket parenthesis "$)$" was been omitted. It has been assmued in the corrected version that it should be this: $\vert\nabla u (x_1,\dots,x_{m-1},t)\vert^2$ but it is not obvious and may require someone knowledgeable to check this out.}}
+Let $u \in \map {C_0^\infty} G$.
+Put $\map u x = 0$ if $x = \tuple {x_1, x_2, \ldots, x_n} \notin G$.
+Then:
+:$u \in \map {C_0^\infty} {\R^n}$
+Denote by $a,b$ the extremes of the last coordinate: $a = \inf \limits_{x \mathop \in G} x_n$ and $b = \sup \limits_{y \mathop \in G} x_n$
+Then for any $x$:
+:$\displaystyle \size {\map u {x_1, x_2, \ldots, x_n} }^2 = \size {\int \limits_a^{x_n} \frac {\partial u} {\partial x_n} \tuple {x_1, x_2, \ldots, x_{n - 1}, t} \rd t}^2$
+By [[Cauchy-Bunyakovsky-Schwarz Inequality]]:
+{{begin-eqn}}
+{{eqn | l = \left\vert{\int \limits_a^{x_n} \frac {\partial u} {\partial x_n} \left({x_1, \ldots, x_{n-1}, t}\right) \rd t}\right\vert^2
+      | o = \le
+      | r = \int \limits_a^{x_n} 1^2 \ \mathrm d t \int \limits_a^{x_n} \left\vert{\frac{\partial u} {\partial x_n} \left({x_1, \ldots, x_{n-1}, t}\right)}\right\vert^2 \rd t
+      | c = 
+}}
+{{eqn | o = \le
+      | r = \map {\operatorname {diam} } G \int \limits_a^b \size {\map {\nabla u} {x_1, \ldots, x_{m - 1}, t} }^2 \rd t
+      | c = 
+}}
+{{end-eqn}}
+Integrating this we get:
+:$\displaystyle \norm u_{\map {L^2} G} \le \map {\operatorname {diam} } G \int \limits_G \paren {\int \limits_a^b \size {\map {\nabla u} {x_1, \ldots, x_{m - 1}, t} }^2 \rd t} \rd x$
+By [[Fubini's Theorem]]:
+{{begin-eqn}}
+{{eqn | l = \int \limits_G \paren {\int \limits_a^b \size {\map {\nabla u} {x_1, \ldots, x_{m - 1}, t} }^2 \rd t} \rd x
+      | r = \int \limits_a^b \rd x_m \int \limits_{\mathbb R^n} \size {\map {\nabla u} {x_1, \ldots, x_{n - 1}, x_m} }^2 \rd x
+      | c = 
+}}
+{{eqn | o = \le
+      | r = \map {\operatorname {diam}^2} G \norm {\nabla u}^2_{\map {L^2} G}
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+=== General case ===
+Let now $u \in \map {W^{2, 1}_0} G$.
+There exists a sequence $\displaystyle \sequence {u_n}_{n \mathop = 1}^\infty \subset \map {C_0^\infty} G$ such that:
+:$\norm {u - u_n}_{\map {W^{2, 1} } G} \to 0$
+as $n \to \infty$.
+Then:
+:$\norm {u - u_n}_{\map {L^2} G} \to 0$
+and:
+:$\norm {\nabla u - \nabla u_n}_{\map {L^2} G} \to 0$
+As $\size {\, \norm {u - u_n} \,} \le \norm {u - u_n}$, it follows that:
+:$\norm {u_n}_{\map {L^2} G} \to \norm u_{\map {L^2} G}$
+and:
+:$\norm {\nabla u_n}_{\map {L^2} G} \to \norm {\nabla u}_{\map {L^2} G}$
+Since the inequality is correct for all $u_n$, it is also correct for $u$.
+{{qed}}
+{{Namedfor|Kurt Otto Friedrichs|cat = Friedrichs}}
+13j07onk2aoc8ytc21fm6g3towbg039
+\end{proof}<|endoftext|>
+\section{Primitive of Reciprocal of Power of a x + b by Power of p x + q}
+Tags: Primitives involving a x + b and p x + q
+
+\begin{theorem}
+:$\displaystyle \int \frac {\mathrm d x} {\left({a x + b}\right)^m \left({p x + q}\right)^n} = \frac {-1} {\left({n - 1}\right) \left({b p - a q}\right)} \left({\frac 1 {\left({a x + b}\right)^{m-1} \left({p x + q}\right)^{n-1} } + a \left({m + n - 2}\right) \int \frac {\mathrm d x} {\left({a x + b}\right)^m \left({p x + q}\right)^{n-1} } }\right)$
+\end{theorem}
+
+\begin{proof}
+From [[Reduction Formula for Primitive of Power of a x + b by Power of p x + q/Increment of Power|Reduction Formula for Primitive of Power of $a x + b$ by Power of $p x + q$: Increment of Power]]:
+:$\displaystyle \int \left({a x + b}\right)^m \left({p x + q}\right)^n \ \mathrm d x = \frac 1 {\left({n + 1}\right) \left({b p - a q}\right)} \left({\left({a x + b}\right)^{m+1} \left({p x + q}\right)^{n+1} - a \left({m + n + 2}\right) \int \left({a x + b}\right)^m \left({p x + q}\right)^{n+1} \ \mathrm d x}\right)$
+Setting $m := -m$ and $n := -n$:
+{{begin-eqn}}
+{{eqn | o = 
+      | r = \int \frac {\mathrm d x} {\left({a x + b}\right)^m \left({p x + q}\right)^n}
+      | c = 
+}}
+{{eqn | r = \int \left({a x + b}\right)^{-m} \left({p x + q}\right)^{-n} \ \mathrm d x
+      | c = 
+}}
+{{eqn | r = \frac 1 {\left({-n + 1}\right) \left({b p - a q}\right)} \left({\left({a x + b}\right)^{-m+1} \left({p x + q}\right)^{-n+1} - a \left({-m - n + 2}\right) \int \left({a x + b}\right)^{-m} \left({p x + q}\right)^{-n+1} \ \mathrm d x}\right)
+      | c = 
+}}
+{{eqn | r = \frac {-1} {\left({n - 1}\right) \left({b p - a q}\right)} \left({\frac 1 {\left({a x + b}\right)^{m-1} \left({p x + q}\right)^{n-1} } + a \left({m + n - 2}\right) \int \frac {\mathrm d x} {\left({a x + b}\right)^m \left({p x + q}\right)^{n-1} } }\right)
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Reduction Formula for Primitive of Power of a x + b by Power of p x + q/Decrement of Power}
+Tags: Primitives involving a x + b and p x + q
+
+\begin{theorem}
+:$\displaystyle \int \left({a x + b}\right)^m \left({p x + q}\right)^n \ \mathrm d x = \frac {\left({a x + b}\right)^{m+1} \left({p x + q}\right)^n} {\left({m + n + 1}\right) a} - \frac {n \left({b p - a q}\right)} {\left({m + n + 1}\right) a} \int \left({a x + b}\right)^m \left({p x + q}\right)^{n-1} \ \mathrm d x$
+\end{theorem}
+
+\begin{proof}
+Aiming for an expression in the form:
+: $\displaystyle \int u \frac {\mathrm d v} {\mathrm d x} \ \mathrm d x = u v - \int v \ \frac {\mathrm d u} {\mathrm d x} \ \mathrm d x$
+in order to use the technique of [[Integration by Parts]], let:
+{{begin-eqn}}
+{{eqn | l = v
+      | r = \left({a x + b}\right)^s
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \frac {\mathrm d v} {\mathrm d x}
+      | r = a s \left({a x + b}\right)^{s-1}
+      | c = [[Derivative of Power]] and [[Derivative of Function of Constant Multiple/Corollary|Derivative of Function of Constant Multiple: Corollary]]
+}}
+{{end-eqn}}
+In order to make $u \dfrac {\mathrm d v} {\mathrm d x}$ equal to the [[Definition:Integrand|integrand]], let:
+{{begin-eqn}}
+{{eqn | l = u
+      | r = \frac {\left({a x + b}\right)^{m - s + 1} } {a s} \left({p x + q}\right)^n
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \frac {\mathrm d u} {\mathrm d x}
+      | r = \frac {a \left({m - s + 1}\right) \left({a x + b}\right)^{m - s} \left({p x + q}\right)^n + p n \left({a x + b}\right)^{m - s + 1} \left({p x + q}\right)^{n-1} } {a s} 
+      | c = [[Product Rule for Derivatives]] and above
+}}
+{{eqn | r = \frac {\left({a x + b}\right)^{m - s} \left({p x + q}\right)^{n-1} } {a s} \left({a \left({m - s + 1}\right) \left({p x + q}\right) + p n \left({a x + b}\right) }\right)
+      | c = extracting common factor
+}}
+{{eqn | r = \frac {\left({a x + b}\right)^{m - s} \left({p x + q}\right)^{n-1} } {a s} \left({a p x \left({m - s + 1 + n}\right) + a q \left({m - s + 1}\right) + p b n}\right)
+      | c = separating out terms in $x$
+}}
+{{end-eqn}}
+Select $s$ such that $m - s + n + 1 = 0$, and so $s = m + n + 1$:
+{{begin-eqn}}
+{{eqn | l = \frac {\mathrm d u} {\mathrm d x}
+      | r = \frac {\left({a x + b}\right)^{m - s} \left({p x + q}\right)^{n-1} } {a \left({m + n + 1}\right)} \left({a q \left({m - \left({m + n + 1}\right) + 1}\right) + p b n}\right)
+      | c = term in $x$ vanishes
+}}
+{{eqn | r = \frac {\left({a x + b}\right)^{m - s} \left({p x + q}\right)^{n-1} } {a \left({m + n + 1}\right)} \left({n \left({p b - a q}\right)}\right)
+      | c = simplifying
+}}
+{{end-eqn}}
+Other instances of $s$ are left as they are, anticipating that they will cancel out later.
+Thus:
+{{begin-eqn}}
+{{eqn | o = 
+      | r = \int \left({a x + b}\right)^m \left({p x + q}\right)^n \ \mathrm d x
+      | c = 
+}}
+{{eqn | r = \int \frac {\left({a x + b}\right)^{m - s + 1} } {a s} \left({p x + q}\right)^n a s \left({a x + b}\right)^{s-1} \ \mathrm d x
+      | c = in the form $\displaystyle \int u \frac {\mathrm d v} {\mathrm d x} \ \mathrm d x$
+}}
+{{eqn | r = \frac {\left({a x + b}\right)^{m - s + 1} } {a s} \left({p x + q}\right)^n \left({a x + b}\right)^s
+      | c = in the form $\displaystyle u v - \int v \frac {\mathrm d u} {\mathrm d x} \ \mathrm d x$
+}}
+{{eqn | o = 
+      | ro= -
+      | r = \int \left({a x + b}\right)^s \frac {\left({a x + b}\right)^{m - s} \left({p x + q}\right)^{n-1} } {a \left({m + n + 1}\right)} \left({n \left({p b - a q}\right)}\right) \ \mathrm d x
+      | c = 
+}}
+{{eqn | r = \frac {\left({a x + b}\right)^{m+1} \left({p x + q}\right)^n} {\left({m + n + 1}\right) a} - \frac {n \left({b p - a q}\right)} {\left({m + n + 1}\right) a} \int \left({a x + b}\right)^m \left({p x + q}\right)^{n-1} \ \mathrm d x
+      | c = [[Primitive of Constant Multiple of Function]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Reduction Formula for Primitive of Power of a x + b by Power of p x + q/Increment of Power}
+Tags: Primitives involving a x + b and p x + q
+
+\begin{theorem}
+:$\displaystyle \int \left({a x + b}\right)^m \left({p x + q}\right)^n \ \mathrm d x = \frac 1 {\left({n + 1}\right) \left({b p - a q}\right)} \left({\left({a x + b}\right)^{m+1} \left({p x + q}\right)^{n+1} - a \left({m + n + 2}\right) \int \left({a x + b}\right)^m \left({p x + q}\right)^{n+1} \ \mathrm d x}\right)$
+\end{theorem}
+
+\begin{proof}
+From [[Reduction Formula for Primitive of Power of a x + b by Power of p x + q/Decrement of Power|Reduction Formula for Primitive of Power of $a x + b$ by Power of $p x + q$: Decrement of Power]]:
+:$\displaystyle \int \left({a x + b}\right)^m \left({p x + q}\right)^n \ \mathrm d x = \frac {\left({a x + b}\right)^{m+1} \left({p x + q}\right)^n} {\left({m + n + 1}\right) a} - \frac {n \left({b p - a q}\right)} {\left({m + n + 1}\right) a} \int \left({a x + b}\right)^m \left({p x + q}\right)^{n-1} \ \mathrm d x$
+Putting $n + 1$ for $n$, this yields:
+{{begin-eqn}}
+{{eqn | l = \int \left({a x + b}\right)^m \left({p x + q}\right)^{n+1} \ \mathrm d x
+      | r = \frac {\left({a x + b}\right)^{m+1} \left({p x + q}\right)^{n+1} } {\left({m + n + 2}\right) a} - \frac {\left({n + 1}\right) \left({b p - a q}\right)} {\left({m + n + 2}\right) a} \int \left({a x + b}\right)^m \left({p x + q}\right)^n \ \mathrm d x
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \int \left({a x + b}\right)^m \left({p x + q}\right)^n
+      | r = \frac {\left({a x + b}\right)^{m+1} \left({p x + q}\right)^{n+1} } {\left({n + 1}\right) \left({b p - a q}\right)} - \frac {\left({m + n + 2}\right) a} {\left({n + 1}\right) \left({b p - a q}\right)} \int \left({a x + b}\right)^m \left({p x + q}\right)^{n+1} \ \mathrm d x
+      | c = 
+}}
+{{eqn | r = \frac 1 {\left({n + 1}\right) \left({b p - a q}\right)} \left({\left({a x + b}\right)^{m+1} \left({p x + q}\right)^{n+1} - a \left({m + n + 2}\right) \int \left({a x + b}\right)^m \left({p x + q}\right)^{n+1} \ \mathrm d x}\right)
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of a x + b over p x + q}
+Tags: Primitives involving a x + b and p x + q
+
+\begin{theorem}
+:$\displaystyle \int \frac {a x + b} {p x + q} \ \mathrm d x  = \frac {a x} p + \frac {b p - a q} {p^2} \ln \left\vert{p x + q}\right\vert + C$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int \frac {a x + b} {p x + q} \ \mathrm d x
+      | r = a \int \frac {x \ \mathrm d x} {p x + q} + b \int \frac {\mathrm d x} {p x + q}
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = a \left({\frac x p - \frac q {p^2} \ln \left\vert{p x + q}\right\vert}\right) + b \int \frac {\mathrm d x} {p x + q} + C
+      | c = [[Primitive of x over a x + b|Primitive of $x$ over $a x + b$]]
+}}
+{{eqn | r = \frac {a x} p - \frac {a q} {p^2} \ln \left\vert{p x + q}\right\vert + b \left({\frac 1 p \ln \left\vert{p x + q}\right\vert}\right) + C
+      | c = [[Primitive of Reciprocal of a x + b|Primitive of Reciprocal of $a x + b$]]
+}}
+{{eqn | r = \frac {a x} p - \frac {a q} {p^2} \ln \left\vert{p x + q}\right\vert + \frac b p \ln \left\vert{p x + q}\right\vert + C
+      | c = 
+}}
+{{eqn | r = \frac {a x} p + \frac {b p - a q} {p^2} \ln \left\vert{p x + q}\right\vert + C
+      | c = common [[Definition:Denominator|denominator]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Reduction Formula for Primitive of Power of x by Power of a x + b/Decrement of Power of a x + b/Proof 1}
+Tags: Primitive of Power of x by Power of a x + b
+
+\begin{theorem}
+:$\displaystyle \int x^m \left({a x + b}\right)^n \rd x = \frac {x^{m+1} \left({a x + b}\right)^n} {m + n + 1} + \frac {n b} {m + n + 1} \int x^m \left({a x + b}\right)^{n - 1} \rd x$
+\end{theorem}
+
+\begin{proof}
+Let $s \in \Z$. 
+{{begin-eqn}}
+{{eqn | l = v
+      | r = x^s
+      | c = 
+}}
+{{eqn | n = 1
+      | ll= \implies
+      | l = \frac {\d v} {\d x}
+      | r = s x^{s - 1}
+      | c = [[Power Rule for Derivatives]]
+}}
+{{end-eqn}}
+Let $u \dfrac {\d v} {\d x} = x^m \left({a x + b}\right)^n$.
+Then:
+{{begin-eqn}}
+{{eqn | l = u
+      | r = \frac {x^m \left({a x + b}\right)^n} {s x^{s - 1} }
+      | c = from $(1)$
+}}
+{{eqn | r = \frac {x^{m - s + 1} \left({a x + b}\right)^n} s
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \frac {\d u} {\d x}
+      | r = \frac {x^{m - s + 1} a n \left({a x + b}\right)^{n - 1} } s + \frac {\left({m - s + 1}\right) x^{m - s} \left({a x + b}\right)^n} s
+      | c = [[Primitive of Power of a x + b|Primitive of Power of $a x + b$]] and [[Product Rule for Derivatives]]
+}}
+{{eqn | r = \frac {x^{m - s} } s \left({a x + b}\right)^{n - 1} \left({a n x + \left({m - s + 1}\right) \left({a x + b}\right) }\right)
+      | c = extracting $\dfrac {x^{m - s} \left({a x + b}\right)^n} s$ as a factor
+}}
+{{eqn | r = \frac {x^{m - s} } s \left({a x + b}\right)^{n - 1} \left({a n x + a m x - a s x + a x + m b - s b + b}\right)
+      | c = multiplying out
+}}
+{{eqn | n = 2
+      | r = \frac {x^{m - s} } s \left({a x + b}\right)^{n - 1} \left({\left({m + n + 1 - s}\right) a x + \left({m - s + 1}\right) b}\right)
+      | c = gathering terms
+}}
+{{end-eqn}}
+Let $s$ be selected such that $m + n + 1 - s = 0$.
+Then $s = m + n + 1$.
+Thus $(2)$ after rearrangement becomes:
+:$\dfrac {\d u} {\d x} = \dfrac {x^{m - s} } {m + n + 1} \left({a x + b}\right)^{n - 1} \left({\left({m - s + 1}\right) b}\right)$
+Then:
+{{begin-eqn}}
+{{eqn | l = u v
+      | r = \frac {x^{m - s + 1} \left({a x + b}\right)^n} {m + n + 1} x^s
+      | c = 
+}}
+{{eqn | r = \frac {x^{m + 1} \left({a x + b}\right)^n} {m + n + 1}
+      | c = 
+}}
+{{end-eqn}}
+and:
+{{begin-eqn}}
+{{eqn | l = v \frac {\d u} {\d x}
+      | r = x^s \dfrac {x^{m - s} } {m + n + 1} \left({a x + b}\right)^{n - 1} \left({\left({m - s + 1}\right) b}\right)
+      | c = 
+}}
+{{eqn | r = \dfrac {x^m} {m + n + 1} \left({a x + b}\right)^{n - 1} \left({\left({m - \left({m + n + 1}\right) + 1}\right) b}\right)
+      | c = 
+}}
+{{eqn | r = \dfrac {x^m} {m + n + 1} \left({a x + b}\right)^{n - 1} \left({-n b}\right)
+      | c = 
+}}
+{{end-eqn}}
+Thus by [[Integration by Parts]]:
+:$\displaystyle \int x^m \left({a x + b}\right)^n \rd x = \frac {x^{m+1} \left({a x + b}\right)^n} {m + n + 1} + \frac {n b} {m + n + 1} \int x^m \left({a x + b}\right)^{n - 1} \rd x$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Power of a x + b over Power of p x + q/Formulation 1}
+Tags: Primitives involving a x + b and p x + q
+
+\begin{theorem}
+:$\displaystyle \int \frac {\left({a x + b}\right)^m} {\left({p x + q}\right)^n} \rd x = \frac {-1} {\left({n - 1}\right) \left({b p - a q}\right)} \left({\frac {\left({a x + b}\right)^{m + 1} } {\left({p x + q}\right)^{n - 1} } + \left({n - m - 2}\right) a \int \frac {\left({a x + b}\right)^m} { \left({p x + q}\right)^{n - 1} } \rd x}\right)$
+\end{theorem}
+
+\begin{proof}
+From [[Reduction Formula for Primitive of Power of a x + b by Power of p x + q/Increment of Power|Reduction Formula for Primitive of Power of $a x + b$ by Power of $p x + q$: Increment of Power]]:
+:$\displaystyle \int \left({a x + b}\right)^m \left({p x + q}\right)^n \rd x = \frac 1 {\left({n + 1}\right) \left({b p - a q}\right)} \left({\left({a x + b}\right)^{m + 1} \left({p x + q}\right)^{n + 1} - \left({m + n + 2}\right) a \int \left({a x + b}\right)^m \left({p x + q}\right)^{n + 1} \rd x}\right)$
+Setting $n := -n$:
+{{begin-eqn}}
+{{eqn | o = 
+      | r = \int \frac {\left({a x + b}\right)^m} {\left({p x + q}\right)^n} \rd x
+      | c = 
+}}
+{{eqn | r = \int \left({a x + b}\right)^m \left({p x + q}\right)^{-n} \rd x
+      | c = 
+}}
+{{eqn | r = \frac 1 {\left({-n + 1}\right) \left({b p - a q}\right)} \left({\left({a x + b}\right)^{m + 1} \left({p x + q}\right)^{-n + 1} - \left({m + -n + 2}\right) a \int \left({a x + b}\right)^m \left({p x + q}\right)^{-n + 1} \rd x}\right)
+      | c = 
+}}
+{{eqn | r = \frac {-1} {\left({n - 1}\right) \left({b p - a q}\right)} \left({\frac {\left({a x + b}\right)^{m + 1} } {\left({p x + q}\right)^{n - 1} } + \left({n - m - 2}\right) a \int \frac {\left({a x + b}\right)^m} {\left({p x + q}\right)^{n - 1} } }\right)
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Power of a x + b over Power of p x + q}
+Tags: Primitives involving a x + b and p x + q
+
+\begin{theorem}
+=== [[Primitive of Power of a x + b over Power of p x + q/Formulation 1|Formulation 1]] ===
+{{:Primitive of Power of a x + b over Power of p x + q/Formulation 1}}
+=== [[Primitive of Power of a x + b over Power of p x + q/Formulation 2|Formulation 2]] ===
+{{:Primitive of Power of a x + b over Power of p x + q/Formulation 2}}
+=== [[Primitive of Power of a x + b over Power of p x + q/Formulation 3|Formulation 3]] ===
+{{:Primitive of Power of a x + b over Power of p x + q/Formulation 3}}
+\end{theorem}<|endoftext|>
+\section{Primitive of Power of a x + b over Power of p x + q/Formulation 2}
+Tags: Primitives involving a x + b and p x + q
+
+\begin{theorem}
+:$\displaystyle \int \frac {\paren {a x + b}^m} {\paren {p x + q}^n} \rd x = \frac {-1} {\paren {n - m - 1} p} \paren {\frac {\paren {a x + b}^m} {\paren {p x + q}^{n - 1} } + m \paren {b p - a q} \int \frac {\paren {a x + b}^{m - 1} } {\paren {p x + q}^n} \rd x}$
+\end{theorem}
+
+\begin{proof}
+From [[Reduction Formula for Primitive of Power of a x + b by Power of p x + q/Decrement of Power|Reduction Formula for Primitive of Power of $a x + b$ by Power of $p x + q$: Decrement of Power]]:
+:$\displaystyle \int \paren {a x + b}^m \paren {p x + q}^n \rd x = \frac {\paren {a x + b}^m \paren {p x + q}^{n + 1} } {\paren {m + n + 1} a} + \frac {m \paren {b p - a q} } {\paren {m + n + 1} p} \int \paren {a x + b}^{m - 1} \paren {p x + q}^n \rd x$
+Setting $n := -n$:
+{{begin-eqn}}
+{{eqn | o = 
+      | r = \int \frac {\paren {a x + b}^m} {\paren {p x + q}^n} \rd x
+      | c = 
+}}
+{{eqn | r = \int \paren {a x + b}^m \paren {p x + q}^{-n} \rd x
+      | c = 
+}}
+{{eqn | r = \frac {\paren {a x + b}^m \paren {p x + q}^{-n + 1} } {\paren {m - n + 1} p} + \frac {m \paren {b p - a q} } {\paren {m - n + 1} p} \int \paren {a x + b}^{m - 1} \paren {p x + q}^{-n} \rd x
+      | c = 
+}}
+{{eqn | r = \frac {-1} {\paren {n - m - 1} p} \paren {\frac {\paren {a x + b}^m} {\paren {p x + q}^{n - 1} } + m  \paren {b p - a q} \int \frac {\paren {a x + b}^{m - 1} } {\paren {p x + q}^n} \rd x}
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Power of a x + b over Power of p x + q/Formulation 3}
+Tags: Primitives involving a x + b and p x + q
+
+\begin{theorem}
+:$\displaystyle \int \frac {\paren {a x + b}^m} {\paren {p x + q}^n} \rd x = \frac {-1} {\paren {n - 1} p} \paren {\frac {\paren {a x + b}^m} {\paren {p x + q}^{n - 1} } - m a \int \frac {\paren {a x + b}^{m - 1} } {\paren {p x + q}^{n - 1}} \rd x}$
+\end{theorem}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = u
+      | r = \paren {a x + b}^m
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d u} {\d x}
+      | r = m a \paren {a x + b}^m
+      | c = [[Power Rule for Derivatives]] and [[Derivatives of Function of a x + b|Derivatives of Function of $a x + b$]]
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = v
+      | r = \frac {\paren {p x + q}^{-\paren {n - 1} } } {-p \paren {n - 1} }
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d v} {\d x}
+      | r = \paren {p x + q}^n
+      | c = [[Power Rule for Derivatives]] and [[Derivatives of Function of a x + b|Derivatives of Function of $a x + b$]]
+}}
+{{end-eqn}}
+Thus:
+{{begin-eqn}}
+{{eqn | o = 
+      | r = \int \frac {\paren {a x + b}^m} {\paren {p x + q}^n} \rd x
+      | c = 
+}}
+{{eqn | r = \paren {a x + b}^m \frac {\paren {p x + q}^{-\paren {n - 1} } } {-p \paren {n - 1} } - \int \frac {\paren {p x + q}^{-\paren {n - 1} } } {-p \paren {n - 1} } m a \paren {a x + b}^m \rd x
+      | c = [[Integration by Parts]]
+}}
+{{eqn | r = \frac {-1} {\paren {n - 1} p} \paren {\frac {\paren {a x + b}^m} {\paren {p x + q}^{n - 1} } - m a \int \frac {\paren {a x + b}^{m-1} } {\paren {p x + q}^{n - 1} } \rd x}
+      | c = simplification
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of p x + q over Root of a x + b}
+Tags: Primitives involving Root of a x + b and p x + q
+
+\begin{theorem}
+:$\displaystyle \int \frac {p x + q} {\sqrt {a x + b} } \rd x = \frac {2 \paren {a p x + 3 a q - 2 b p} } {3 a^2} \sqrt{a x + b}$
+\end{theorem}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = u
+      | r = \sqrt{a x + b}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x
+      | r = \frac {u^2 - b} a
+      | c = 
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = \int \frac {p x + q} {\sqrt {a x + b} } \rd x
+      | r = \int \paren {p \paren {\frac {u^2 - b} a} + q} \frac {2 u} {a u} \rd x
+      | c = [[Primitive of Function of Root of a x + b|Primitive of Function of Root of $a x + b$]]
+}}
+{{eqn | r = \frac 2 {a^2} \int \paren {p u^2 - b p + a q} \rd x
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \frac 2 {a^2} \paren {\frac {p u^3} 3 - b p z + a q z} + C
+      | c = [[Primitive of Power]]
+}}
+{{eqn-intertext|Note that at this point we can assume $u > 0$, otherwise $\sqrt {a x + b}$ would not have been defined.}}
+{{eqn | r = \frac 2 {a^2} \paren {\frac {p \paren {\sqrt {a x + b} }^3} 3 - b p \sqrt {a x + b} + a q \sqrt {a x + b} } + C
+      | c = substituting for $u$
+}}
+{{eqn | r = \frac {2 \paren {a p x - 2 b p + 3 a q} } {3 a^2} \sqrt {a x + b} + C
+      | c = simplification
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Reciprocal of p x + q by Root of a x + b}
+Tags: Primitives involving Root of a x + b and p x + q
+
+\begin{theorem}
+:$\displaystyle \int \frac {\d x} {\paren {p x + q} \sqrt {a x + b} } = \begin {cases}
+\dfrac 1 {\sqrt {b p - a q} \sqrt p} \ln \size {\dfrac {\sqrt {p \paren {a x + b} } - \sqrt {b p - a q} } {\sqrt {p \paren {a x + b} } + \sqrt {b p - a q} } } & : b p - a q > 0 \\
+\dfrac 2 {\sqrt {a q - b p} \sqrt p} \arctan \sqrt {\dfrac {p \paren {a x + b} } {a q - b p} } & : b p - a q < 0 \\
+\end {cases}$
+\end{theorem}
+
+\begin{proof}
+{{finish|Please deal with positive and negative}}
+Let:
+{{begin-eqn}}
+{{eqn | l = u
+      | r = \sqrt {a x + b}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x
+      | r = \frac {u^2 - b} a
+      | c = 
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = \int \frac {\d x} {\paren {p x + q} \sqrt{a x + b} }
+      | r = \int \frac {2 u} {p \paren {\paren {\dfrac {u^2 - b} a} + q} a u} \rd u
+      | c = [[Primitive of Function of Root of a x + b|Primitive of Function of $\sqrt{a x + b}$]]
+}}
+{{eqn | r = \int \frac {2 a u} {\paren {p \paren {u^2 - b} + a q} a u} \rd u
+      | c = simplifying
+}}
+{{eqn | r = 2 \int \frac {\d u} {p u^2 - b p + a q}
+      | c = simplifying
+}}
+{{eqn | n = 1
+      | r = 2 \int \frac {\d u} {\paren {\sqrt p u}^2 - \paren {\sqrt {b p - a q} }^2}
+      | c = putting [[Definition:Denominator|denominator]] into the form $z^2 - a^2$
+}}
+{{end-eqn}}
+Now let $z = \sqrt p u$
+{{begin-eqn}}
+{{eqn | l = z
+      | r = \sqrt p u
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d z} {\d u}
+      | r = \sqrt p
+      | c = 
+}}
+{{end-eqn}}
+Thus:
+{{begin-eqn}}
+{{eqn | l = \int \frac {\d x} {\paren {p x + q} \sqrt {a x + b} }
+      | r = 2 \int \frac {\d u} {\paren {\sqrt p u}^2 - \paren {\sqrt {b p - a q} }^2}
+      | c = from $(1)$
+}}
+{{eqn | n = 2
+      | r = \frac 2 {\sqrt p} \int \frac {\d z} {z^2 - \paren {\sqrt {b p - a q} }^2}
+      | c = simplifying
+}}
+{{end-eqn}}
+Let $b p - a q > 0$.
+Let $d = \sqrt {b p - a q}$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \int \frac {\d x} {\paren {p x + q} \sqrt{a x + b} }
+      | r = \frac 2 {\sqrt p} \int \frac {\d z} {z^2 - d^2}
+      | c = from $(2)$
+}}
+{{eqn | r = \frac 2 {\sqrt p} \frac 1 {2 d} \ln \size {\frac {z - d} {z + d} } + C
+      | c = [[Primitive of Reciprocal of x squared minus a squared/Logarithm Form|Primitive of $\dfrac 1 {x^2 - a^2}$: Logarithm Form]]
+}}
+{{eqn | r = \frac 1 {\sqrt p d} \ln \size {\frac {\sqrt p u - d} {\sqrt p u + d} } + C
+      | c = substituting for $z$
+}}
+{{eqn | r = \frac 1 {\sqrt p \sqrt {b p - a q} } \ln \size {\frac {\sqrt {p \paren {a x + b} } - \sqrt {b p - a q} } {\sqrt {p \paren {a x + b} } + \sqrt {b p - a q} } } + C
+      | c = substituting for $u$ and $d$
+}}
+{{end-eqn}}
+{{qed|lemma}}
+Let $b p - a q < 0$.
+Let $d = \sqrt {-\paren {b p - a q} } = \sqrt {a q - b p}$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \int \frac {\d x} {\paren {p x + q} \sqrt{a x + b} }
+      | r = \frac 2 {\sqrt p} \int \frac {\d z} {z^2 + d^2}
+      | c = from $(2)$
+}}
+{{eqn | r = \frac 2 {\sqrt p} \frac 1 d \arctan {\frac z d} + C
+      | c = [[Primitive of Reciprocal of x squared plus a squared/Arctangent Form|Primitive of $\dfrac 1 {x^2 + a^2}$]]
+}}
+{{eqn | r = \frac 2 {\sqrt p d} \arctan {\frac {\sqrt p u} d} + C
+      | c = substituting for $z$
+}}
+{{eqn | r = \frac 2 {\sqrt {a q - b p} \sqrt p} \arctan {\sqrt {\frac {p \paren {a x + b} } {a q - b p} } } + C
+      | c = substituting for $u$ and $d$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Root of a x + b over p x + q}
+Tags: Primitives involving Root of a x + b and p x + q
+
+\begin{theorem}
+:$\displaystyle \int \frac {\sqrt{a x + b} } {p x + q} \rd x = \begin{cases}
+\dfrac {2 \sqrt{a x + b} } p + \dfrac {\sqrt {b p - a q} } {p \sqrt p} \ln \size {\dfrac {\sqrt {p \paren {a x + b} } - \sqrt {b p - a q} } {\sqrt {p \paren {a x + b} } + \sqrt {b p - a q} } } & : b p - a q > 0 \\
+\dfrac {2 \sqrt{a x + b} } p - \dfrac {\sqrt {a q - b p} } {p \sqrt p} \arctan \sqrt {\dfrac {p \paren {a x + b} } {a q - b p} } & : b p - a q < 0 \\
+\end{cases}$
+\end{theorem}
+
+\begin{proof}
+From [[Primitive of Power of a x + b over Power of p x + q/Formulation 2|Primitive of Power of $a x + b$ over Power of $p x + q$: Formulation 2]]:
+:$\displaystyle \int \frac {\paren {a x + b}^m} {\paren {p x + q}^n} \rd x = \frac {-1} {\paren {n - m - 1} p} \paren {\frac {\paren {a x + b}^m} {\paren {p x + q}^{n - 1} } + m \paren {b p - a q} \int \frac {\paren {a x + b}^{m - 1} } {\paren {p x + q}^n} \rd x}$
+Setting $m := \dfrac 1 2$ and $n = 1$:
+{{begin-eqn}}
+{{eqn | l = \int \frac {\paren {a x + b}^{1/2} } {\paren {p x + q}^n} \rd x
+      | r = \frac {-1} {\paren {1 - \frac 1 2 - 1} p} \paren {\frac {\paren {a x + b}^{1/2} } {\paren {p x + q}^0} + \frac 1 2 \paren {b p - a q} \int \frac {\paren {a x + b}^{1/2 - 1} } {\paren {p x + q} } \rd x}
+      | c = 
+}}
+{{eqn | r = \frac {2 \sqrt {a x + b} } p + \frac {b p - a q} p \int \frac {\d x} {\sqrt {a x + b} \paren {p x + q} }
+      | c = 
+}}
+{{end-eqn}}
+From [[Primitive of Reciprocal of p x + q by Root of a x + b|Primitive of Reciprocal of $p x + q$ by Root of $a x + b$]]:
+:$\displaystyle \int \frac {\d x} {\paren {p x + q} \sqrt{a x + b} } = \begin{cases}
+\dfrac 1 {\sqrt {b p - a q} \sqrt p} \ln \size {\dfrac {\sqrt {p \paren {a x + b} } - \sqrt {b p - a q} } {\sqrt {p \paren {a x + b} } + \sqrt {b p - a q} } } & : b p - a q > 0 \\
+\dfrac 2 {\sqrt {a q - b p} \sqrt p} \arctan \sqrt {\dfrac {p \paren {a x + b} } {a q - b p} } & : b p - a q < 0 \\
+\end{cases}$
+The result follows by substitution.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Power of p x + q by Root of a x + b}
+Tags: Primitives involving Root of a x + b and p x + q
+
+\begin{theorem}
+:$\displaystyle \int \paren {p x + q}^n \sqrt {a x + b} \rd x = \frac {2 \paren {p x + q}^{n + 1} \sqrt {a x + b} } {\paren {2 n + 3} p} + \frac {b p - a q} {\paren {2 n + 3} p} \int \frac {\paren {p x + q}^n} {\sqrt{a x + b} } \rd x$
+\end{theorem}
+
+\begin{proof}
+From [[Reduction Formula for Primitive of Power of a x + b by Power of p x + q/Decrement of Power|Reduction Formula for Primitive of Power of $a x + b$ by Power of $p x + q$: Decrement of Power]]:
+:$\displaystyle \int \paren {a x + b}^m \paren {p x + q}^n \rd x = \frac {\paren {a x + b}^m \paren {p x + q}^{n + 1} } {\paren {m + n + 1} p} + \frac {m \paren {b p - a q} } {\paren {m + n + 1} p} \int \paren {a x + b}^{m - 1} \paren {p x + q}^n \rd x$
+Setting $m := \dfrac 1 2$:
+{{begin-eqn}}
+{{eqn | l = \int \paren {a x + b}^{1 / 2} \paren {p x + q}^n \rd x
+      | r = \frac {\paren {a x + b}^{1 / 2} \paren {p x + q}^{n + 1} } {\paren {\frac 1 2 + n + 1} p} + \frac {\dfrac 1 2 \paren {b p - a q} } {\paren {\frac 1 2 + n + 1} p} \int \paren {a x + b}^{1 / 2 - 1} \paren {p x + q}^n \rd x
+      | c = 
+}}
+{{eqn | r = \frac {2 \paren {p x + q}^{n + 1} \sqrt {a x + b} } {\paren {2 n + 3} p} + \frac {b p - a q} {\paren {2 n + 3} p} \int \frac {\paren {p x + q}^n} {\sqrt {a x + b} } \rd x
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Reciprocal of Power of p x + q by Root of a x + b}
+Tags: Primitives involving Root of a x + b and p x + q
+
+\begin{theorem}
+:$\displaystyle \int \frac {\mathrm d x} {\left({p x + q}\right)^n \sqrt{a x + b} } = \frac {\sqrt{a x + b} } {\left({n - 1}\right) \left({a q - b p}\right) \left({p x + q}\right)^{n-1} } + \frac {\left({2 n - 3}\right) a} {2 \left({n - 1}\right)  \left({a q - b p}\right)} \int \frac {\mathrm d x} {\left({p x + q}\right)^{n-1} \sqrt{a x + b} }$
+\end{theorem}
+
+\begin{proof}
+From [[Primitive of Reciprocal of Power of a x + b by Power of p x + q|Primitive of Reciprocal of $\dfrac 1 {\left({a x + b}\right)^m \left({p x + q}\right)^n}$]]:
+:$\displaystyle \int \frac {\mathrm d x} {\left({a x + b}\right)^m \left({p x + q}\right)^n} = \frac {-1} {\left({n - 1}\right) \left({b p - a q}\right)} \left({\frac 1 {\left({a x + b}\right)^{m-1} \left({p x + q}\right)^{n-1} } + a \left({m + n - 2}\right) \int \frac {\mathrm d x} {\left({a x + b}\right)^m \left({p x + q}\right)^{n-1} } }\right)$
+Setting $m := \dfrac 1 2$:
+{{begin-eqn}}
+{{eqn | l = \int \frac {\mathrm d x} {\left({p x + q}\right)^n \sqrt{a x + b} }
+      | r = \frac {-1} {\left({n - 1}\right) \left({b p - a q}\right)} \left({\frac 1 {\left({a x + b}\right)^{1/2 - 1} \left({p x + q}\right)^{n-1} } + a \left({\frac 1 2 + n - 2}\right) \int \frac {\mathrm d x} {\left({a x + b}\right)^{1/2} \left({p x + q}\right)^{n-1} } }\right)
+      | c = 
+}}
+{{eqn | r = \frac {\sqrt{a x + b} } {\left({n - 1}\right) \left({a q - b p}\right) \left({p x + q}\right)^{n-1} } + \frac {\left({2 n - 3}\right) a} {2 \left({n - 1}\right)  \left({a q - b p}\right)} \int \frac {\mathrm d x} {\left({p x + q}\right)^{n-1} \sqrt{a x + b} }
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Power of p x + q over Root of a x + b}
+Tags: Primitives involving Root of a x + b and p x + q
+
+\begin{theorem}
+:$\displaystyle \int \frac {\left({p x + q}\right)^n} {\sqrt{a x + b} } \ \mathrm d x = \frac {2 \left({p x + q}\right)^n \sqrt{a x + b} } {\left({2 n + 1}\right) a} + \frac {2 n \left({a q - b p}\right)} {\left({2 n + 1}\right) a} \int \frac {\left({p x + q}\right)^{n-1} } {\sqrt{a x + b} } \ \mathrm d x$
+\end{theorem}
+
+\begin{proof}
+From [[Reduction Formula for Primitive of Power of a x + b by Power of p x + q/Decrement of Power|Reduction Formula for Primitive of Power of $a x + b$ by Power of $p x + q$: Decrement of Power]]:
+:$\displaystyle \int \left({a x + b}\right)^m \left({p x + q}\right)^n \ \mathrm d x = \frac {\left({a x + b}\right)^{m+1} \left({p x + q}\right)^n} {\left({m + n + 1}\right) a} - \frac {n \left({b p - a q}\right)} {\left({m + n + 1}\right) a} \int \left({a x + b}\right)^m \left({p x + q}\right)^{n-1} \ \mathrm d x$
+Setting $m := -\dfrac 1 2$:
+{{begin-eqn}}
+{{eqn | l = \int \frac {\left({p x + q}\right)^n} {\sqrt{a x + b} } \ \mathrm d x
+      | r = \frac {\left({a x + b}\right)^{-1/2 + 1} \left({p x + q}\right)^n} {\left({-\frac 1 2 + n + 1}\right) a} - \frac {n \left({b p - a q}\right)} {\left({-\frac 1 2 + n + 1}\right) a} \int \left({a x + b}\right)^{-1/2} \left({p x + q}\right)^{n-1} \ \mathrm d x
+      | c = 
+}}
+{{eqn | r = \frac {2 \left({p x + q}\right)^n \sqrt{a x + b} } {\left({2 n + 1}\right) a} + \frac {2 n \left({a q - b p}\right)} {\left({2 n + 1}\right) a} \int \frac {\left({p x + q}\right)^{n-1} } {\sqrt{a x + b} } \ \mathrm d x
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Root of a x + b over Power of p x + q}
+Tags: Primitives involving Root of a x + b and p x + q
+
+\begin{theorem}
+:$\displaystyle \int \frac {\sqrt{a x + b} } {\left({p x + q}\right)^n} \ \mathrm d x = \frac {- \sqrt{a x + b} } {\left({n - 1}\right) p \left({p x + q}\right)^{n-1} } + \frac a {2 \left({n - 1}\right) p} \int \frac {\mathrm d x} {\left({p x + q}\right)^{n-1} \sqrt{a x + b} }$
+\end{theorem}
+
+\begin{proof}
+From [[Primitive of Power of a x + b over Power of p x + q/Formulation 3|Primitive of Power of $a x + b$ over Power of $p x + q$: Formulation 3]]:
+:$\displaystyle \int \frac {\left({a x + b}\right)^m} {\left({p x + q}\right)^n} \ \mathrm d x = \frac {-1} {\left({n - 1}\right) p} \left({\frac {\left({a x + b}\right)^m} {\left({p x + q}\right)^{n-1} } + m a \int \frac {\left({a x + b}\right)^{m-1} } { \left({p x + q}\right)^{n-1} } \ \mathrm d x}\right)$
+Setting $m := \dfrac 1 2$:
+{{begin-eqn}}
+{{eqn | l = \int \frac {\sqrt{a x + b} } {\left({p x + q}\right)^n} \ \mathrm d x
+      | r = \frac {-1} {\left({n - 1}\right) p} \left({\frac {\left({a x + b}\right)^{1/2} } {\left({p x + q}\right)^{n-1} } + \frac 1 2 a \int \frac {\left({a x + b}\right)^{1/2-1} } { \left({p x + q}\right)^{n-1} } \ \mathrm d x}\right)
+      | c = 
+}}
+{{eqn | r =\frac {- \sqrt{a x + b} } {\left({n - 1}\right) p \left({p x + q}\right)^{n-1} } + \frac a {2 \left({n - 1}\right) p} \int \frac {\mathrm d x} {\left({p x + q}\right)^{n-1} \sqrt{a x + b} }
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Reciprocal of Root of a x + b by Root of p x + q}
+Tags: Primitives involving Root of a x + b and Root of p x + q
+
+\begin{theorem}
+:$\displaystyle \int \frac {\d x} {\sqrt {\paren {a x + b} \paren {p x + q} } } = \begin {cases}
+\dfrac 2 {\sqrt {a p} } \map \ln {\sqrt {p \paren {a x + b} } + \sqrt {a \paren {p x + q} } } + C & : \dfrac {b p - a q} p > 0 \\
+\dfrac 2 {\sqrt {a p} } \sinh^{-1} \sqrt {\dfrac {p \paren {a x + b} } {b p - a q} } + C & : \dfrac {b p - a q} p < 0 \\
+\end {cases}$
+\end{theorem}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = u
+      | r = \sqrt {a x + b}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x
+      | r = \frac {u^2 - b} a
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \sqrt {p x + q}
+      | r = \sqrt {p \paren {\frac {u^2 - b} a} + q}
+      | c = 
+}}
+{{eqn | r = \sqrt {\frac {p \paren {u^2 - b} + a q} a}
+      | c = 
+}}
+{{eqn | r = \sqrt {\frac {p u^2 - b p + a q} a}
+      | c = 
+}}
+{{eqn | n = 1
+      | r = \sqrt {\frac p a} \sqrt {u^2 - \paren {\frac {b p - a q} p} }
+      | c = 
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = \int \frac {\d x} {\sqrt {\paren {a x + b} \paren {p x + q} } }
+      | r = \int \frac {2 u \rd u} {a \sqrt {\frac p a} \sqrt {u^2 - \paren {\frac {b p - a q} p} } u}
+      | c = [[Primitive of Function of Root of a x + b|Primitive of Function of Root of $a x + b$]]
+}}
+{{eqn | r = \frac 2 {\sqrt {a p} } \int \frac {\d u} {\sqrt {u^2 - \paren {\frac {b p - a q} p} } }
+      | c = [[Primitive of Constant Multiple of Function]]
+}}
+{{eqn-intertext|Setting $c^2 :{{=}} \dfrac {b p - a q} p$:}}
+{{eqn | n = 2
+      | r = \frac 2 {\sqrt {a p} } \int \frac {\rd u} {\sqrt{u^2 - c^2} }
+      | c = 
+}}
+{{end-eqn}}
+Let $c^2 > 0$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \int \frac {\d x} {\sqrt {\paren {a x + b} \paren {p x + q} } }
+      | r = \frac 2 {\sqrt {a p} } \int \frac {\d u} {\sqrt{u^2 - c^2} }
+      | c = from $(2)$
+}}
+{{eqn | r = \frac 2 {\sqrt {a p} } \ln \size {u + \sqrt {u^2 - c^2} } + C
+      | c = [[Primitive of Reciprocal of Root of x squared minus a squared/Logarithm Form|Primitive of Reciprocal of $\sqrt {x^2 - a^2}$: Logarithm Form]]
+}}
+{{eqn | r = \frac 2 {\sqrt {a p} } \ln \size {\sqrt {a x + b} + \sqrt {u^2 - \frac {b p - a q} p} } + C
+      | c = substituting for $u$ and $c$
+}}
+{{eqn | r = \frac 2 {\sqrt {a p} } \map \ln {\sqrt {a x + b} + \sqrt {u^2 - \frac {b p - a q} p} } + C
+      | c = as both those [[Definition:Square Root|square roots]] are [[Definition:Positive Real Number|positive]]
+}}
+{{eqn | r = \frac 2 {\sqrt {a p} } \map \ln {\sqrt {a x + b} + \sqrt {\frac a p} \sqrt {p x + q} } + C
+      | c = substituting from $(1)$
+}}
+{{eqn | r = \frac 2 {\sqrt {a p} } \map \ln {\frac {\sqrt {p \paren {a x + b} } + \sqrt {a \paren {p x + q} } } {\sqrt p} } + C
+      | c = simplifying
+}}
+{{eqn | r = \frac 2 {\sqrt {a p} } \map \ln {\sqrt {p \paren {a x + b} } + \sqrt {a \paren {p x + q} } } - \ln {\sqrt p} + C
+      | c = [[Difference of Logarithms]]
+}}
+{{eqn | r = \frac 2 {\sqrt {a p} } \map \ln {\sqrt {p \paren {a x + b} } + \sqrt {a \paren {p x + q} } } + C
+      | c = subsuming $-\ln {\sqrt p}$ into the [[Definition:Arbitrary Constant (Calculus)|arbitrary constant]]
+}}
+{{end-eqn}}
+{{qed|lemma}}
+Let $c^2 < 0$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \int \frac {\d x} {\sqrt {\paren {a x + b} \paren {p x + q} } }
+      | r = \frac 2 {\sqrt {a p} } \int \frac {\d u} {\sqrt {u^2 + c^2} }
+      | c = from $(2)$
+}}
+{{eqn | r = \frac 2 {\sqrt {a p} } \sinh^{-1} {\frac u c} + C
+      | c = [[Primitive of Reciprocal of Root of x squared plus a squared/Inverse Hyperbolic Sine Form|Primitive of Reciprocal of $\sqrt {x^2 + a^2}$: $\sinh^{-1}$ form]]
+}}
+{{eqn | r = \frac 2 {\sqrt {a p} } \sinh^{-1} \frac {\sqrt {a x + b} } {\sqrt {\frac {b p - a q} p} } + C
+      | c = substituting for $u$ and $c$
+}}
+{{eqn | r = \frac 2 {\sqrt {a p} } \sinh^{-1} \sqrt {\frac {p \paren {a x + b} } {b p - a q} } + C
+      | c = simplifying
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Derivative of Laplace Transform}
+Tags: Laplace Transforms
+
+\begin{theorem}
+Let $f: \R \to \R$ or $\R \to \C$ be a [[Definition:Continuous|continuous]] [[Definition:Function|function]], [[Definition:Differentiable on Interval|differentiable]] on any [[Definition:Closed Real Interval|closed interval]] $\closedint 0 a$.
+Let $\laptrans f = F$ denote the [[Definition:Laplace Transform|Laplace transform]] of $f$.
+Then, everywhere that $\dfrac \d {\d s} \laptrans f$ exists:
+:$\dfrac \d {\d s} \laptrans {\map f t} = -\laptrans {t \, \map f t}$
+\end{theorem}
+
+\begin{proof}
+{{mistake|see talk page}}
+{{begin-eqn}}
+{{eqn | l = \frac \d {\d s} \laptrans {\map f t}
+      | r = \frac \d {\d s} \int_0^{\to +\infty} \map f t \, e^{-s t} \rd t
+      | c = {{Defof|Laplace Transform}}
+}}
+{{eqn | r = \int_0^{\to +\infty} \map {\frac {\partial} {\partial s} } {\map f t \, e^{-s t} } \rd t
+      | c = [[Definite Integral of Partial Derivative]]
+}}
+{{eqn | r = \int_0^{\to +\infty} \map f t \, \map {\frac {\partial} {\partial s} } {e^{-st} } \rd t
+}}
+{{eqn | r = -\int_0^{\to +\infty} t \, \map f t \, e^{-s t} \rd t
+      | c = [[Derivative of Exponential Function]]
+}}
+{{eqn | r = -\laptrans {t \, \map f t}
+      | c = {{Defof|Laplace Transform}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of x over Root of a x + b by Root of p x + q}
+Tags: Primitives involving Root of a x + b and Root of p x + q
+
+\begin{theorem}
+:$\displaystyle \int \frac {x \ \mathrm d x} {\sqrt{\left({a x + b}\right) \left({p x + q}\right)} } = \frac {\sqrt{\left({a x + b}\right) \left({p x + q}\right)} } {a p} - \frac {b p + a q} {2 a p} \int \frac {\mathrm d x} {\sqrt{\left({a x + b}\right) \left({p x + q}\right)} }$
+\end{theorem}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = u
+      | r = \sqrt{a x + b}
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = x
+      | r = \frac {u^2 - b} a
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \sqrt{p x + q}
+      | r = \sqrt{p \left({\frac {u^2 - b} a}\right) + q}
+      | c = 
+}}
+{{eqn | r = \sqrt{\frac {p \left({u^2 - b}\right) + a q} a}
+      | c = 
+}}
+{{eqn | r = \sqrt{\frac {p u^2 - b p + a q} a}
+      | c = 
+}}
+{{eqn | n = 1
+      | r = \sqrt {\frac p a} \sqrt{u^2 - \left({\frac {b p - a q} p}\right)}
+      | c = 
+}}
+{{end-eqn}}
+Let $c^2 = \dfrac {b p - a q} p$.
+Then:
+{{begin-eqn}}
+{{eqn | o = 
+      | r = \int \frac {x \ \mathrm d x} {\sqrt{\left({a x + b}\right) \left({p x + q}\right)} }
+}}
+{{eqn | r = \int \frac {2 u} a \left({\frac {u^2 - b} a}\right) \frac {\mathrm d u} {u \sqrt {\frac p a} \sqrt{u^2 - c^2} }
+      | c = [[Primitive of Function of Root of a x + b|Primitive of Function of Root of $a x + b$]]
+}}
+{{eqn | r = \frac 2 {a \sqrt {a p} } \int \frac {u^2 \ \mathrm d u} {\sqrt{u^2 - c^2} } - \frac {2 b} {a \sqrt {a p} } \int \frac {\mathrm d u} {\sqrt{u^2 - c^2} }
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \frac 2 {a \sqrt {a p} } \left({\frac {u \sqrt{u^2 - c^2} } 2 + \frac {c^2} 2 \ln \left({u + \sqrt {u^2 - c^2} }\right) }\right)
+      | c = [[Primitive of x squared over Root of x squared minus a squared|Primitive of $\dfrac {x^2}{\sqrt{x^2 - a^2} }$]]
+}}
+{{eqn | ro= -
+      | r = \frac {2 b} {a \sqrt {a p} } \ln \left({u + \sqrt {u^2 - c^2} }\right)
+      | c = ... and [[Primitive of Reciprocal of Root of x squared minus a squared/Logarithm Form|Primitive of $\dfrac 1 {\sqrt{x^2 - a^2} }$: Logarithm Form]]
+}}
+{{eqn | r = \frac 2 {a \sqrt {a p} } \left({\frac {u \sqrt{u^2 - c^2} } 2 + \left({\frac {c^2} 2 - b}\right) \ln \left({u + \sqrt {u^2 - c^2} }\right) }\right)
+      | c = common factor
+}}
+{{eqn | r = \frac 2 {a \sqrt {a p} } \left({\frac {\sqrt{a x + b} \sqrt {\frac a p} \sqrt{p x + q} } 2 + \left({\frac {c^2} 2 - b}\right) \ln \left({\sqrt{a x + b} + \sqrt {\frac a p} \sqrt{p x + q} }\right)}\right)
+      | c = substituting for $u$, noting that $\sqrt{p x + q} = \sqrt {\frac p a} \sqrt{u^2 - c^2}$
+}}
+{{eqn | r = \frac {\sqrt{\left({a x + b}\right) \left({p x + q}\right)} } {a p} + \left({\frac {b p - a q} {a p \sqrt {a p} }  - \frac {2 b} {a \sqrt {a p} } }\right) \ln \left({\frac {\sqrt {p \left({a x + b}\right)} + \sqrt {a \left({p x + q}\right)} } {\sqrt p} }\right)
+      | c = substituting for $c^2$ and simplifying
+}}
+{{eqn | r = \frac {\sqrt{\left({a x + b}\right) \left({p x + q}\right)} } {a p} + \left({\frac {b p - a q - 2 b p} {a p \sqrt {a p} } }\right) \ln \left({\frac {\sqrt {p \left({a x + b}\right)} + \sqrt {a \left({p x + q}\right)} } {\sqrt p} }\right)
+      | c = common [[Definition:Denominator|denominator]]
+}}
+{{eqn | r = \frac {\sqrt{\left({a x + b}\right) \left({p x + q}\right)} } {a p} - \left({\frac {b p + a q} {a p \sqrt {a p} } }\right) \frac {\sqrt {a p} } 2 \int \frac {\mathrm d x} {\sqrt{\left({a x + b}\right) \left({p x + q}\right)} }
+      | c = [[Primitive of Reciprocal of Root of a x + b by Root of p x + q|Primitive of Reciprocal of $\sqrt{\left({a x + b}\right) \left({p x + q}\right)}$]]
+}}
+{{eqn | r = \frac {\sqrt{\left({a x + b}\right) \left({p x + q}\right)} } {a p} - \left({\frac {b p + a q} {2 a p} }\right) \int \frac {\mathrm d x} {\sqrt{\left({a x + b}\right) \left({p x + q}\right)} }
+      | c = simplifying
+}}
+{{end-eqn}}
+{{qed}}
+{{improve|This is messy -- open to anyone else to write a better version}}
+\end{proof}<|endoftext|>
+\section{Higher Order Derivatives of Laplace Transform}
+Tags: Laplace Transforms
+
+\begin{theorem}
+:$\dfrac {\d^n} {\d s^n} \laptrans {\map f t} = \paren {-1}^n \laptrans {t^n \, \map f t}$
+\end{theorem}
+
+\begin{proof}
+The proof proceeds by [[Principle of Mathematical Induction|induction]] on $n$, the [[Definition:Order of Derivative|order of the derivative]] of $\laptrans f$.
+The case $n = 0$ is verified as follows:
+{{begin-eqn}}
+{{eqn | l = \frac {\d^0} {\d s^0} \laptrans {\map f t}
+      | r = \laptrans {\map f t}
+      | c = {{Defof|Zeroth Derivative}}
+}}
+{{eqn | r = \paren {-1}^0 \laptrans {t^0 \, \map f t}
+      | c = {{Defof|Power of Natural Number|Zeroth Power}}
+}}
+{{end-eqn}}
+=== Basis for the Induction ===
+The case $n = 1$ is demonstrated in [[Derivative of Laplace Transform]]:
+:$\dfrac \d {\d s} \laptrans {\map f t} = -\laptrans {t \, \map f t}$
+This is the [[Definition:Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Fix $n \in \N$ with $n \ge 0$.
+Assume:
+:$\dfrac {\d^n} {\d s^n} \laptrans {\map f t} = \paren {-1}^n \laptrans {t^n \, \map f t}$
+This is our [[Definition:Induction Hypothesis|induction hypothesis]].
+=== Induction Step ===
+This is our [[Definition:Induction Step|induction step]]:
+{{begin-eqn}}
+{{eqn | l = \frac {\d^{n + 1} } {\d s^{n + 1} } \laptrans {\map f t}
+      | r = \frac {\d}{\d s} \frac {\d^n} {\d s^n} \laptrans {\map f t}
+}}
+{{eqn | r = \frac {\d}{\d s} \paren {\paren {-1}^n \laptrans {t^n \, \map f t} }
+      | c = [[Higher Order Derivatives of Laplace Transform#Induction Hypothesis|Induction Hypothesis]]
+}}
+{{eqn | r = \paren {-1}^n \paren {-\laptrans {t \times t^n \, \map f t} }
+      | c = [[Derivative of Laplace Transform]]
+}}
+{{eqn | r = \paren {-1}^{n + 1} \laptrans {t^{n + 1} \, \map f t}
+      | c = simplification
+}}
+{{end-eqn}}
+The result follows by the [[Principle of Mathematical Induction]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Root of a x + b by Root of p x + q}
+Tags: Primitives involving Root of a x + b and Root of p x + q
+
+\begin{theorem}
+:$\displaystyle \int \sqrt {\paren {a x + b} \paren {p x + q} } \rd x = \frac {2 a p x + b p + a q} {4 a p} \sqrt {\paren {a x + b} \paren {p x + q} } - \frac {\paren {b p - a q}^2} {8 a p} \int \frac {\d x} {\sqrt {\paren {a x + b} \paren {p x + q} } }$
+\end{theorem}
+
+\begin{proof}
+From [[Primitive of Power of p x + q by Root of a x + b|Primitive of $\paren {p x + q}^n \sqrt {a x + b}$]]:
+:$\displaystyle \int \paren {p x + q}^n \sqrt {a x + b} \rd x = \frac {2 \paren {p x + q}^{n + 1} \sqrt {a x + b} } {\paren {2 n + 3} p} + \frac {b p - a q} {\paren {2 n + 3} p} \int \frac {\paren {p x + q}^n} {\sqrt {a x + b} } \rd x$
+Putting $n = \dfrac 1 2$:
+{{begin-eqn}}
+{{eqn | o = 
+      | r = \int \sqrt {\paren {a x + b} \paren {p x + q} } \rd x
+      | c = 
+}}
+{{eqn | r = \frac {2 \paren {p x + q}^{1 / 2 + 1} \sqrt {a x + b} } {\paren {2 \cdot \frac 1 2 + 3} p} + \frac {b p - a q} {\paren {2 \cdot \frac 1 2 + 3} p} \int \frac {\paren {p x + q}^{1 / 2} } {\sqrt {a x + b} } \rd x
+      | c = 
+}}
+{{eqn | r = \frac {2 \paren {p x + q} \sqrt {p x + q} \sqrt {a x + b} } {4 p} + \frac {b p - a q} {4 p} \int \frac {\sqrt {p x + q} } {\sqrt {a x + b} } \rd x
+      | c = 
+}}
+{{eqn | r = \frac {2 \paren {p x + q} \sqrt {p x + q} \sqrt {a x + b} } {4 p} + \frac {b p - a q} {4 p} \paren {\frac {\sqrt {\paren {a x + b} \paren {p x + q} } } a + \frac {a q - b p} {2 a} \int \frac {\d x} {\sqrt {\paren {a x + b} \paren {p x + q} } } }
+      | c = [[Primitive of Root of p x + q over Root of a x + b|Primitive of $\dfrac {\sqrt {p x + q} } {\sqrt {a x + b} }$]]
+}}
+{{eqn | r = \frac {\paren {2 a \paren {p x + q} + \paren {b p - a q} } \sqrt {\paren {a x + b} \paren {p x + q} } } {4 a p} - \frac {\paren {b p - a q}^2 } {8 a p} \int \frac {\d x} {\sqrt {\paren {a x + b} \paren {p x + q} } }
+      | c = extracting common factor
+}}
+{{eqn | r = \frac {2 a p x + b p + a q} {4 a p} \sqrt {\paren {a x + b} \paren {p x + q} } - \frac {\paren {b p - a q}^2} {8 a p} \int \frac {\d x} {\sqrt {\paren {a x + b} \paren {p x + q} } }
+      | c = simplifying
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Root of p x + q over Root of a x + b}
+Tags: Primitives involving Root of a x + b and Root of p x + q
+
+\begin{theorem}
+:$\displaystyle \int \sqrt {\frac {p x + q} {a x + b} } \rd x = \frac {\sqrt {\paren {a x + b} \paren {p x + q} } } a + \frac {a q - b p} {2 a} \int \frac {\d x} {\sqrt {\paren {a x + b} \paren {p x + q} } }$
+\end{theorem}
+
+\begin{proof}
+From [[Primitive of Power of p x + q over Root of a x + b|Primitive of $\dfrac {\paren {p x + q}^n} {\sqrt {a x + b} }$]]:
+:$\displaystyle \int \frac {\paren {p x + q}^n} {\sqrt {a x + b} } \rd x = \frac {2 \paren {p x + q}^n \sqrt {a x + b} } {\paren {2 n + 1} a} + \frac {2 n \paren {a q - b p} } {\paren {2 n + 1} a} \int \frac {\paren {p x + q}^{n - 1} } {\sqrt {a x + b} } \rd x$
+Putting $n = \dfrac 1 2$:
+{{begin-eqn}}
+{{eqn | l = \int \sqrt {\frac {p x + q} {a x + b} } \rd x
+      | r = \frac {2 \paren {p x + q}^{1/2} \sqrt {a x + b} } {\paren {2 \cdot \frac 1 2 + 1} a} + \frac {2 \cdot \frac 1 2 \paren {a q - b p} } {\paren {2 \cdot \frac 1 2 + 1} a} \int \frac {\paren {p x + q}^{1/2 - 1} } {\sqrt {a x + b} } \rd x
+      | c = 
+}}
+{{eqn | r = \frac {\sqrt {\paren {a x + b} \paren {p x + q} } } a + \frac {a q - b p} {2 a} \int \frac {\d x} {\sqrt {\paren {a x + b} \paren {p x + q} } }
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Reciprocal of p x + q by Root of a x + b by Root of p x + q}
+Tags: Primitives involving Root of a x + b and Root of p x + q
+
+\begin{theorem}
+:$\displaystyle \int \frac {\d x} {\paren {p x + q} \sqrt {\paren {a x + b} \paren {p x + q} } } = \frac {2 \sqrt{a x + b} } {\paren {a q - b p} \sqrt {p x + q} } + C$
+\end{theorem}
+
+\begin{proof}
+From [[Primitive of Reciprocal of Power of p x + q by Root of a x + b|Primitive of $\paren {p x + q}^n \sqrt {a x + b}$]]:
+:$\displaystyle \int \frac {\d x} {\paren {p x + q}^n \sqrt{a x + b} } = \frac {\sqrt{a x + b} } {\paren {n - 1} \paren {a q - b p} \paren {p x + q}^{n - 1} } + \frac {\paren {2 n - 3} a} {2 \paren {n - 1} \paren {a q - b p} } \int \frac {\d x} {\paren {p x + q}^{n - 1} } {\sqrt {a x + b} }$
+Putting $n = \dfrac 3 2$:
+{{begin-eqn}}
+{{eqn | l = \int \frac {\d x} {\paren {p x + q} \sqrt {\paren {a x + b} \paren {p x + q} } }
+      | r = \frac {\sqrt {a x + b} } {\paren {\frac 3 2 - 1} \paren {a q - b p} \paren {p x + q}^{3 / 2 - 1} } + \frac {\paren {2 \cdot \frac 3 2 - 3} a} {2 \paren {\frac 3 2 - 1} \paren {a q - b p} } \int \frac {\d x} {\paren {p x + q}^{3 / 2 - 1} \sqrt {a x + b} }
+      | c = 
+}}
+{{eqn | r = \frac {2 \sqrt {a x + b} } {\paren {a q - b p} \sqrt {p x + q} } + \frac {\paren {3 - 3} a} {a q - b p} \int \frac {\d x} {\sqrt {p x + q} \sqrt {a x + b} } + C
+      | c = 
+}}
+{{eqn | r = \frac {2 \sqrt {a x + b} } {\paren {a q - b p} \sqrt {p x + q} } + C
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of x over x squared plus a squared}
+Tags: Primitive of x over x squared plus a squared, Primitives involving x squared plus a squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {x \rd x} {x^2 + a^2} = \frac 1 2 \map \ln {x^2 + a^2} + C$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = u
+      | r = x^2 + a^2
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d u} {\d x}
+      | r = 2 x 
+      | c = [[Power Rule for Derivatives]] and [[Derivative of Constant]]
+}}
+{{eqn | ll= \leadsto
+      | l = \int \frac {x \rd x} {x^2 + a^2}
+      | r = \frac 1 2 \ln \size {x^2 + a^2} + C
+      | c = [[Primitive of Function under its Derivative]]
+}}
+{{eqn | r = \frac 1 2 \, \map \ln {x^2 + a^2} + C
+      | c = [[Absolute Value of Even Power]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+From [[Primitive of Power of x less one over Power of x plus Power of a]]:
+:$\displaystyle \int \frac {x^{n - 1} \rd x} {x^n + a^n} = \frac 1 n \ln \size {x^n + a^n} + C$
+So:
+{{begin-eqn}}
+{{eqn | l = \int \frac {x \rd x} {x^2 + a^2}
+      | r = \frac 1 2 \ln \size {x^2 + a^2} + C
+      | c = [[Primitive of Power of x less one over Power of x plus Power of a|Primitive of $\dfrac {x^{n - 1} } {\paren {x^n + a^n} }$]] with $n = 2$
+}}
+{{eqn | r = \frac 1 2 \, \map \ln {x^2 + a^2} + C
+      | c = [[Absolute Value of Even Power]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of x squared over x squared plus a squared}
+Tags: Primitives involving x squared plus a squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {x^2 \rd x} {x^2 + a^2} = x - a \arctan {\frac x a} + C$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int \frac {x^2 \rd x} {x^2 + a^2}
+      | r = \int \paren {1 - \frac {a^2} {x^2 + a^2} } \rd x
+      | c = long division
+}}
+{{eqn | r = \int \d x - a^2 \int \frac {\d x} {x^2 + a^2} 
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = x - a^2 \int \frac {\d x} {x^2 + a^2} + C
+      | c = [[Primitive of Constant]]
+}}
+{{eqn | r = x - a^2 \frac 1 a \arctan {\frac x a} + C
+      | c = [[Primitive of Reciprocal of x squared plus a squared/Arctangent Form|Primitive of Reciprocal of $x^2 + a^2$]]
+}}
+{{eqn | r = x - a \arctan {\frac x a} + C
+      | c = simplifying
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = x
+      | r = a \tan \theta
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d x} {\d \theta}
+      | r = a \sec^2 \theta
+      | c = [[Derivative of Tangent Function]]
+}}
+{{eqn | ll= \leadsto
+      | l = \int \frac {x^2 \rd x} {\paren {x^2 + a^2}^2}
+      | r = \int \frac {a^2 \tan^2 \theta a \sec^2 \theta} {\paren {a^2 \tan^2 \theta + a^2}^2} \rd \theta
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \int \frac {a^3 \tan^2 \theta \sec^2 \theta} {a^4 \sec^4 \theta} \rd \theta
+      | c = [[Difference of Squares of Secant and Tangent]]
+}}
+{{eqn | r = \frac 1 a \int \frac {\tan^2 \theta} {\sec^2 \theta} \rd \theta
+      | c = simplification
+}}
+{{eqn | r = \frac 1 a \int \tan^2 \theta \cos^2 \theta \rd \theta
+      | c = {{Defof|Real Secant Function}}
+}}
+{{eqn | r = \frac 1 a \int \frac {\sin^2 \theta} {\cos^2 \theta} \cos^2 \theta \rd \theta
+      | c = {{Defof|Real Tangent Function}}
+}}
+{{eqn | r = \frac 1 a \int \sin^2 \theta \rd \theta
+      | c = simplification
+}}
+{{eqn | r = \frac \theta {2 a} - \frac {\sin 2 \theta} {4 a} + C
+      | c = [[Primitive of Square of Sine Function]]
+}}
+{{eqn | r = \frac \theta {2 a}  - \frac 1 {2 a} \frac {\tan \theta} {1 + \tan^2 \theta} + C
+      | c = [[Tangent Half-Angle Substitution for Sine]]
+}}
+{{eqn | r = -\frac x {2 \paren {x^2 + a^2} } + \frac 1 {2 a} \arctan \frac x a + C
+      | c = substituting for $\theta$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int \frac {x^2 \rd x} {\paren {x^2 + a^2}^2}
+      | r = \int \frac {x^2 + a^2 - a^2} {\paren {x^2 + a^2}^2} \rd x
+      | c = 
+}}
+{{eqn | r = \int \frac {x^2 + a^2} {\paren {x^2 + a^2}^2} \rd x - a^2 \int \frac {\d x} {\paren {x^2 + a^2}^2}
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \int \frac {\d x} {x^2 + a^2} + a^2 \int \frac {\d x} {\paren {x^2 + a^2}^2}
+      | c = simplification
+}}
+{{eqn | r = \frac 1 a \arctan {\frac x a} + a^2 \int \frac {\d x} {\paren {x^2 + a^2}^2} + C
+      | c = [[Primitive of Reciprocal of x squared plus a squared/Arctangent Form|Primitive of Reciprocal of $x^2 + a^2$]]
+}}
+{{eqn | r = \frac 1 a \arctan {\frac x a} + a^2 \paren {\frac x {2 a^2 \paren {x^2 + a^2} } + \frac 1 {2 a^3} \arctan \frac x a} + C
+      | c = [[Primitive of Reciprocal of x squared plus a squared squared|Primitive of Reciprocal of $\paren {x^2 + a^2}^2$]]
+}}
+{{eqn | r = \frac {-x} {2 \paren {x^2 + a^2} } + \frac 1 {2 a} \arctan \frac x a + C
+      | c = simplifying
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of x cubed over x squared plus a squared}
+Tags: Primitives involving x squared plus a squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {x^3 \rd x} {x^2 + a^2} = \frac {x^2} 2 - \frac {a^2} 2 \map \ln {x^2 + a^2} + C$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int \frac {x^3 \rd x} {x^2 + a^2}
+      | r = \int \paren {x - \frac {a^2 x} {x^2 + a^2} } \rd x
+      | c = long division
+}}
+{{eqn | r = \int x \rd x - a^2 \int \frac {x \rd x} {x^2 + a^2} 
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \frac {x^2} 2 - a^2 \int \frac {x \rd x} {x^2 + a^2} + C
+      | c = [[Primitive of Power]]
+}}
+{{eqn | r = \frac {x^2} 2 - a^2 \paren {\frac 1 2 \map \ln {x^2 + a^2} } + C
+      | c = [[Primitive of x over x squared plus a squared|Primitive of $\dfrac x {x^2 + a^2}$]]
+}}
+{{eqn | r = \frac {x^2} 2 - \frac {a^2} 2 \map \ln {x^2 + a^2} + C
+      | c = simplifying
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = z
+      | r = x^2 + a^2
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d z} {\d x}
+      | r = 2 x
+      | c = [[Derivative of Power]]
+}}
+{{eqn | ll= \leadsto
+      | l = \int \frac {x^3 \rd x} {\paren {x^2 + a^2}^2}
+      | r = \int \frac {\paren {z - a^2} } {z^2} \frac {\d z} 2
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \frac 1 2 \int \frac {\d z} z - \frac {a^2} 2 \int \frac {\d z} {z^2}
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \frac 1 2 \int \frac {\d z} z + \frac {a^2} {2 z} + C
+      | c = [[Primitive of Power]]
+}}
+{{eqn | r = \frac 1 2 \ln z + \frac {a^2} {2 z} + C
+      | c = [[Primitive of Reciprocal]]
+}}
+{{eqn | r = \frac {a^2} {2 \paren {x^2 + a^2} } + \frac 1 2 \map \ln {x^2 + a^2} + C
+      | c = substituting for $z$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int \frac {x^3 \ \mathrm d x} {\left({x^2 + a^2}\right)^2}
+      | r = \int \frac {x \left({x^2 + a^2 - a^2}\right)} {\left({x^2 + a^2}\right)^2} \ \mathrm d x
+      | c = 
+}}
+{{eqn | r = \int \frac {x \left({x^2 + a^2}\right)} {\left({x^2 + a^2}\right)^2} \ \mathrm d x - a^2 \int \frac {x \ \mathrm d x} {\left({x^2 + a^2}\right)^2}
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \int \frac {x \ \mathrm d x} {x^2 + a^2} - a^2 \int \frac {x \ \mathrm d x} {\left({x^2 + a^2}\right)^2}
+      | c = simplification
+}}
+{{eqn | r = \frac 1 2 \ln \left({x^2 + a^2}\right) - a^2 \int \frac {x \ \mathrm d x} {\left({x^2 + a^2}\right)^2} + C
+      | c = [[Primitive of x over x squared plus a squared|Primitive of $\dfrac x {x^2 + a^2}$]]
+}}
+{{eqn | r = \frac 1 2 \ln \left({x^2 + a^2}\right) - a^2 \left({\frac {-1} {2 \left({x^2 - a^2}\right)} }\right) + C
+      | c = [[Primitive of x over x squared minus a squared squared|Primitive of $\dfrac x {\left({x^2 + a^2}\right)^2}$]]
+}}
+{{eqn | r = \frac {a^2} {2 \left({x^2 + a^2}\right)} + \frac 1 2 \ln \left({x^2 + a^2}\right) + C
+      | c = simplifying
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Reciprocal of x by x squared plus a squared}
+Tags: Primitive of Reciprocal of x by x squared plus a squared
+
+\begin{theorem}
+:$\ds \int \frac {\rd x} {x \paren {x^2 + a^2} } = \frac 1 {2 a^2} \map \ln {\frac {x^2} {x^2 + a^2} } + C$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int \frac {\d x} {x \paren {x^2 + a^2} }
+      | r = \int \paren {\frac 1 {a^2 x} - \frac x {a^2 \paren {x^2 + a^2} } } \rd x
+      | c = [[Primitive of Reciprocal of x by x squared plus a squared/Partial Fraction Expansion|Partial Fraction Expansion]]
+}}
+{{eqn | r = \frac 1 {a^2} \int \frac {\d x} x - \frac 1 {a^2} \int \frac {x \rd x} {x^2 + a^2}
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \frac 1 {a^2} \ln \size x - \frac 1 {a^2} \int \frac {x \rd x} {x^2 + a^2} + C
+      | c = [[Primitive of Reciprocal]]
+}}
+{{eqn | r = \frac 1 {a^2} \ln \size x - \frac 1 {a^2} \paren {\frac 1 2 \, \map \ln {x^2 + a^2} } + C
+      | c = [[Primitive of x over x squared plus a squared|Primitive of $\dfrac x {x^2 + a^2}$]]
+}}
+{{eqn | r = \frac 1 {2 a^2} \map \ln {\frac {x^2} {x^2 + a^2} } + C
+      | c = [[Difference of Logarithms]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int \frac {\d x} {x \paren {x^2 + a^2} }
+      | r = \int \frac {a^2 \rd x} {a^2 x \paren {x^2 + a^2} }
+      | c = multiplying [[Definition:Numerator|top]] and [[Definition:Denominator|bottom]] by $a^2$
+}}
+{{eqn | r = \int \frac {\paren {x^2 + a^2 - x^2} \rd x} {a^2 x \paren {x^2 + a^2} }
+      | c = adding and subtracting $x^2$
+}}
+{{eqn | r = \frac 1 {a^2} \int \frac {\paren {x^2 + a^2} \rd x} {x \paren {x^2 + a^2} } - \frac 1 {a^2} \int \frac {x^2 \rd x} {x \paren {x^2 + a^2} }
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \frac 1 {a^2} \int \frac {\d x} x - \frac 1 {a^2} \int \frac {x \rd x} {x^2 + a^2}
+      | c = simplifying
+}}
+{{eqn | r = \frac 1 {a^2} \ln \size x - \frac 1 {a^2} \int \frac {x \rd x} {x^2 + a^2} + C
+      | c = [[Primitive of Reciprocal]]
+}}
+{{eqn | r = \frac 1 {a^2} \ln \size x - \frac 1 {a^2} \paren {\frac 1 2 \map \ln {x^2 + a^2} } + C
+      | c = [[Primitive of x over x squared plus a squared|Primitive of $\dfrac x {x^2 + a^2}$]]
+}}
+{{eqn | r = \frac 1 {2 a^2} \map \ln {\frac {x^2} {x^2 + a^2} } + C
+      | c = [[Difference of Logarithms]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+From [[Primitive of Reciprocal of x by Power of x plus Power of a]]:
+:$\displaystyle \int \frac {\mathrm d x} {x \left({x^n + a^n}\right)} = \frac 1 {n a^n} \ln \left\vert{\frac {x^n} {x^n + a^n} }\right\vert + C$
+So:
+{{begin-eqn}}
+{{eqn | l = \int \frac {\mathrm d x} {x \left({x^2 + a^2}\right)}
+      | r = \frac 1 {2 a^2} \ln \left\vert{\frac {x^2} {x^2 + a^2} }\right\vert + C
+      | c = [[Primitive of Reciprocal of x by Power of x plus Power of a|Primitive of $\dfrac 1 {x \left({x^n + a^n}\right)}$]] with $n = 2$
+}}
+{{eqn | r = \frac 1 {2 a^2} \ln \left({\frac {x^2} {x^2 + a^2} }\right) + C
+      | c = [[Absolute Value of Even Power]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Reciprocal of x squared by x squared plus a squared}
+Tags: Primitives involving x squared plus a squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {\d x} {x^2 \paren {x^2 + a^2} } = -\frac 1 {a^2 x} - \frac 1 {a^3} \arctan \frac x a + C$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int \frac {\d x} {x^2 \paren {x^2 + a^2} }
+      | r = \int \paren {\frac 1 {a^2 x^2} - \frac 1 {a^2 \paren {x^2 + a^2} } } \rd x
+      | c = [[Primitive of Reciprocal of x squared by x squared plus a squared/Partial Fraction Expansion|Partial Fraction Expansion]]
+}}
+{{eqn | r = \frac 1 {a^2} \int \frac {\d x} {x^2} - \frac 1 {a^2} \int \frac {1 \rd x} {x^2 + a^2}
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = -\frac 1 {a^2 x} - \frac 1 {a^2} \int \frac {x \rd x} {x^2 + a^2} + C
+      | c = [[Primitive of Power]]
+}}
+{{eqn | r = -\frac 1 {a^2 x} - \frac 1 {a^2} \paren {\frac 1 a \arctan \frac x a} + C
+      | c = [[Primitive of Reciprocal of x squared plus a squared/Arctangent Form|Primitive of Reciprocal of $x^2 + a^2$]]
+}}
+{{eqn | r = -\frac 1 {a^2 x} - \frac 1 {a^3} \arctan \frac x a + C
+      | c = simplifying
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Reciprocal of x cubed by x squared plus a squared}
+Tags: Primitives involving x squared plus a squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {\mathrm d x} {x^3 \left({x^2 + a^2}\right)} = -\frac 1 {2 a^2 x^2} - \frac 1 {2 a^4} \ln \left({\frac {x^2 + a^2} {x^2} }\right) + C$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int \frac {\mathrm d x} {x^3 \left({x^2 + a^2}\right)}
+      | r = \int \left({\frac 1 {a^2 x^3} - \frac 1 {a^4 x} + \frac x {a^4 \left({x^2 + a^2}\right)} }\right) \ \mathrm d x
+      | c = [[Primitive of Reciprocal of x cubed by x squared plus a squared/Partial Fraction Expansion|Partial Fraction Expansion]]
+}}
+{{eqn | r = \frac 1 {a^2} \int \frac {\mathrm d x} {x^3} - \frac 1 {a^4} \int \frac {\mathrm d x} x + \frac 1 {a^4} \int \frac {x \ \mathrm d x} {x^2 + a^2}
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \frac {-1} {a^2 x^2} - \frac 1 {a^4} \int \frac {\mathrm d x} x + \frac 1 {a^4} \int \frac {x \ \mathrm d x} {x^2 + a^2} + C
+      | c = [[Primitive of Power]]
+}}
+{{eqn | r = \frac {-1} {a^2 x^2} - \frac 1 {a^4} \ln \left\vert{x}\right\vert + \frac 1 {a^4} \int \frac {x \ \mathrm d x} {x^2 + a^2} + C
+      | c = [[Primitive of Reciprocal]]
+}}
+{{eqn | r = \frac {-1} {a^2 x^2} - \frac 1 {a^4} \ln \left\vert{x}\right\vert + \frac 1 {a^4} \left({\frac 1 2 \ln \left ({x^2 + a^2}\right)}\right) + C
+      | c = [[Primitive of x over x squared plus a squared|Primitive of $\dfrac x {x^2 + a^2}$]]
+}}
+{{eqn | r = -\frac 1 {a^2 x^2} - \frac 1 {2 a^4} \ln \left({x^2}\right) + \frac 1 {2 a^4} \ln \left ({x^2 + a^2}\right) + C
+      | c = [[Logarithm of Power]]
+}}
+{{eqn | r = -\frac 1 {2 a^2 x^2} - \frac 1 {2 a^4} \ln \left({\frac {x^2 + a^2} {x^2} }\right) + C
+      | c = [[Difference of Logarithms]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Reciprocal of x squared plus a squared squared}
+Tags: Primitives involving x squared plus a squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {\d x} {\paren {x^2 + a^2}^2} = \frac x {2 a^2 \paren {x^2 + a^2} } + \frac 1 {2 a^3} \arctan \frac x a + C$
+\end{theorem}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = x
+      | r = a \tan \theta
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d x} {\d \theta}
+      | r = a \sec^2 \theta
+      | c = [[Derivative of Tangent Function]]
+}}
+{{eqn | ll= \leadsto
+      | l = \int \frac {\d x} {\paren {x^2 + a^2}^2}
+      | r = \int \frac {a \sec^2 \theta} {\paren {a^2 \tan^2 \theta + a^2}^2} \rd \theta
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \int \frac {a \sec^2 \theta} {\paren {a^2 \sec^2 \theta}^2} \rd \theta
+      | c = [[Difference of Squares of Secant and Tangent]]
+}}
+{{eqn | r = \frac 1 {a^3} \int \frac {\d \theta} {\sec^2 \theta}
+      | c = simplification
+}}
+{{eqn | r = \frac 1 {a^3} \int \cos^2 \theta \rd \theta
+      | c = {{Defof|Real Secant Function}}
+}}
+{{eqn | r = \frac 1 {a^3} \paren {\frac \theta 2 + \frac {\sin 2 \theta} 4} + C
+      | c = [[Primitive of Square of Cosine Function]]
+}}
+{{eqn | r = \frac \theta {2 a^3} + \frac 1 {2 a^3} \frac {\tan \theta} {1 + \tan^2 \theta} + C
+      | c = [[Tangent Half-Angle Substitution for Sine]]
+}}
+{{eqn | r = \frac x {2 a^2 \paren {x^2 + a^2} } + \frac 1 {2 a^3} \arctan \frac x a + C
+      | c = substituting for $\theta$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of x over x squared plus a squared squared}
+Tags: Primitives involving x squared plus a squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {x \ \mathrm d x} {\left({x^2 + a^2}\right)^2} = -\frac 1 {2 \left({x^2 + a^2}\right)} + C$
+\end{theorem}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = z
+      | r = x^2 + a^2
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \frac {\mathrm d z} {\mathrm d x}
+      | r = 2 x
+      | c = [[Derivative of Power]]
+}}
+{{eqn | ll= \implies
+      | l = \int \frac {x \ \mathrm d x} {\left({x^2 + a^2}\right)^2}
+      | r = \int \frac {\mathrm d z} {2 z^2}
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = -\frac 1 {2 z} + C
+      | c = [[Primitive of Power]]
+}}
+{{eqn | r = -\frac 1 {2 \left({x^2 + a^2}\right)} + C
+      | c = substituting for $z$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of x squared over x squared plus a squared squared}
+Tags: Primitive of x squared over x squared plus a squared squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {x^2 \ \mathrm d x} {\left({x^2 + a^2}\right)^2} = \frac {-x} {2 \left({x^2 + a^2}\right)} + \frac 1 {2 a} \arctan \frac x a + C$
+\end{theorem}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = x
+      | r = a \tan \theta
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d x} {\d \theta}
+      | r = a \sec^2 \theta
+      | c = [[Derivative of Tangent Function]]
+}}
+{{eqn | ll= \leadsto
+      | l = \int \frac {x^2 \rd x} {\paren {x^2 + a^2}^2}
+      | r = \int \frac {a^2 \tan^2 \theta a \sec^2 \theta} {\paren {a^2 \tan^2 \theta + a^2}^2} \rd \theta
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \int \frac {a^3 \tan^2 \theta \sec^2 \theta} {a^4 \sec^4 \theta} \rd \theta
+      | c = [[Difference of Squares of Secant and Tangent]]
+}}
+{{eqn | r = \frac 1 a \int \frac {\tan^2 \theta} {\sec^2 \theta} \rd \theta
+      | c = simplification
+}}
+{{eqn | r = \frac 1 a \int \tan^2 \theta \cos^2 \theta \rd \theta
+      | c = {{Defof|Real Secant Function}}
+}}
+{{eqn | r = \frac 1 a \int \frac {\sin^2 \theta} {\cos^2 \theta} \cos^2 \theta \rd \theta
+      | c = {{Defof|Real Tangent Function}}
+}}
+{{eqn | r = \frac 1 a \int \sin^2 \theta \rd \theta
+      | c = simplification
+}}
+{{eqn | r = \frac \theta {2 a} - \frac {\sin 2 \theta} {4 a} + C
+      | c = [[Primitive of Square of Sine Function]]
+}}
+{{eqn | r = \frac \theta {2 a}  - \frac 1 {2 a} \frac {\tan \theta} {1 + \tan^2 \theta} + C
+      | c = [[Tangent Half-Angle Substitution for Sine]]
+}}
+{{eqn | r = -\frac x {2 \paren {x^2 + a^2} } + \frac 1 {2 a} \arctan \frac x a + C
+      | c = substituting for $\theta$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int \frac {x^2 \rd x} {\paren {x^2 + a^2}^2}
+      | r = \int \frac {x^2 + a^2 - a^2} {\paren {x^2 + a^2}^2} \rd x
+      | c = 
+}}
+{{eqn | r = \int \frac {x^2 + a^2} {\paren {x^2 + a^2}^2} \rd x - a^2 \int \frac {\d x} {\paren {x^2 + a^2}^2}
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \int \frac {\d x} {x^2 + a^2} + a^2 \int \frac {\d x} {\paren {x^2 + a^2}^2}
+      | c = simplification
+}}
+{{eqn | r = \frac 1 a \arctan {\frac x a} + a^2 \int \frac {\d x} {\paren {x^2 + a^2}^2} + C
+      | c = [[Primitive of Reciprocal of x squared plus a squared/Arctangent Form|Primitive of Reciprocal of $x^2 + a^2$]]
+}}
+{{eqn | r = \frac 1 a \arctan {\frac x a} + a^2 \paren {\frac x {2 a^2 \paren {x^2 + a^2} } + \frac 1 {2 a^3} \arctan \frac x a} + C
+      | c = [[Primitive of Reciprocal of x squared plus a squared squared|Primitive of Reciprocal of $\paren {x^2 + a^2}^2$]]
+}}
+{{eqn | r = \frac {-x} {2 \paren {x^2 + a^2} } + \frac 1 {2 a} \arctan \frac x a + C
+      | c = simplifying
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of x cubed over x squared plus a squared squared}
+Tags: Primitive of x cubed over x squared plus a squared squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {x^3 \ \mathrm d x} {\left({x^2 + a^2}\right)^2} = \frac {a^2} {2 \left({x^2 + a^2}\right)} + \frac 1 2 \ln \left({x^2 + a^2}\right) + C$
+\end{theorem}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = z
+      | r = x^2 + a^2
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d z} {\d x}
+      | r = 2 x
+      | c = [[Derivative of Power]]
+}}
+{{eqn | ll= \leadsto
+      | l = \int \frac {x^3 \rd x} {\paren {x^2 + a^2}^2}
+      | r = \int \frac {\paren {z - a^2} } {z^2} \frac {\d z} 2
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \frac 1 2 \int \frac {\d z} z - \frac {a^2} 2 \int \frac {\d z} {z^2}
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \frac 1 2 \int \frac {\d z} z + \frac {a^2} {2 z} + C
+      | c = [[Primitive of Power]]
+}}
+{{eqn | r = \frac 1 2 \ln z + \frac {a^2} {2 z} + C
+      | c = [[Primitive of Reciprocal]]
+}}
+{{eqn | r = \frac {a^2} {2 \paren {x^2 + a^2} } + \frac 1 2 \map \ln {x^2 + a^2} + C
+      | c = substituting for $z$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int \frac {x^3 \ \mathrm d x} {\left({x^2 + a^2}\right)^2}
+      | r = \int \frac {x \left({x^2 + a^2 - a^2}\right)} {\left({x^2 + a^2}\right)^2} \ \mathrm d x
+      | c = 
+}}
+{{eqn | r = \int \frac {x \left({x^2 + a^2}\right)} {\left({x^2 + a^2}\right)^2} \ \mathrm d x - a^2 \int \frac {x \ \mathrm d x} {\left({x^2 + a^2}\right)^2}
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \int \frac {x \ \mathrm d x} {x^2 + a^2} - a^2 \int \frac {x \ \mathrm d x} {\left({x^2 + a^2}\right)^2}
+      | c = simplification
+}}
+{{eqn | r = \frac 1 2 \ln \left({x^2 + a^2}\right) - a^2 \int \frac {x \ \mathrm d x} {\left({x^2 + a^2}\right)^2} + C
+      | c = [[Primitive of x over x squared plus a squared|Primitive of $\dfrac x {x^2 + a^2}$]]
+}}
+{{eqn | r = \frac 1 2 \ln \left({x^2 + a^2}\right) - a^2 \left({\frac {-1} {2 \left({x^2 - a^2}\right)} }\right) + C
+      | c = [[Primitive of x over x squared minus a squared squared|Primitive of $\dfrac x {\left({x^2 + a^2}\right)^2}$]]
+}}
+{{eqn | r = \frac {a^2} {2 \left({x^2 + a^2}\right)} + \frac 1 2 \ln \left({x^2 + a^2}\right) + C
+      | c = simplifying
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Reciprocal of x by x squared plus a squared squared}
+Tags: Primitives involving x squared plus a squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {\d x} {x \paren {x^2 + a^2}^2} = \frac 1 {2 a^2 \paren {x^2 + a^2} } + \frac 1 {2 a^4} \map \ln {\frac {x^2} {x^2 + a^2} } + C$
+\end{theorem}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = \int \frac {\d x} {x \paren {x^2 + a^2}^2}
+      | r = \int \paren {\frac 1 {a^4 x} - \frac x {a^4 \paren {x^2 + a^2} } - \frac x {a^2 \paren {x^2 + a^2}^2} } \rd x
+      | c = [[Primitive of Reciprocal of x by x squared plus a squared squared/Partial Fraction Expansion|Partial Fraction Expansion]]
+}}
+{{eqn | r = \frac 1 {a^4} \int \frac {\d x} x - \frac 1 {a^4} \int \frac {x \rd x} {x^2 + a^2} - \frac 1 {a^2} \int \frac {x \rd x} {\paren {x^2 + a^2}^2}
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \frac 1 {a^4} \ln x - \frac 1 {a^4} \int \frac {x \rd x} {x^2 + a^2} - \frac 1 {a^2} \int \frac {x \rd x} {\paren {x^2 + a^2}^2} + C
+      | c = [[Primitive of Reciprocal]]
+}}
+{{eqn | r = \frac 1 {a^4} \ln x - \frac 1 {a^4} \paren {\frac 1 2 \map \ln {x^2 + a^2} } - \frac 1 {a^2} \int \frac {x \rd x} {\paren {x^2 + a^2}^2} + C
+      | c = [[Primitive of x over x squared plus a squared|Primitive of $\dfrac x {x^2 + a^2}$]]
+}}
+{{eqn | r = \frac 1 {a^4} \ln x - \frac 1 {a^4} \paren {\frac 1 2 \map \ln {x^2 + a^2} } - \frac 1 {a^2} \paren {-\frac 1 {2 \paren {x^2 + a^2} } } + C
+      | c = [[Primitive of x over x squared plus a squared squared|Primitive of $\dfrac x {\left({x^2 + a^2}\right)^2}$]]
+}}
+{{eqn | r = \frac 1 {2 a^2 \paren {x^2 + a^2} } + \frac 1 {2 a^4} \map \ln {\frac {x^2} {x^2 + a^2} } + C
+      | c = simplification
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Reciprocal of x squared by x squared plus a squared squared}
+Tags: Primitives involving x squared plus a squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {\d x} {x^2 \paren {x^2 + a^2}^2} = -\frac 1 {a^4 x} - \frac x {2 a^4 \paren {x^2 + a^2} } - \frac 3 {2 a^5} \arctan \frac x a + C$
+\end{theorem}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = \int \frac {\d x} {x^2 \paren {x^2 + a^2}^2}
+      | r = \int \paren {\frac 1 {a^4 x^2} - \frac 1 {a^4 \paren {x^2 + a^2} } - \frac 1 {a^2 \paren {x^2 + a^2}^2} } \rd x
+      | c = [[Primitive of Reciprocal of x squared by x squared plus a squared squared/Partial Fraction Expansion|Partial Fraction Expansion]]
+}}
+{{eqn | r = \frac 1 {a^4} \int \frac {\d x} {x^2} - \frac 1 {a^4} \int \frac {\d x} {x^2 + a^2} - \frac 1 {a^2} \int \frac {\d x} {\paren {x^2 + a^2}^2}
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \frac 1 {a^4} \frac {-1} x - \frac 1 {a^4} \int \frac {\d x} {x^2 + a^2} - \frac 1 {a^2} \int \frac {\d x} {\paren {x^2 + a^2}^2} + C
+      | c = [[Primitive of Power]]
+}}
+{{eqn | r = -\frac 1 {a^4 x} - \frac 1 {a^4} \paren {\frac 1 a \arctan \frac x a} - \frac 1 {a^2} \int \frac {\d x} {\paren {x^2 + a^2}^2} + C
+      | c = [[Primitive of Reciprocal of x squared plus a squared/Arctangent Form|Primitive of Reciprocal of $x^2 + a^2$]]
+}}
+{{eqn | r = -\frac 1 {a^4 x} - \frac 1 {a^5} \arctan \frac x a - \frac 1 {a^2} \paren {\frac x {2 a^2 \paren {x^2 + a^2} } + \frac 1 {2 a^3} \arctan \frac x a} + C
+      | c = [[Primitive of Reciprocal of x squared plus a squared squared|Primitive of Reciprocal of $\paren {x^2 + a^2}^2$]]
+}}
+{{eqn | r = -\frac 1 {a^4 x} - \frac x {2 a^4 \paren {x^2 + a^2} } - \frac 3 {2 a^5} \arctan \frac x a + C
+      | c = simplification
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Reciprocal of x cubed by x squared plus a squared squared}
+Tags: Primitives involving x squared plus a squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {\d x} {x^3 \paren {x^2 + a^2}^2} = -\frac 1 {2 a^4 x^2} - \frac 1 {2 a^4 \paren {x^2 + a^2} } - \frac 1 {a^6} \map \ln {\frac {x^2} {x^2 + a^2} } + C$
+\end{theorem}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = \int \frac {\d x} {x^3 \paren {x^2 + a^2}^2}
+      | r = \int \paren {-\frac 2 {a^6 x} + \frac 1 {a^4 x^3} + \frac {2 x} {a^6 \paren {x^2 + a^2} } + \frac x {a^4 \paren {x^2 + a^2}^2} } \rd x
+      | c = [[Primitive of Reciprocal of x cubed by x squared plus a squared squared/Partial Fraction Expansion|Partial Fraction Expansion]]
+}}
+{{eqn | r = -\frac 2 {a^6} \int \frac {\d x} x + \frac 1 {a^4} \int \frac {\d x} {x^3} + \frac 2 {a^6} \int \frac {x \rd x} {x^2 + a^2} + \frac 1 {a^4} \int \frac {x \rd x} {\paren {x^2 + a^2}^2}
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = -\frac 2 {a^6} \ln x + \frac 1 {a^4} \int \frac {\d x} {x^3} + \frac 2 {a^6} \int \frac {x \rd x} {x^2 + a^2} + \frac 1 {a^4} \int \frac {x \rd x} {\paren {x^2 + a^2}^2} + C
+      | c = [[Primitive of Reciprocal]]
+}}
+{{eqn | r = -\frac 2 {a^6} \ln x - \frac 1 {2 a^4 x^2} + \frac 2 {a^6} \int \frac {x \rd x} {x^2 + a^2} + \frac 1 {a^4} \int \frac {x \rd x} {\paren {x^2 + a^2}^2} + C
+      | c = [[Primitive of Power]]
+}}
+{{eqn | r = -\frac 2 {a^6} \ln x - \frac 1 {2 a^4 x^2} + \frac 2 {a^6} \paren {\frac 1 2 \map \ln {x^2 + a^2} } + \frac 1 {a^4} \int \frac {x \rd x} {\paren {x^2 + a^2}^2} + C
+      | c = [[Primitive of x over x squared plus a squared|Primitive of $\dfrac x {x^2 + a^2}$]]
+}}
+{{eqn | r = -\frac 2 {a^6} \ln x - \frac 1 {2 a^4 x^2} + \frac 1 {a^6} \map \ln {x^2 + a^2} + \frac 1 {a^4} \paren {-\frac 1 {2 \paren {x^2 + a^2} } } + C
+      | c = [[Primitive of x over x squared plus a squared squared|Primitive of $\dfrac x {\paren {x^2 + a^2}^2}$]]
+}}
+{{eqn | r = -\frac 1 {2 a^4 x^2} - \frac 1 {2 a^4 \paren {x^2 + a^2} } - \frac 1 {a^6} \map \ln {\frac {x^2} {x^2 + a^2} } + C
+      | c = simplification
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Reciprocal of Power of x squared plus a squared}
+Tags: Primitives involving x squared plus a squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {\mathrm d x} {\left({x^2 + a^2}\right)^n} = \frac x {2 \left({n - 1}\right) a^2 \left({x^2 + a^2}\right)^{n - 1} } + \frac {2 n - 3} {\left({2 n - 2}\right) a^2} \int \frac {\mathrm d x} {\left({x^2 + a^2}\right)^{n - 1} }$
+\end{theorem}
+
+\begin{proof}
+Aiming for an expression in the form:
+: $\displaystyle \int u \frac {\mathrm d v} {\mathrm d x} \ \mathrm d x = u v - \int v \ \frac {\mathrm d u} {\mathrm d x} \ \mathrm d x$
+in order to use the technique of [[Integration by Parts]], let:
+{{begin-eqn}}
+{{eqn | l = v
+      | r = x
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \frac {\mathrm d v} {\mathrm d x}
+      | r = 1
+      | c = [[Power Rule for Derivatives]]
+}}
+{{end-eqn}}
+Thus:
+{{begin-eqn}}
+{{eqn | l = u
+      | r = \frac 1 {\left({x^2 + a^2}\right)^{n - 1} }
+      | c = 
+}}
+{{eqn | r = \left({x^2 + a^2}\right)^{- \left({n - 1}\right)}
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \frac {\mathrm d u} {\mathrm d x}
+      | r = - 2 \left({n - 1}\right) x \left({x^2 + a^2}\right)^{- \left({n - 1}\right) - 1}
+      | c = [[Power Rule for Derivatives]] and [[Chain Rule for Derivatives]]
+}}
+{{eqn | r = - 2 \left({n - 1}\right) x \left({x^2 + a^2}\right)^{-n}
+      | c = 
+}}
+{{eqn | r = \frac {- \left({2 n - 2}\right) x} {\left({x^2 + a^2}\right)^n}
+      | c = 
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = \int \frac {\mathrm d x} {\left({x^2 + a^2}\right)^{n - 1} }
+      | r = \frac 1 {\left({x^2 + a^2}\right)^{n - 1} } x - \int x \frac {- \left({2 n - 2}\right) x} {\left({x^2 + a^2}\right)^n} \ \mathrm d x
+      | c = [[Integration by Parts]]
+}}
+{{eqn | r = \frac x {\left({x^2 + a^2}\right)^{n - 1} } + \left({2 n - 2}\right) \int \frac {x^2} {\left({x^2 + a^2}\right)^n} \ \mathrm d x
+      | c = simplifying
+}}
+{{eqn | r = \frac x {\left({x^2 + a^2}\right)^{n - 1} } + \left({2 n - 2}\right) \int \frac {x^2 + a^2 - a^2} {\left({x^2 + a^2}\right)^n} \ \mathrm d x
+      | c = 
+}}
+{{eqn | r = \frac x {\left({x^2 + a^2}\right)^{n - 1} } + \left({2 n - 2}\right) \int \frac {x^2 + a^2} {\left({x^2 + a^2}\right)^n} \ \mathrm d x - \left({2 n - 2}\right) a^2 \int \frac {\mathrm d x} {\left({x^2 + a^2}\right)^n}
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \frac x {\left({x^2 + a^2}\right)^{n - 1} } + \left({2 n - 2}\right) \int \frac {\mathrm d x} {\left({x^2 + a^2}\right)^{n - 1} } - \left({2 n - 2}\right) a^2 \int \frac {\mathrm d x} {\left({x^2 + a^2}\right)^n} 
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \left({1 - \left({2 n - 2}\right)}\right) \int \frac {\mathrm d x} {\left({x^2 + a^2}\right)^{n - 1} }
+      | r = \frac x {\left({x^2 + a^2}\right)^{n - 1} } - \left({2 n - 2}\right) a^2 \int \frac {\mathrm d x} {\left({x^2 + a^2}\right)^n}
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \int \frac {\mathrm d x} {\left({x^2 + a^2}\right)^n}
+      | r = \frac x {2 \left({n - 1}\right) a^2 \left({x^2 + a^2}\right)^{n - 1} } + \frac {2 n - 3} {\left({2 n - 2}\right) a^2} \int \frac {\mathrm d x} {\left({x^2 + a^2}\right)^{n - 1} }
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of x over Power of x squared plus a squared}
+Tags: Primitives involving x squared plus a squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {x \ \mathrm d x} {\left({x^2 + a^2}\right)^n} = \frac {-1} {2 \left({n - 1}\right) \left({x^2 + a^2}\right)^{n - 1} }$
+\end{theorem}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = z
+      | r = x^2 + a^2
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \frac {\mathrm d z} {\mathrm d x}
+      | r = 2 x
+      | c = [[Power Rule for Derivatives]]
+}}
+{{eqn | ll= \implies
+      | l = \int \frac {x \ \mathrm d x} {\left({x^2 + a^2}\right)^n}
+      | r = \int \frac {\mathrm d z} {2 z^n}
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \frac 1 2 \int z^{-n} \ \mathrm d z
+      | c = [[Primitive of Constant Multiple of Function]]
+}}
+{{eqn | r = \frac 1 2 \frac {z^{-n + 1} } {-n + 1} + C
+      | c = [[Primitive of Power]]
+}}
+{{eqn | r = \frac {-1} {2 \left({n - 1}\right) z^{n - 1} } + C
+      | c = simplifying
+}}
+{{eqn | r = \frac {-1} {2 \left({n - 1}\right) \left({x^2 + a^2}\right)^{n - 1} } + C
+      | c = substituting for $z$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Reciprocal of x by Power of x squared plus a squared}
+Tags: Primitives involving x squared plus a squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {\mathrm d x} {x \left({x^2 + a^2}\right)^n} = \frac 1 {2 \left({n - 1}\right) a^2 \left({x^2 + a^2}\right)^{n - 1} } + \frac 1 {a^2} \int \frac {\mathrm d x} {x \left({x^2 + a^2}\right)^{n - 1} }$
+\end{theorem}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = z
+      | r = x^2 + a^2
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \frac {\mathrm d z} {\mathrm d x}
+      | r = 2 x
+      | c = [[Power Rule for Derivatives]]
+}}
+{{eqn | ll= \implies
+      | l = \int \frac {\mathrm d x} {x \left({x^2 + a^2}\right)^n}
+      | r = \int \frac {\mathrm d z} {2 x^2 z^n}
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \frac 1 2 \int \frac {\mathrm d z} {\left({z - a^2}\right) z^n}
+      | c = [[Primitive of Constant Multiple of Function]] and substituting for $x^2$
+}}
+{{end-eqn}}
+From [[Reduction Formula for Primitive of Power of x by Power of a x + b/Increment of Power of x|Reduction Formula for Primitive of $x^m \left({a x + b}\right)^n$: Increment of Power of $x$]]:
+:$\displaystyle \int x^m \left({a x + b}\right)^n \ \mathrm d x = \frac {x^{m+1} \left({a x + b}\right)^{n + 1} } {\left({m + 1}\right) b} - \frac {\left({m + n + 2}\right) a} {\left({m + 1}\right) b} \int x^{m + 1} \left({a x + b}\right)^n \ \mathrm d x$
+Let:
+{{begin-eqn}}
+{{eqn | l = a
+      | o = :=
+      | r = 1
+      | c = 
+}}
+{{eqn | l = b
+      | o = :=
+      | r = -a^2
+      | c = 
+}}
+{{eqn | l = x
+      | o = :=
+      | r = z
+      | c = 
+}}
+{{eqn | l = m
+      | o = :=
+      | r = -n
+      | c = 
+}}
+{{eqn | l = n
+      | o = :=
+      | r = -1
+      | c = 
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = \int \frac {\mathrm d x} {x \left({x^2 + a^2}\right)^n}
+      | r = \frac 1 2 \int \frac {\mathrm d z} {\left({z - a^2}\right) z^n}
+      | c = 
+}}
+{{eqn | r = \frac 1 2 \left({\frac {z^{-n+1} \left({z - a^2}\right)^{-1 + 1} } {-\left({-n + 1}\right) a^2} - \frac {\left({-n + \left({-1}\right) + 2}\right)} {-\left({-n + 1}\right) a^2} \int z^{-n + 1} \left({z - a^2}\right)^{-1} \ \mathrm d z}\right)
+      | c = 
+}}
+{{eqn | r = \frac 1 {2 \left({n - 1}\right) a^2 z^{n-1} } + \frac 1 {2 a^2} \int \frac {\mathrm d z} {z^{n - 1} \left({z - a^2}\right)}
+      | c = simplification
+}}
+{{eqn | r = \frac 1 {2 \left({n - 1}\right) a^2 \left({x^2 + a^2}\right)^{n-1} } + \frac 1 {2 a^2} \int \frac {2 x \ \mathrm d x} {x^2 \left({x^2 + a^2}\right)^{n - 1} }
+      | c = substituting for $z$ and $\mathrm d z$
+}}
+{{eqn | r = \frac 1 {2 \left({n - 1}\right) a^2 \left({x^2 + a^2}\right)^{n - 1} } + \frac 1 {a^2} \int \frac {\mathrm d x} {x \left({x^2 + a^2}\right)^{n - 1} }
+      | c = simplifying
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Reciprocal of Power of x by Power of x squared plus a squared}
+Tags: Primitives involving x squared plus a squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {\d x} {x^m \paren {x^2 + a^2}^n} = \frac 1 {a^2} \int \frac {\d x} {x^m \paren {x^2 + a^2}^{n - 1} } - \frac 1 {a^2} \int \frac {\d x} {x^{m - 2} \paren {x^2 + a^2}^n}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int \frac {\d x} {x^m \paren {x^2 + a^2}^{n - 1} }
+      | r = \int \frac {\paren {x^2 + a^2} \rd x} {x^m \paren {x^2 + a^2}^{n - 1} \paren {x^2 + a^2} }
+      | c = 
+}}
+{{eqn | r = \int \frac {\paren {x^2 + a^2} \rd x} {x^m \paren {x^2 + a^2}^{\left({n - 1}\right) + 1} }
+      | c = 
+}}
+{{eqn | r = \int \frac {x^2 \rd x} {x^m \paren {x^2 + a^2}^n} + a^2 \int \frac {\d x} {x^m \paren {x^2 + a^2}^n}
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \int \frac {\d x} {x^{m - 2} \paren {x^2 + a^2}^n} + a^2 \int \frac {\d x} {x^m \paren {x^2 + a^2}^n}
+      | c = simplifying
+}}
+{{eqn | ll= \leadsto
+      | l = a^2 \int \frac {\d x} {x^m \paren {x^2 + a^2}^n} 
+      | r = \int \frac {\d x} {x^m \paren {x^2 + a^2}^{n - 1} } - \int \frac {\d x} {x^{m - 2} \paren {x^2 + a^2}^n}
+      | c = changing sides
+}}
+{{eqn | ll= \leadsto
+      | l = \int \frac {\d x} {x^m \paren {x^2 + a^2}^n} 
+      | r = \frac 1 {a^2} \int \frac {\d x} {x^m \paren {x^2 + a^2}^{n - 1} } - \frac 1 {a^2} \int \frac {\d x} {x^{m - 2} \paren {x^2 + a^2}^n}
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of x over x squared minus a squared}
+Tags: Primitives involving x squared minus a squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {x \rd x} {x^2 - a^2} = \frac 1 2 \map \ln {x^2 - a^2} + C$
+for $x^2 > a^2$.
+\end{theorem}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = z
+      | r = x^2 - a^2
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d z} {\d x}
+      | r = 2 x
+      | c = [[Power Rule for Derivatives]]
+}}
+{{eqn | ll= \leadsto
+      | l = \int \frac {\d x} {x \paren {x^2 - a^2} }
+      | r = \int \frac {\d z} {2 z}
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \frac 1 2 \int \frac {\d z} z
+      | c = [[Primitive of Constant Multiple of Function]]
+}}
+{{eqn | r = \frac 1 2 \ln z + C
+      | c = [[Primitive of Reciprocal/Corollary|Primitive of Reciprocal: Corollary]] as $z > 0$
+}}
+{{eqn | r = \frac 1 2 \map \ln {x^2 - a^2} + C
+      | c = substituting for $z$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of x cubed over x squared minus a squared}
+Tags: Primitives involving x squared minus a squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {x^3 \ \mathrm d x} {x^2 - a^2} = \frac {x^2} 2 + \frac {a^2} 2 \ln \left({x^2 - a^2}\right) + C$
+for $x^2 > a^2$.
+\end{theorem}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = \int \frac {x^3 \ \mathrm d x} {x^2 - a^2}
+      | r = \int \frac {x \left({x^2 - a^2 + a^2}\right)} {x^2 - a^2} \ \mathrm d x
+      | c = 
+}}
+{{eqn | r = \int \frac {x \left({x^2 - a^2}\right)} {x^2 - a^2} \ \mathrm d x + \int \frac {a^2 x} {x^2 - a^2} \ \mathrm d x
+      | c = 
+}}
+{{eqn | r = \int x \ \mathrm d x + a^2 \int \frac {x \ \mathrm d x} {x^2 - a^2}
+      | c = [[Primitive of Constant Multiple of Function]]
+}}
+{{eqn | r = \frac {x^2} 2 + a^2 \int \frac {x \ \mathrm d x} {x^2 - a^2} + C
+      | c = [[Primitive of Power]]
+}}
+{{eqn | r = \frac {x^2} 2 + a^2 \left({\frac 1 2 \ln \left({x^2 - a^2}\right)}\right) + C
+      | c = [[Primitive of x over x squared minus a squared|Primitive of $\dfrac x {x^2 - a^2}$]]
+}}
+{{eqn | r = \frac {x^2} 2 + \frac {a^2} 2 \ln \left({x^2 - a^2}\right) + C
+      | c = simplifying
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Reciprocal of x by x squared minus a squared}
+Tags: Primitives involving x squared minus a squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {\d x} {x \paren {x^2 - a^2} } = \frac 1 {2 a^2} \, \map \ln {\frac {x^2 - a^2} {x^2} } + C$
+for $x^2 > a^2$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int \frac {\d x} {x \paren {x^2 - a^2} }
+      | r = \int \paren {\frac x {a^2 \paren {x^2 - a^2} }  - \frac 1 {a^2 x} } \rd x
+      | c = [[Primitive of Reciprocal of x by x squared minus a squared/Partial Fraction Expansion|Partial Fraction Expansion]]
+}}
+{{eqn | r = \frac 1 {a^2} \int \frac {x \rd x} {x^2 - a^2} - \frac 1 {a^2} \int \frac {\d x} x
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \frac 1 {a^2} \int \frac {x \rd x} {x^2 - a^2} - \frac 1 {a^2} \ln \size x + C
+      | c = [[Primitive of Reciprocal]]
+}}
+{{eqn | r = \frac 1 {a^2} \paren {\frac 1 2 \map \ln {x^2 - a^2} } - \frac 1 {a^2} \ln \size x + C
+      | c = [[Primitive of x over x squared minus a squared|Primitive of $\dfrac x {x^2 - a^2}$]]
+}}
+{{eqn | r = \frac 1 {2 a^2} \map \ln {x^2 - a^2} - \frac 1 {2 a^2} \ln \size {x^2} + C
+      | c = [[Logarithm of Power]]
+}}
+{{eqn | r = \frac 1 {2 a^2} \map \ln {x^2 - a^2} - \frac 1 {2 a^2} \map \ln {x^2} + C
+      | c = as $x^2 > 0$
+}}
+{{eqn | r = \frac 1 {2 a^2} \map \ln {\frac {x^2 - a^2} {x^2} } + C
+      | c = [[Difference of Logarithms]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Reciprocal of x squared by x squared minus a squared}
+Tags: Primitives involving x squared minus a squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {\mathrm d x} {x^2 \left({x^2 - a^2}\right)} = \frac 1 {a^2 x} + \frac 1 {2 a^3} \ln \left({\frac {x - a} {x + a} }\right) + C$
+for $x^2 > a^2$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int \frac {\mathrm d x} {x^2 \left({x^2 - a^2}\right)}
+      | r = \int \left({\frac 1 {a^2 \left({x^2 - a^2}\right)}  - \frac 1 {a^2 x^2} }\right) \ \mathrm d x
+      | c = [[Primitive of Reciprocal of x squared by x squared minus a squared/Partial Fraction Expansion|Partial Fraction Expansion]]
+}}
+{{eqn | r = \frac 1 {a^2} \int \frac {\mathrm d x} {x^2 - a^2} - \frac 1 {a^2} \int \frac {\mathrm d x} {x^2}
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \frac 1 {a^2} \int \frac {\mathrm d x} {x^2 - a^2} + \frac 1 {a^2 x} + C
+      | c = [[Primitive of Power]]
+}}
+{{eqn | r = \frac 1 {a^2} \left({\frac 1 2 \ln \left ({\frac {x - a} {x + a} }\right) }\right) + \frac 1 {a^2 x} + C
+      | c = [[Primitive of Reciprocal of x squared minus a squared/Logarithm Form|Primitive of $\dfrac 1 {x^2 - a^2}$]]
+}}
+{{eqn | r = \frac 1 {a^2 x} + \frac 1 {2 a^3} \ln \left({\frac {x - a} {x + a} }\right) + C
+      | c = simplifying
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Reciprocal of x cubed by x squared minus a squared}
+Tags: Primitives involving x squared minus a squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {\mathrm d x} {x^3 \left({x^2 - a^2}\right)} = \frac 1 {2 a^2 x^2} + \frac 1 {2 a^4} \ln \left({\frac {x^2 - a^2} {x^2} }\right) + C$
+for $x^2 > a^2$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int \frac {\mathrm d x} {x^3 \left({x^2 - a^2}\right)}
+      | r = \int \left({\frac {-1} {a^2 x^3} - \frac 1 {a^4 x} + \frac x {a^4 \left({x^2 - a^2}\right)} }\right) \ \mathrm d x
+      | c = [[Primitive of Reciprocal of x cubed by x squared minus a squared/Partial Fraction Expansion|Partial Fraction Expansion]]
+}}
+{{eqn | r = \frac {-1} {a^2} \int \frac {\mathrm d x} {x^3} - \frac 1 {a^4} \int \frac {\mathrm d x} x + \frac 1 {a^4} \int \frac {x \ \mathrm d x} {x^2 - a^2}
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \frac 1 {2 a^2 x^2} - \frac 1 {a^4} \int \frac {\mathrm d x} x + \frac 1 {a^4} \int \frac {x \ \mathrm d x} {x^2 - a^2} + C
+      | c = [[Primitive of Power]]
+}}
+{{eqn | r = \frac 1 {2 a^2 x^2} - \frac 1 {a^4} \ln \left\vert{x}\right\vert + \frac 1 {a^4} \int \frac {x \ \mathrm d x} {x^2 - a^2} + C
+      | c = [[Primitive of Reciprocal]]
+}}
+{{eqn | r = \frac 1 {2 a^2 x^2} - \frac 1 {a^4} \ln \left\vert{x}\right\vert + \frac 1 {a^4} \left({\frac 1 2 \ln \left ({x^2 -a^2}\right)}\right) + C
+      | c = [[Primitive of x over x squared minus a squared|Primitive of $\dfrac x {x^2 - a^2}$]]
+}}
+{{eqn | r = \frac 1 {2 a^2 x^2} - \frac 1 {2 a^4} \ln \left({x^2}\right) + \frac 1 {2 a^4} \ln \left ({x^2 - a^2}\right) + C
+      | c = [[Logarithm of Power]]
+}}
+{{eqn | r = \frac 1 {2 a^2 x^2} + \frac 1 {2 a^4} \ln \left({\frac {x^2 - a^2} {x^2} }\right) + C
+      | c = [[Difference of Logarithms]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Reciprocal of x squared minus a squared squared}
+Tags: Primitives involving x squared minus a squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {\d x} {\left({x^2 - a^2}\right)^2} = \frac {-x} {2 a^2 \left({x^2 - a^2}\right)} + \frac 1 {4 a^3} \ln \left({\frac {x + a} {x - a} }\right) + C$
+for $x^2 > a^2$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int \frac {\d x} {\left({x^2 - a^2}\right)^2}
+      | r = \int \left({\frac 1 {4 a^3 \left({x + a}\right)} - \frac 1 {4 a^3 \left({x - a}\right)} + \frac 1 {4 a^2 \left({x + a}\right)^2} + \frac 1 {4 a^2 \left({x - a}\right)^2} }\right) \rd x
+      | c = [[Primitive of Reciprocal of x squared minus a squared squared/Partial Fraction Expansion|Partial Fraction Expansion]]
+}}
+{{eqn | r = \frac 1 {4 a^3} \int \frac {\d x} {x + a} - \frac 1 {4 a^3} \int \frac {\d x} {x - a} + \frac 1 {4 a^2} \int \frac {\d x} {\left({x + a}\right)^2} + \frac 1 {4 a^2} \int \frac {\d x} {\left({x - a}\right)^2}
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \frac 1 {4 a^3} \ln \left\vert{x + a}\right\vert - \frac 1 {4 a^3} \ln \left\vert{x - a}\right\vert + \frac 1 {4 a^2} \int \frac {\d x} {\left({x + a}\right)^2} + \frac 1 {4 a^2} \int \frac {\d x} {\left({x - a}\right)^2} + C
+      | c = [[Primitive of Function of a x + b|Primitive of Function of $a x + b$]] and [[Primitive of Reciprocal]]
+}}
+{{eqn | r = \frac 1 {4 a^3} \ln \left\vert{\frac {x + a} {x - a} }\right\vert + \frac 1 {4 a^2} \int \frac {\d x} {\left({x + a}\right)^2} + \frac 1 {4 a^2} \int \frac {\d x} {\left({x - a}\right)^2} + C
+      | c = [[Difference of Logarithms]]
+}}
+{{eqn | r = \frac 1 {4 a^3} \ln \left({\frac {x + a} {x - a} }\right) + \frac 1 {4 a^2} \int \frac {\d x} {\left({x + a}\right)^2} + \frac 1 {4 a^2} \int \frac {\d x} {\left({x - a}\right)^2} + C
+      | c = [[Sign of Quotient of Factors of Difference of Squares]]
+}}
+{{eqn | r = \frac 1 {4 a^3} \ln \left({\frac {x + a} {x - a} }\right) + \frac 1 {4 a^2} \frac {-1} {\left({x + a}\right)} + \frac 1 {4 a^2} \frac {-1} {\left({x - a}\right)} + C
+      | c = [[Primitive of Function of a x + b|Primitive of Function of $a x + b$]] and [[Primitive of Power]]
+}}
+{{eqn | r = \frac 1 {4 a^3} \ln \left({\frac {x + a} {x - a} }\right) - \left({\frac 1 {4 a^2 \left({x + a}\right)} + \frac 1  {4 a^2 \left({x - a}\right)} }\right) + C
+      | c = simplifying
+}}
+{{eqn | r = \frac 1 {4 a^3} \ln \left({\frac {x + a} {x - a} }\right) - \left({\frac {\left({x - a}\right) + \left({x + a}\right)} {4 a^2 \left({x + a}\right) \left({x - a}\right)} }\right) + C
+      | c = simplifying
+}}
+{{eqn | r = \frac {-x} {2 a^2 \left({x^2 - a^2}\right)} + \frac 1 {4 a^3} \ln \left({\frac {x + a} {x - a} }\right) + C
+      | c = [[Difference of Two Squares]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Chapman-Kolmogorov Equation}
+Tags: Markov Chains
+
+\begin{theorem}
+Let $X$ be a [[Definition:Discrete State-Space|discrete state-space]] [[Definition:Markov Chain|Markov chain]] with $n$-step [[Definition:Transition Probability Matrix|transition probability matrix]]:
+:$\mathbf P^{\paren n} = \sqbrk {\map {p^{\paren n} } {j, k} }_{j, k \mathop \in S}$
+where:
+:$\map {p^{\paren n} } {j, k} = \condprob {X_{m + n} = k} {X_m = j} = {p_{j k} }^{\paren n}$ is the $n$-step [[Definition:Transition Probability|transition probability]].
+Then:
+:$\mathbf P^{\paren {n + m} } = \mathbf P^{\paren n} \mathbf P^{\paren m}$
+or equivalently:
+:$\displaystyle {p_{i j} }^{\paren {n + m} } = \sum_{k \mathop \in S} {p_{i k} }^{\paren n} {p_{k j} }^{\paren m}$
+\end{theorem}
+
+\begin{proof}
+We consider the [[Definition:Conditional Probability|conditional probability]] on the {{LHS}}:
+{{begin-eqn}}
+{{eqn | r = \map \Pr {X_{m + n} = j \mid X_0 = i}
+      | o = 
+}}
+{{eqn | r = \condprob {\paren {\bigcup_{k \mathop \in S} \sqbrk {X_{n + m} = j, X_n = k} } } {X_0 = i}
+}}
+{{eqn | r = \sum_{k \mathop \in S} \condprob {X_{n + m} = j, X_n = k} {X_0 = i}
+      | c = {{Defof|Countable Additivity}}
+}}
+{{eqn | r = \sum_{k \mathop \in S} \condprob {X_{n + m} = j} {X_n = k, X_0 = i} \times \condprob {X_n = k} {X_0 = i}
+      | c = [[Chain Rule for Probability]]
+}}
+{{eqn | r = \sum_{k \mathop \in S} \condprob {X_{n + m} = j} {X_n = k} \times \condprob {X_n = k} {X_0 = i}
+      | c = [[Markov Property]]
+}}
+{{eqn | r = \sum_{k \mathop \in S} {p_{i k} }^{\paren n} {p_{k j} }^{\paren m}
+      | c = {{Defof|Stationary Transition Probabilities}}
+}}
+{{end-eqn}}
+{{qed}}
+{{Namedfor|Sydney Chapman|name2 = Andrey Nikolaevich Kolmogorov|cat = Chapman|cat2 = Kolmogorov}}
+[[Category:Markov Chains]]
+1hzcrk065noab0yoi3pnovz3kcvdkp8
+\end{proof}<|endoftext|>
+\section{Primitive of x over x squared minus a squared squared}
+Tags: Primitives involving x squared minus a squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {x \ \mathrm d x} {\left({x^2 - a^2}\right)^2} = \frac {-1} {2 \left({x^2 - a^2}\right)} + C$
+for $x^2 > a^2$.
+\end{theorem}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = z
+      | r = x^2 - a^2
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \frac {\mathrm d z} {\mathrm d x}
+      | r = 2 x
+      | c = [[Derivative of Power]]
+}}
+{{eqn | ll= \implies
+      | l = \int \frac {x \ \mathrm d x} {\left({x^2 - a^2}\right)^2}
+      | r = \int \frac {\mathrm d z} {2 z^2}
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \frac {-1} {2 z} + C
+      | c = [[Primitive of Power]]
+}}
+{{eqn | r = \frac {-1} {2 \left({x^2 - a^2}\right)} + C
+      | c = substituting for $z$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of x squared over x squared minus a squared squared}
+Tags: Primitives involving x squared minus a squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {x^2 \ \mathrm d x} {\left({x^2 - a^2}\right)^2} = \frac {-x} {2 \left({x^2 - a^2}\right)} + \frac 1 {4 a} \ln \left({\frac {x - a} {x + a} }\right) + C$
+for $x^2 > a^2$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int \frac {x^2 \ \mathrm d x} {\left({x^2 - a^2}\right)^2}
+      | r = \int \frac {x^2 - a^2 + a^2} {\left({x^2 - a^2}\right)^2} \ \mathrm d x
+      | c = 
+}}
+{{eqn | r = \int \frac {x^2 - a^2} {\left({x^2 - a^2}\right)^2} \ \mathrm d x + a^2 \int \frac {\mathrm d x} {\left({x^2 - a^2}\right)^2}
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \int \frac {\mathrm d x} {x^2 - a^2} + a^2 \int \frac {\mathrm d x} {\left({x^2 - a^2}\right)^2}
+      | c = simplification
+}}
+{{eqn | r = \frac 1 {2 a} \left({\frac {x - a} {x + a} }\right) + a^2 \int \frac {\mathrm d x} {\left({x^2 - a^2}\right)^2} + C
+      | c = [[Primitive of Reciprocal of x squared minus a squared/Logarithm Form|Primitive of Reciprocal of $x^2 - a^2$]]
+}}
+{{eqn | r = \frac 1 {2 a} \left({\frac {x - a} {x + a} }\right) + a^2 \left({\frac {-x} {2 a^2 \left({x^2 - a^2}\right)} - \frac 1 {4 a^3} \left({\frac {x - a} {x + a} }\right)}\right) + C
+      | c = [[Primitive of Reciprocal of x squared minus a squared squared|Primitive of Reciprocal of $\left({x^2 - a^2}\right)^2$]]
+}}
+{{eqn | r = \frac {-x} {2 \left({x^2 - a^2}\right)} + \frac 1 {4 a} \ln \left({\frac {x - a} {x + a} }\right) + C
+      | c = simplifying
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of x cubed over x squared minus a squared squared}
+Tags: Primitives involving x squared minus a squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {x^3 \ \mathrm d x} {\left({x^2 - a^2}\right)^2} = \frac {-a^2} {2 \left({x^2 - a^2}\right)} + \frac 1 2 \ln \left({x^2 - a^2}\right) + C$
+for $x^2 > a^2$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int \frac {x^3 \ \mathrm d x} {\left({x^2 - a^2}\right)^2}
+      | r = \int \frac {x \left({x^2 - a^2 + a^2}\right)} {\left({x^2 - a^2}\right)^2} \ \mathrm d x
+      | c = 
+}}
+{{eqn | r = \int \frac {x \left({x^2 - a^2}\right)} {\left({x^2 - a^2}\right)^2} \ \mathrm d x + a^2 \int \frac {x \ \mathrm d x} {\left({x^2 - a^2}\right)^2}
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \int \frac {x \ \mathrm d x} {x^2 - a^2} + a^2 \int \frac {x \ \mathrm d x} {\left({x^2 - a^2}\right)^2}
+      | c = simplification
+}}
+{{eqn | r = \frac 1 2 \ln \left({x^2 - a^2}\right) + a^2 \int \frac {x \ \mathrm d x} {\left({x^2 - a^2}\right)^2} + C
+      | c = [[Primitive of x over x squared minus a squared|Primitive of $\dfrac x {x^2 - a^2}$]]
+}}
+{{eqn | r = \frac 1 2 \ln \left({x^2 - a^2}\right) + a^2 \left({-\frac 1 {2 \left({x^2 - a^2}\right)} }\right) + C
+      | c = [[Primitive of x over x squared minus a squared squared|Primitive of Reciprocal of $\dfrac x {\left({x^2 - a^2}\right)^2}$]]
+}}
+{{eqn | r = \frac {-a^2} {2 \left({x^2 - a^2}\right)} + \frac 1 2 \ln \left({x^2 - a^2}\right) + C
+      | c = simplifying
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Reciprocal of x by x squared minus a squared squared}
+Tags: Primitives involving x squared minus a squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {\d x} {x \paren {x^2 - a^2}^2} = \frac {-1} {2 a^2 \left({x^2 - a^2}\right)} + \frac 1 {2 a^4} \ln \left({\frac {x^2} {x^2 - a^2} }\right) + C$
+for $x^2 > a^2$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int \frac {\d x} {x \paren {x^2 - a^2}^2}
+      | r = \int \left({\frac 1 {a^4 x} + \frac {-x} {a^4 \paren {x^2 - a^2} } + \frac x {a^2 \paren {x^2 - a^2}^2} }\right) \rd x
+      | c = [[Primitive of Reciprocal of x by x squared minus a squared squared/Partial Fraction Expansion|Partial Fraction Expansion]]
+}}
+{{eqn | r = \frac 1 {a^4} \int \frac {\d x} x + \frac {-1} {a^4} \int \frac {x \rd x} {x^2 - a^2} + \frac 1 {a^2} \int \frac {x \rd x} {\paren {x^2 - a^2}^2}
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \frac 1 {a^4} \ln \size x + \frac {-1} {a^4} \int \frac {x \rd x} {x^2 - a^2} + \frac 1 {a^2} \int \frac {x \rd x} {\paren {x^2 - a^2}^2} + C
+      | c = [[Primitive of Reciprocal]]
+}}
+{{eqn | r = \frac 1 {a^4} \ln \size x + \frac {-1} {a^4} \paren {\frac 1 2 \map \ln {x^2 - a^2} } + \frac 1 {a^2} \int \frac {x \rd x} {\paren {x^2 - a^2}^2} + C
+      | c = [[Primitive of x over x squared minus a squared|Primitive of $\dfrac x {x^2 - a^2}$]]
+}}
+{{eqn | r = \frac 1 {a^4} \ln \size x  + \frac {-1} {2 a^4} \map \ln {x^2 - a^2} + \frac 1 {a^2} \paren {\frac {-1} {2 \paren {x^2 - a^2} } } + C
+      | c = [[Primitive of x over x squared minus a squared squared|Primitive of $\dfrac x {\paren {x^2 - a^2}^2}$]]
+}}
+{{eqn | r = \frac 1 {2 a^4} \map \ln {x^2} + \frac {-1} {2 a^4} \map \ln {x^2 - a^2} + \frac 1 {a^2} \paren {\frac {-1} {2 \paren {x^2 - a^2} } } + C
+      | c = [[Logarithm of Power]] and $x^2 > 0$
+}}
+{{eqn | r = \frac {-1} {2 a^2 \paren {x^2 - a^2} } + \frac 1 {2 a^4} \map \ln {\frac {x^2} {x^2 - a^2} } + C
+      | c = [[Difference of Logarithms]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Reciprocal of x squared by x squared minus a squared squared}
+Tags: Primitives involving x squared minus a squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {\mathrm d x} {x^2 \left({x^2 - a^2}\right)^2} = \frac {-1} {a^4 x} - \frac x {2 a^4 \left({x^2 - a^2}\right)} + \frac 3 {4 a^5} \ln \left({\frac {x + a} {x - a} }\right) + C$
+for $x^2 > a^2$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int \frac {\mathrm d x} {x^2 \left({x^2 - a^2}\right)^2}
+      | r = \int \left({\frac 1 {a^4 x^2} + \frac 3 {4 a^5 \left({x + a}\right)} - \frac 3 {4 a^5 \left({x - a}\right)} + \frac 1 {4 a^4 \left({x + a}\right)^2} + \frac 1 {4 a^4 \left({x - a}\right)^2} }\right) \ \mathrm d x
+      | c = [[Primitive of Reciprocal of x squared by x squared minus a squared squared/Partial Fraction Expansion|Partial Fraction Expansion]]
+}}
+{{eqn | r = \frac 1 {a^4} \int \frac {\mathrm d x} {x^2} + \frac 3 {4 a^5} \int \frac {\mathrm d x} {x + a} - \frac 3 {4 a^5} \int \frac {\mathrm d x} {x - a} + \frac 1 {4 a^4} \int \frac {\mathrm d x} {\left({x + a}\right)^2} + \frac 1 {4 a^4} \int \frac {\mathrm d x} {\left({x - a}\right)^2}
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \frac 1 {a^4} \frac {-1} x + \frac 3 {4 a^5} \int \frac {\mathrm d x} {x + a} - \frac 3 {4 a^5} \int \frac {\mathrm d x} {x - a} + \frac 1 {4 a^4} \int \frac {\mathrm d x} {\left({x + a}\right)^2} + \frac 1 {4 a^4} \int \frac {\mathrm d x} {\left({x - a}\right)^2} + C
+      | c = [[Primitive of Power]]
+}}
+{{eqn | r = \frac {-1} {a^4 x} + \frac 3 {4 a^5} \ln \left\vert{x + a}\right\vert - \frac 3 {4 a^5} \ln \left\vert{x - a}\right\vert + \frac 1 {4 a^4} \int \frac {\mathrm d x} {\left({x + a}\right)^2} + \frac 1 {4 a^4} \int \frac {\mathrm d x} {\left({x - a}\right)^2} + C
+      | c = [[Primitive of Reciprocal]]
+}}
+{{eqn | r = \frac {-1} {a^4 x} + \frac 3 {4 a^5} \ln \left\vert{x + a}\right\vert - \frac 3 {4 a^5} \ln \left\vert{x - a}\right\vert + \frac 1 {4 a^4} \frac {-1} {x + a} + \frac 1 {4 a^4} \frac {-1} {x - a} + C
+      | c = [[Primitive of Power]]
+}}
+{{eqn | r = \frac {-1} {a^4 x} + \frac 3 {4 a^5} \ln \left\vert{\frac {x + a} {x - a} }\right\vert + \frac {-1} {4 a^4 \left({ x + a}\right)} + \frac {-1} {4 a^4 \left({x - a}\right)} + C
+      | c = [[Difference of Logarithms]]
+}}
+{{eqn | r = \frac {-1} {a^4 x} + \frac 3 {4 a^5} \ln \left({\frac {x + a} {x - a} }\right) + \frac {-1} {4 a^4 \left({ x + a}\right)} + \frac {-1} {4 a^4 \left({x - a}\right)} + C
+      | c = [[Sign of Quotient of Factors of Difference of Squares]]
+}}
+{{eqn | r = \frac {-1} {a^4 x} - \frac x {2 a^4 \left({x^2 - a^2}\right)} + \frac 3 {4 a^5} \ln \left({\frac {x + a} {x - a} }\right) + C
+      | c = simplification
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Reciprocal of x cubed by x squared minus a squared squared}
+Tags: Primitives involving x squared minus a squared
+
+\begin{theorem}
+:$\ds \int \frac {\d x} {x^3 \paren {x^2 - a^2}^2} = \frac {-1} {2 a^4 x^2} - \frac 1 {2 a^4 \paren {x^2 - a^2} } + \frac 1 {a^6} \map \ln {\frac {x^2} {x^2 - a^2} } + C$
+for $x^2 > a^2$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int \frac {\d x} {x^3 \paren {x^2 - a^2}^2}
+      | r = \int \paren {\frac 1 {a^4 x^3} + \frac 2 {a^6 x} - \frac {2 x} {a^6 \paren {x^2 - a^2} } + \frac x {a^4 \paren {x^2 - a^2}^2} } \rd x
+      | c = [[Primitive of Reciprocal of x cubed by x squared minus a squared squared/Partial Fraction Expansion|Partial Fraction Expansion]]
+}}
+{{eqn | r = \frac 1 {a^4} \int \frac {\d x} {x^3} + \frac 2 {a^6} \int \frac {\d x} x - \frac 2 {a^6} \int \frac {x \rd x} {x^2 - a^2} + \frac 1 {a^4} \int \frac {x \rd x} {\paren {x^2 - a^2}^2}
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \frac 1 {a^4} \frac {-1} {2 x^2} + \frac 2 {a^6} \int \frac {\d x} x - \frac 2 {a^6} \int \frac {x \rd x} {x^2 - a^2} + \frac 1 {a^4} \int \frac {x \rd x} {\paren {x^2 - a^2}^2} + C
+      | c = [[Primitive of Power]]
+}}
+{{eqn | r = \frac {-1} {2 a^4 x^2} + \frac 2 {a^6} \ln \size x - \frac 2 {a^6} \int \frac {x \rd x} {x^2 - a^2} + \frac 1 {a^4} \int \frac {x \rd x} {\paren {x^2 - a^2}^2} + C
+      | c = [[Primitive of Reciprocal]]
+}}
+{{eqn | r = \frac {-1} {2 a^4 x^2} + \frac 2 {a^6} \ln \size x - \frac 2 {a^6} \frac 1 2 \map \ln {x^2 - a^2} + \frac 1 {a^4} \int \frac {x \rd x} {\paren {x^2 - a^2}^2} + C
+      | c = [[Primitive of x over x squared minus a squared|Primitive of $\dfrac x {x^2 - a^2}$]]
+}}
+{{eqn | r = \frac {-1} {2 a^4 x^2} + \frac 2 {a^6} \ln \size x - \frac 1 {a^6} \map \ln {x^2 - a^2} + \frac 1 {a^4} \frac {-1} {2 \paren {x^2 - a^2} }  + C
+      | c = [[Primitive of x over x squared minus a squared squared|Primitive of $\dfrac x {\paren {x^2 - a^2}^2}$]]
+}}
+{{eqn | r = \frac {-1} {2 a^4 x^2} + \frac 1 {a^6} \map \ln {x^2} - \frac 1 {a^6} \map \ln {x^2 - a^2} + \frac 1 {a^4} \frac {-1} {2 \paren {x^2 - a^2} }  + C
+      | c = [[Logarithm of Power]] and $x^2 > 0$
+}}
+{{eqn | r = \frac {-1} {2 a^4 x^2} - \frac 1 {2 a^4 \paren {x^2 - a^2} } + \frac 1 {a^6} \map \ln {\frac {x^2} {x^2 - a^2} } + C
+      | c = [[Difference of Logarithms]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Reciprocal of Power of x squared minus a squared}
+Tags: Primitives involving x squared minus a squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {\mathrm d x} {\left({x^2 - a^2}\right)^n} = \frac {-x} {2 \left({n - 1}\right) a^2 \left({x^2 - a^2}\right)^{n - 1} } - \frac {2 n - 3} {\left({2 n - 2}\right) a^2} \int \frac {\mathrm d x} {\left({x^2 - a^2}\right)^{n - 1} }$
+for $x^2 > a^2$.
+\end{theorem}
+
+\begin{proof}
+Aiming for an expression in the form:
+: $\displaystyle \int u \frac {\mathrm d v} {\mathrm d x} \ \mathrm d x = u v - \int v \ \frac {\mathrm d u} {\mathrm d x} \ \mathrm d x$
+in order to use the technique of [[Integration by Parts]], let:
+{{begin-eqn}}
+{{eqn | l = v
+      | r = x
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \frac {\mathrm d v} {\mathrm d x}
+      | r = 1
+      | c = [[Power Rule for Derivatives]]
+}}
+{{end-eqn}}
+Thus:
+{{begin-eqn}}
+{{eqn | l = u
+      | r = \frac 1 {\left({x^2 - a^2}\right)^{n - 1} }
+      | c = 
+}}
+{{eqn | r = \left({x^2 - a^2}\right)^{- \left({n - 1}\right)}
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \frac {\mathrm d u} {\mathrm d x}
+      | r = - \left({n - 1}\right) \left({2 x}\right) \left({x^2 - a^2}\right)^{- \left({n - 1}\right) - 1}
+      | c = [[Power Rule for Derivatives]] and [[Chain Rule for Derivatives]]
+}}
+{{eqn | r = - 2 \left({n - 1}\right) x \left({x^2 - a^2}\right)^{-n}
+      | c = 
+}}
+{{eqn | r = \frac {- \left({2 n - 2}\right) x} {\left({x^2 - a^2}\right)^n}
+      | c = 
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = \int \frac {\mathrm d x} {\left({x^2 - a^2}\right)^{n - 1} }
+      | r = \frac 1 {\left({x^2 - a^2}\right)^{n - 1} } x - \int x \frac {- \left({2 n - 2}\right) x} {\left({x^2 - a^2}\right)^n} \ \mathrm d x
+      | c = [[Integration by Parts]]
+}}
+{{eqn | r = \frac x {\left({x^2 - a^2}\right)^{n - 1} } + \left({2 n - 2}\right) \int \frac {x^2} {\left({x^2 - a^2}\right)^n} \ \mathrm d x
+      | c = simplifying
+}}
+{{eqn | r = \frac x {\left({x^2 - a^2}\right)^{n - 1} } + \left({2 n - 2}\right) \int \frac {x^2 - a^2 + a^2} {\left({x^2 - a^2}\right)^n} \ \mathrm d x
+      | c = 
+}}
+{{eqn | r = \frac x {\left({x^2 - a^2}\right)^{n - 1} } + \left({2 n - 2}\right) \int \frac {x^2 - a^2} {\left({x^2 - a^2}\right)^n} \ \mathrm d x + \left({2 n - 2}\right) a^2 \int \frac {\mathrm d x} {\left({x^2 - a^2}\right)^n}
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \frac x {\left({x^2 - a^2}\right)^{n - 1} } + \left({2 n - 2}\right) \int \frac {\mathrm d x} {\left({x^2 - a^2}\right)^{n - 1} } + \left({2 n - 2}\right) a^2 \int \frac {\mathrm d x} {\left({x^2 - a^2}\right)^n} 
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \left({1 - \left({2 n - 2}\right)}\right) \int \frac {\mathrm d x} {\left({x^2 - a^2}\right)^{n - 1} }
+      | r = \frac x {\left({x^2 - a^2}\right)^{n - 1} } + \left({2 n - 2}\right) a^2 \int \frac {\mathrm d x} {\left({x^2 - a^2}\right)^n}
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \int \frac {\mathrm d x} {\left({x^2 - a^2}\right)^n}
+      | r = \frac {-x} {2 \left({n - 1}\right) a^2 \left({x^2 - a^2}\right)^{n - 1} } - \frac {2 n - 3} {\left({2 n - 2}\right) a^2} \int \frac {\mathrm d x} {\left({x^2 - a^2}\right)^{n - 1} }
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of x over Power of x squared minus a squared}
+Tags: Primitives involving x squared minus a squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {x \rd x} {\paren {x^2 - a^2}^n} = \frac {-1} {2 \paren {n - 1} \paren {x^2 - a^2}^{n - 1} }$
+for $x^2 > a^2$.
+\end{theorem}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = z
+      | r = x^2 - a^2
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d z} {\d x}
+      | r = 2 x
+      | c = [[Power Rule for Derivatives]]
+}}
+{{eqn | ll= \leadsto
+      | l = \int \frac {x \rd x} {\paren {x^2 - a^2}^n}
+      | r = \int \frac {\d z} {2 z^n}
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \frac 1 2 \int z^{-n} \rd z
+      | c = [[Primitive of Constant Multiple of Function]]
+}}
+{{eqn | r = \frac 1 2 \frac {z^{-n + 1} } {-n + 1} + C
+      | c = [[Primitive of Power]]
+}}
+{{eqn | r = \frac {-1} {2 \paren {n - 1} z^{n - 1} } + C
+      | c = simplifying
+}}
+{{eqn | r = \frac {-1} {2 \paren {n - 1} \paren {x^2 - a^2}^{n - 1} } + C
+      | c = substituting for $z$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Reciprocal of x by Power of x squared minus a squared}
+Tags: Primitives involving x squared minus a squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {\mathrm d x} {x \left({x^2 - a^2}\right)^n} = \frac {-1} {2 \left({n - 1}\right) a^2 \left({x^2 - a^2}\right)^{n - 1} } - \frac 1 {a^2} \int \frac {\mathrm d x} {x \left({x^2 - a^2}\right)^{n - 1} }$
+for $x^2 > a^2$.
+\end{theorem}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = z
+      | r = x^2 - a^2
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \frac {\mathrm d z} {\mathrm d x}
+      | r = 2 x
+      | c = [[Power Rule for Derivatives]]
+}}
+{{eqn | ll= \implies
+      | l = \int \frac {\mathrm d x} {x \left({x^2 - a^2}\right)^n}
+      | r = \int \frac {\mathrm d z} {2 x^2 z^n}
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \frac 1 2 \int \frac {\mathrm d z} {\left({z + a^2}\right) z^n}
+      | c = [[Primitive of Constant Multiple of Function]] and substituting for $x^2$
+}}
+{{end-eqn}}
+From [[Reduction Formula for Primitive of Power of x by Power of a x + b/Increment of Power of x|Reduction Formula for Primitive of $x^m \left({a x + b}\right)^n$: Increment of Power of $x$]]:
+:$\displaystyle \int x^m \left({a x + b}\right)^n \ \mathrm d x = \frac {x^{m+1} \left({a x + b}\right)^{n + 1} } {\left({m + 1}\right) b} - \frac {\left({m + n + 2}\right) a} {\left({m + 1}\right) b} \int x^{m + 1} \left({a x + b}\right)^n \ \mathrm d x$
+Let:
+{{begin-eqn}}
+{{eqn | l = a
+      | o = :=
+      | r = 1
+      | c = 
+}}
+{{eqn | l = b
+      | o = :=
+      | r = a^2
+      | c = 
+}}
+{{eqn | l = x
+      | o = :=
+      | r = z
+      | c = 
+}}
+{{eqn | l = m
+      | o = :=
+      | r = -n
+      | c = 
+}}
+{{eqn | l = n
+      | o = :=
+      | r = -1
+      | c = 
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = \int \frac {\mathrm d x} {x \left({x^2 - a^2}\right)^n}
+      | r = \frac 1 2 \int \frac {\mathrm d z} {\left({z + a^2}\right) z^n}
+      | c = 
+}}
+{{eqn | r = \frac 1 2 \left({\frac {z^{-n+1} \left({z + a^2}\right)^{-1 + 1} } {\left({-n + 1}\right) a^2} - \frac {\left({-n + \left({-1}\right) + 2}\right)} {\left({-n + 1}\right) a^2} \int z^{-n + 1} \left({z + a^2}\right)^{-1} \ \mathrm d z}\right)
+      | c = 
+}}
+{{eqn | r = \frac {-1} {2 \left({n - 1}\right) a^2 z^{n-1} } - \frac 1 {2 a^2} \int \frac {\mathrm d z} {z^{n - 1} \left({z + a^2}\right)}
+      | c = simplification
+}}
+{{eqn | r = \frac {-1} {2 \left({n - 1}\right) a^2 \left({x^2 - a^2}\right)^{n-1} } - \frac 1 {2 a^2} \int \frac {2 x \ \mathrm d x} {x^2 \left({x^2 - a^2}\right)^{n - 1} }
+      | c = substituting for $z$ and $\mathrm d z$
+}}
+{{eqn | r = \frac {-1} {2 \left({n - 1}\right) a^2 \left({x^2 - a^2}\right)^{n - 1} } - \frac 1 {a^2} \int \frac {\mathrm d x} {x \left({x^2 - a^2}\right)^{n - 1} }
+      | c = simplifying
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Power of x over Power of x squared minus a squared}
+Tags: Primitives involving x squared minus a squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {x^m \rd x} {\paren {x^2 - a^2}^n} = \int \frac {x^{m - 2} \rd x} {\paren {x^2 - a^2}^{n - 1} } + a^2 \int \frac {x^{m - 2} \rd x} {\paren {x^2 - a^2}^n}$
+for $x^2 > a^2$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int \frac {x^m \rd x} {\paren {x^2 - a^2}^n}
+      | r = \int \frac {x^{m - 2} \paren {x^2} \rd x} {\paren {x^2 - a^2}^n}
+      | c = 
+}}
+{{eqn | r = \int \frac {x^{m - 2} \paren {x^2 - a^2 + a^2} \rd x} {\paren {x^2 - a^2}^n}
+      | c = 
+}}
+{{eqn | r = \int \frac {x^{m - 2} \paren {x^2 - a^2} \rd x} {\paren {x^2 - a^2}^n} + a^2 \int \frac {x^{m - 2} \rd x} {\paren {x^2 - a^2}^n}
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \int \frac {x^{m - 2} \rd x} {\paren {x^2 - a^2}^{n - 1} } + a^2 \int \frac {x^{m - 2} \rd x} {\paren {x^2 - a^2}^n}
+      | c = simplifying
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Reciprocal of Power of x by Power of x squared minus a squared}
+Tags: Primitives involving x squared minus a squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {\d x} {x^m \paren {x^2 - a^2}^n} = \frac 1 {a^2} \int \frac {\d x} {x^{m - 2} \paren {x^2 - a^2}^n} - \frac 1 {a^2} \int \frac {\d x} {x^m \paren {x^2 - a^2}^{n - 1} }$
+for $x^2 > a^2$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int \frac {\d x} {x^m \paren {x^2 - a^2}^{n - 1} }
+      | r = \int \frac {\paren {x^2 - a^2} \rd x} {x^m \paren {x^2 - a^2}^{n - 1} \paren {x^2 - a^2} }
+      | c = 
+}}
+{{eqn | r = \int \frac {\paren {x^2 - a^2} \rd x} {x^m \paren {x^2 - a^2}^{\paren {n - 1} + 1} }
+      | c = 
+}}
+{{eqn | r = \int \frac {x^2 \rd x} {x^m \paren {x^2 - a^2}^n} - a^2 \int \frac {\d x} {x^m \paren {x^2 - a^2}^n}
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \int \frac {\d x} {x^{m - 2} \paren {x^2 - a^2}^n} - a^2 \int \frac {\d x} {x^m \paren {x^2 - a^2}^n}
+      | c = simplifying
+}}
+{{eqn | ll= \leadsto
+      | l = a^2 \int \frac {\d x} {x^m \paren {x^2 - a^2}^n} 
+      | r = \int \frac {\d x} {x^{m - 2} \paren {x^2 - a^2}^n} - \int \frac {\d x} {x^m \paren {x^2 - a^2}^{n - 1} }
+      | c = changing sides
+}}
+{{eqn | ll= \leadsto
+      | l = \int \frac {\d x} {x^m \paren {x^2 - a^2}^n} 
+      | r = \frac 1 {a^2} \int \frac {\d x} {x^{m - 2} \paren {x^2 - a^2}^n} - \frac 1 {a^2} \int \frac {\d x} {x^m \paren {x^2 - a^2}^{n - 1} }
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sign of Quotient of Factors of Difference of Squares}
+Tags: Real Analysis, Signum Function
+
+\begin{theorem}
+Let $a, b \in \R$ such that $a \ne b$.
+Then
+:$\map \sgn {a^2 - b^2} = \map \sgn {\dfrac {a + b} {a - b} } = \map \sgn {\dfrac {a - b} {a + b} }$
+where $\sgn$ denotes the [[Definition:Signum Function|signum]] of a [[Definition:Real Number|real number]].
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map \sgn {\frac {a - b} {a + b} }
+      | r = \map \sgn {a - b} \frac 1 {\map \sgn {a + b} }
+      | c = [[Signum Function is Completely Multiplicative]]
+}}
+{{eqn | r = \map \sgn {a - b} \map \sgn {a + b}
+      | c = [[Signum Function of Reciprocal]]
+}}
+{{eqn | r = \map \sgn {\paren {a - b} \paren {a + b} }
+      | c = [[Signum Function is Completely Multiplicative]]
+}}
+{{eqn | r = \map \sgn {a^2 - b^2}
+      | c = [[Difference of Two Squares]]
+}}
+{{eqn | r = \map \sgn {\paren {a + b} \paren {a - b} }
+      | c = [[Difference of Two Squares]]
+}}
+{{eqn | r = \map \sgn {a + b} \map \sgn {a - b}
+      | c = [[Signum Function is Completely Multiplicative]]
+}}
+{{eqn | r = \map \sgn {a + b} \frac 1 {\map \sgn {a - b} }
+      | c = [[Signum Function of Reciprocal]]
+}}
+{{eqn | r = \map \sgn {\frac {a + b} {a - b} }
+      | c = [[Signum Function is Completely Multiplicative]]
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Real Analysis]]
+[[Category:Signum Function]]
+ss7k87h3t6ty6ypqnd7cqi9f1xktmko
+\end{proof}<|endoftext|>
+\section{Signum Function is Completely Multiplicative}
+Tags: Signum Function, Completely Multiplicative Functions
+
+\begin{theorem}
+The [[Definition:Signum Function|signum function]] on the set of [[Definition:Real Number|real numbers]] is a [[Definition:Completely Multiplicative Function|completely multiplicative function]]:
+:$\forall x, y \in \R: \map \sgn {x y} = \map \sgn x \map \sgn y$
+\end{theorem}
+
+\begin{proof}
+Let $x = 0$ or $y = 0$.
+Then:
+{{begin-eqn}}
+{{eqn | l = x y
+      | r = 0
+      | c = 
+}}
+{{eqn | n = 1
+      | ll= \leadsto
+      | l = \map \sgn {x y}
+      | r = 0
+      | c = 
+}}
+{{end-eqn}}
+and either $\map \sgn x = 0$ or $\map \sgn y = 0$ and so:
+{{begin-eqn}}
+{{eqn | l = \map \sgn x \map \sgn y
+      | r = 0
+      | c = 
+}}
+{{eqn | r = \map \sgn {x y}
+      | c = from $(1)$ above
+}}
+{{end-eqn}}
+{{qed|lemma}}
+Let $x > 0$ and $y > 0$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \map \sgn x
+      | r = 1
+      | c = 
+}}
+{{eqn | lo= \land
+      | l = \map \sgn y
+      | r = 1
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \map \sgn x \map \sgn y
+      | r = 1
+      | c = 
+}}
+{{end-eqn}}
+and:
+{{begin-eqn}}
+{{eqn | l = x y
+      | o = >
+      | r = 0
+      | c = 
+}}
+{{eqn | l = \map \sgn {x y}
+      | r = 1
+      | c ={{Defof|Signum Function}}
+}}
+{{eqn | r = \map \sgn x \map \sgn y
+      | c = 
+}}
+{{end-eqn}}
+{{qed|lemma}}
+Let $x < 0$ and $y < 0$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \map \sgn x
+      | r = -1
+      | c = 
+}}
+{{eqn | lo= \land
+      | l = \map \sgn y
+      | r = -1
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \map \sgn x \map \sgn y
+      | r = 1
+      | c = 
+}}
+{{end-eqn}}
+and:
+{{begin-eqn}}
+{{eqn | l = x y
+      | o = >
+      | r = 0
+      | c = 
+}}
+{{eqn | l = \map \sgn {x y}
+      | r = 1
+      | c = {{Defof|Signum Function}}
+}}
+{{eqn | r = \map \sgn x \map \sgn y
+      | c = 
+}}
+{{end-eqn}}
+{{qed|lemma}}
+Let $x < 0$ and $y > 0$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \map \sgn x
+      | r = -1
+      | c = 
+}}
+{{eqn | lo= \land
+      | l = \map \sgn y
+      | r = 1
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \map \sgn x \map \sgn y
+      | r = -1
+      | c = 
+}}
+{{end-eqn}}
+and:
+{{begin-eqn}}
+{{eqn | l = x y
+      | o = <
+      | r = 0
+      | c = 
+}}
+{{eqn | l = \map \sgn {x y}
+      | r = -1
+      | c = {{Defof|Signum Function}}
+}}
+{{eqn | r = \map \sgn x \map \sgn y
+      | c = 
+}}
+{{end-eqn}}
+{{qed|lemma}}
+The same argument, [[Definition:Mutatis Mutandis|mutatis mutandis]], covers the case where $x > 0$ and $y < 0$.
+{{qed}}
+[[Category:Signum Function]]
+[[Category:Completely Multiplicative Functions]]
+5e7i1rjnmnjqtw4i4x1qgjun1m0cife
+\end{proof}<|endoftext|>
+\section{Signum Function is Quotient of Number with Absolute Value}
+Tags: Signum Function, Absolute Value Function
+
+\begin{theorem}
+Let $x \in \R_{\ne 0}$ be a [[Definition:Zero (Number)|non-zero]] [[Definition:Real Number|real number]].
+Then:
+:$\operatorname{sgn} \left({x}\right) = \dfrac x {\left\vert{x}\right\vert} = \dfrac {\left\vert{x}\right\vert} x$
+where:
+:$\operatorname{sgn} \left({x}\right)$ denotes the [[Definition:Signum Function|signum function]] of $x$
+:$\left\vert{x}\right\vert$ denotes the [[Definition:Absolute Value|absolute value]] of $x$.
+\end{theorem}
+
+\begin{proof}
+Let $x \in \R_{\ne 0}$.
+Then either $x > 0$ or $x < 0$.
+Let $x > 0$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \frac x {\left\vert{x}\right\vert}
+      | r = \frac x x
+      | c = Definition of [[Definition:Absolute Value|Absolute Value]], as $x > 0$
+}}
+{{eqn | r = 1
+      | c = 
+}}
+{{eqn | r = \operatorname{sgn} \left({x}\right)
+      | c = Definition of [[Definition:Signum Function|Signum Function]], as $x > 0$
+}}
+{{end-eqn}}
+Similarly:
+{{begin-eqn}}
+{{eqn | l = \frac {\left\vert{x}\right\vert} x
+      | r = \frac x x
+      | c = Definition of [[Definition:Absolute Value|Absolute Value]], as $x > 0$
+}}
+{{eqn | r = 1
+      | c = 
+}}
+{{eqn | r = \operatorname{sgn} \left({x}\right)
+      | c = Definition of [[Definition:Signum Function|Signum Function]], as $x > 0$
+}}
+{{end-eqn}}
+{{qed|lemma}}
+Let $x < 0$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \frac x {\left\vert{x}\right\vert}
+      | r = \frac x {-x}
+      | c = Definition of [[Definition:Absolute Value|Absolute Value]], as $x < 0$
+}}
+{{eqn | r = -1
+      | c = 
+}}
+{{eqn | r = \operatorname{sgn} \left({x}\right)
+      | c = Definition of [[Definition:Signum Function|Signum Function]], as $x < 0$
+}}
+{{end-eqn}}
+Similarly:
+{{begin-eqn}}
+{{eqn | l = \frac {\left\vert{x}\right\vert} x
+      | r = \frac {-x} x
+      | c = Definition of [[Definition:Absolute Value|Absolute Value]], as $x < 0$
+}}
+{{eqn | r = -1
+      | c = 
+}}
+{{eqn | r = \operatorname{sgn} \left({x}\right)
+      | c = Definition of [[Definition:Signum Function|Signum Function]], as $x < 0$
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Signum Function]]
+[[Category:Absolute Value Function]]
+ijvb82dxkam36lfkeikjdgawoymdy7y
+\end{proof}<|endoftext|>
+\section{Absolute Value Function is Completely Multiplicative}
+Tags: Absolute Value Function, Completely Multiplicative Functions, Absolute Value Function is Completely Multiplicative
+
+\begin{theorem}
+The [[Definition:Absolute Value|absolute value function]] on the [[Definition:Real Number|real numbers]] $\R$ is [[Definition:Completely Multiplicative Function|completely multiplicative]]:
+:$\forall x, y \in \R: \left\vert{x y}\right\vert = \left\vert{x}\right\vert \, \left\vert{y}\right\vert$
+where $\left \vert{a}\right \vert$ denotes the [[Definition:Absolute Value|absolute value]] of $a$.
+\end{theorem}
+
+\begin{proof}
+Let $x = 0$ or $y = 0$.
+Then:
+{{begin-eqn}}
+{{eqn | l = x y
+      | r = 0
+      | c = 
+}}
+{{eqn | n = 1
+      | ll= \implies
+      | l = \left\vert{x y}\right\vert
+      | r = 0
+      | c = 
+}}
+{{end-eqn}}
+and either $\left\vert{x}\right\vert = 0$ or $\left\vert{y}\right\vert = 0$ and so:
+{{begin-eqn}}
+{{eqn | l = \left\vert{x}\right\vert \, \left\vert{y}\right\vert
+      | r = 0
+      | c = 
+}}
+{{eqn | r = \left\vert{x y}\right\vert
+      | c = from $(1)$ above
+}}
+{{end-eqn}}
+{{qed|lemma}}
+Let $x > 0$ and $y > 0$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \left\vert{x}\right\vert
+      | r = x
+      | c = 
+}}
+{{eqn | lo= \land
+      | l = \left\vert{y}\right\vert
+      | r = y
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \left\vert{x}\right\vert \, \left\vert{y}\right\vert
+      | r = x y
+      | c = 
+}}
+{{end-eqn}}
+and:
+{{begin-eqn}}
+{{eqn | l = x y
+      | o = >
+      | r = 0
+      | c = 
+}}
+{{eqn | l = \left\vert{x y}\right\vert
+      | r = x y
+      | c = Definition of [[Definition:Absolute Value|Absolute Value]]
+}}
+{{eqn | r = \left\vert{x}\right\vert \, \left\vert{y}\right\vert
+      | c = 
+}}
+{{end-eqn}}
+{{qed|lemma}}
+Let $x < 0$ and $y < 0$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \left\vert{x}\right\vert
+      | r = -x
+      | c = 
+}}
+{{eqn | lo= \land
+      | l = \left\vert{y}\right\vert
+      | r = -y
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \left\vert{x}\right\vert \, \left\vert{y}\right\vert
+      | r = \left({-x}\right) \left({-y}\right)
+      | c = 
+}}
+{{eqn | r = x y
+      | c = 
+}}
+{{end-eqn}}
+and:
+{{begin-eqn}}
+{{eqn | l = x y
+      | o = >
+      | r = 0
+      | c = 
+}}
+{{eqn | l = \left\vert{x y}\right\vert
+      | r = x y
+      | c = {{Defof|Absolute Value}}
+}}
+{{eqn | r = \left\vert{x}\right\vert \, \left\vert{y}\right\vert
+      | c = 
+}}
+{{end-eqn}}
+{{qed|lemma}}
+Let $x < 0$ and $y > 0$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \left\vert{x}\right\vert
+      | r = - x
+      | c = 
+}}
+{{eqn | lo= \land
+      | l = \left\vert{y}\right\vert
+      | r = y
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \left\vert{x}\right\vert \, \left\vert{y}\right\vert
+      | r = - x y
+      | c = 
+}}
+{{end-eqn}}
+and:
+{{begin-eqn}}
+{{eqn | l = x y
+      | o = <
+      | r = 0
+      | c = 
+}}
+{{eqn | l = \left\vert{x y}\right\vert
+      | r = - x y
+      | c = {{Defof|Absolute Value}}
+}}
+{{eqn | r = \left\vert{x}\right\vert \, \left\vert{y}\right\vert
+      | c = 
+}}
+{{end-eqn}}
+{{qed|lemma}}
+The same argument, [[Definition:Mutatis Mutandis|mutatis mutandis]], covers the case where $x > 0$ and $y < 0$.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Follows directly from:
+: [[Real Numbers form Ordered Integral Domain]]
+: [[Product of Absolute Values on Ordered Integral Domain]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \size {x y}
+      | r = \sqrt {\paren {x y}^2}
+      | c = {{Defof|Absolute Value|index = 2}}
+}}
+{{eqn | r = \sqrt {x^2 y^2}
+      | c = 
+}}
+{{eqn | r = \sqrt {x^2} \sqrt{y^2}
+      | c = 
+}}
+{{eqn | r = \size x \cdot \size y
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+We have the result [[Multiplicative Group of Reals is Subgroup of Complex]].
+Therefore, any result applying to all [[Definition:Complex Number|complex numbers]] will also hold for all [[Definition:Real Number|real numbers]].
+The result follows from [[Complex Modulus of Product of Complex Numbers]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Completely Multiplicative Function of Quotient}
+Tags: Multiplicative Functions
+
+\begin{theorem}
+Let $f: \R \to \R$ be a [[Definition:Completely Multiplicative Function|completely multiplicative function]].
+Then:
+:$\forall x, y \in \R, y \ne 0: f \left({\dfrac x y}\right) = \dfrac {f \left({x}\right)} {f \left({y}\right)}$
+whenever $f \left({y}\right) \ne 0$.
+\end{theorem}
+
+\begin{proof}
+Let $z = \dfrac x y$.
+Then:
+{{begin-eqn}}
+{{eqn | l = z
+      | r = \dfrac x y
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = x
+      | r = y z
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = f \left({x}\right)
+      | r = f \left({y}\right) f \left({z}\right)
+      | c = as $f$ is [[Definition:Completely Multiplicative Function|completely multiplicative]]
+}}
+{{eqn | ll= \implies
+      | l = \frac {f \left({x}\right)} {f \left({y}\right)}
+      | r = f \left({z}\right)
+      | c = dividing by $f \left({y}\right)$
+}}
+{{eqn | ll= \implies
+      | l = \frac {f \left({x}\right)} {f \left({y}\right)}
+      | r = f \left({\frac x y}\right)
+      | c = by definition of $z$
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Multiplicative Functions]]
+qkds76zo6dwiwedvaofz6vsj1h1kkca
+\end{proof}<|endoftext|>
+\section{Signum Function of Reciprocal}
+Tags: Signum Function
+
+\begin{theorem}
+Let $x \in \R$ such that $x \ne 0$.
+Then:
+:$\map \sgn x = \map \sgn {\dfrac 1 x}$
+where $\map \sgn x$ denotes the [[Definition:Signum Function|signum]] of $x$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map \sgn x
+      | r = 1
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = x
+      | o = >
+      | r = 0
+      | c = {{Defof|Signum Function}}
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \frac 1 x
+      | o = >
+      | r = 0
+      | c = [[Reciprocal of Strictly Positive Real Number is Strictly Positive]]
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \map \sgn {\dfrac 1 x}
+      | r = 1
+      | c = {{Defof|Signum Function}}
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Signum Function]]
+3fktxynn8ed1viz3c18ts1pma7lwj8f
+\end{proof}<|endoftext|>
+\section{Primitive of x over a squared minus x squared}
+Tags: Primitives involving a squared minus x squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {x \rd x} {a^2 - x^2} = -\frac 1 2 \, \map \ln {a^2 - x^2} + C$
+for $x^2 < a^2$.
+\end{theorem}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = z
+      | r = a^2 - x^2
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d z} {\d x}
+      | r = -2 x
+      | c = [[Power Rule for Derivatives]]
+}}
+{{eqn | ll= \leadsto
+      | l = \int \frac {\d x} {x \paren {a^2 - x^2} }
+      | r = \int \frac {\d z} {-2 z}
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = -\frac 1 2 \int \frac {\d z} z
+      | c = [[Primitive of Constant Multiple of Function]]
+}}
+{{eqn | r = -\frac 1 2 \ln z + C
+      | c = [[Primitive of Reciprocal/Corollary|Primitive of Reciprocal: Corollary]] as $z > 0$
+}}
+{{eqn | r = -\frac 1 2 \, \map \ln {a^2 - x^2} + C
+      | c = substituting for $z$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Linear Combination of Laplace Transforms}
+Tags: Laplace Transforms
+
+\begin{theorem}
+Then:
+:$\laptrans {\lambda \, \map f t + \mu \, \map g t} = \lambda \laptrans {\map f t} + \mu \laptrans {\map g t}$
+everywhere all the above expressions are defined.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \laptrans {\lambda \, \map f t + \mu \, \map g t}
+      | r = \int_0^{\to +\infty} e^{-s t} \paren {\lambda \, \map f t + \mu \, \map g t} \rd t
+      | c = {{Defof|Laplace Transform}}
+}}
+{{eqn | r = \lim_{A \mathop \to +\infty} \paren {\int_0^A e^{-s t} \paren {\lambda \, \map f t + \mu \, \map g t} \rd t}
+      | c = {{Defof|Improper Integral}}
+}}
+{{eqn | r = \lim_{A \mathop \to +\infty} \paren {\lambda \int_0^A e^{-s t} \map f t \rd t + \mu \int_0^A e^{-s t} \map g t \rd t}
+      | c = distributing $e^{-s t}$, [[Linear Combination of Complex Integrals]]
+}}
+{{eqn | r = \lambda \lim_{A \mathop \to +\infty} \int_0^A e^{-s t} \map f t \rd t + \mu \lim_{A \mathop \to +\infty} \int_0^A e^{-s t} \map g t \rd t
+      | c = [[Combination Theorem for Limits at Infinity]]
+}}
+{{eqn | r = \lambda \int_0^{\to +\infty} e^{-s t} \map f t \rd t + \mu \int_0^{\to +\infty} e^{-st} \map g t \rd t
+      | c = {{Defof|Improper Integral}}
+}}
+{{eqn | r = \lambda \laptrans {\map f t} + \mu \laptrans {\map g t}
+      | c = {{Defof|Laplace Transform}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of x cubed over a squared minus x squared}
+Tags: Primitives involving a squared minus x squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {x^3 \ \mathrm d x} {a^2 - x^2} = -\frac {x^2} 2 - \frac {a^2} 2 \ln \left({a^2 - x^2}\right) + C$
+for $x^2 < a^2$.
+\end{theorem}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = \int \frac {x^3 \ \mathrm d x} {a^2 - x^2}
+      | r = \int \frac {x \left({x^2 - a^2 + a^2}\right)} {a^2 - x^2} \ \mathrm d x
+      | c = 
+}}
+{{eqn | r = \int \frac {-x \left({a^2 - x^2}\right)} {a^2 - x^2} \ \mathrm d x + \int \frac {a^2 x} {a^2 - x^2} \ \mathrm d x
+      | c = 
+}}
+{{eqn | r = -\int x \ \mathrm d x + a^2 \int \frac {x \ \mathrm d x} {a^2 - x^2}
+      | c = [[Primitive of Constant Multiple of Function]]
+}}
+{{eqn | r = -\frac {x^2} 2 + a^2 \int \frac {x \ \mathrm d x} {a^2 - x^2} + C
+      | c = [[Primitive of Power]]
+}}
+{{eqn | r = -\frac {x^2} 2 + a^2 \left({-\frac 1 2 \ln \left({a^2 - x^2}\right)}\right) + C
+      | c = [[Primitive of x over a squared minus x squared|Primitive of Reciprocal of $\dfrac x {a^2 - x^2}$]]
+}}
+{{eqn | r = -\frac {x^2} 2 - \frac {a^2} 2 \ln \left({a^2 - x^2}\right) + C
+      | c = simplifying
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Reciprocal of x squared by a squared minus x squared}
+Tags: Primitives involving a squared minus x squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {\d x} {x^2 \paren {a^2 - x^2} } = \frac {-1} {a^2 x} + \frac 1 {2 a^3} \map \ln {\frac {a + x} {a - x} } + C$
+for $x^2 < a^2$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int \frac {\d x} {x^2 \paren {a^2 - x^2} }
+      | r = \int \paren {\frac 1 {a^2 \paren {a^2 - x^2} } + \frac 1 {a^2 x^2} } \rd x
+      | c = [[Primitive of Reciprocal of x squared by a squared minus x squared/Partial Fraction Expansion|Partial Fraction Expansion]]
+}}
+{{eqn | r = \frac 1 {a^2} \int \frac {\d x} {a^2 - x^2} + \frac 1 {a^2} \int \frac {\d x} {x^2}
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \frac 1 {a^2} \int \frac {\d x} {a^2 - x^2} + \frac 1 {a^2 x} + C
+      | c = [[Primitive of Power]]
+}}
+{{eqn | r = \frac 1 {a^2} \paren {\frac 1 2 \map \ln {\frac {a + x} {a - x} } } - \frac 1 {a^2 x} + C
+      | c = [[Primitive of Reciprocal of a squared minus x squared/Logarithm Form|Primitive of $\dfrac 1 {a^2 - x^2}$]]
+}}
+{{eqn | r = \frac {-1} {a^2 x} + \frac 1 {2 a^3} \map \ln {\frac {a + x} {a - x} } + C
+      | c = simplifying
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of x over a squared minus x squared squared}
+Tags: Primitives involving a squared minus x squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {x \ \mathrm d x} {\left({a^2 - x^2}\right)^2} = \frac 1 {2 \left({a^2 - x^2}\right)} + C$
+for $x^2 < a^2$.
+\end{theorem}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = z
+      | r = a^2 - x^2
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \frac {\mathrm d z} {\mathrm d x}
+      | r = -2 x
+      | c = [[Derivative of Power]]
+}}
+{{eqn | ll= \implies
+      | l = \int \frac {x \ \mathrm d x} {\left({a^2 - x^2}\right)^2}
+      | r = \int \frac {\mathrm d z} {-2 z^2}
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = -\frac 1 2 \int \frac {\mathrm d z} {z^2}
+      | c = [[Primitive of Constant Multiple of Function]]
+}}
+{{eqn | r = -\frac 1 2 \left({\frac {-1} z}\right) + C
+      | c = [[Primitive of Power]]
+}}
+{{eqn | r = \frac 1 {2 \left({a^2 - x^2}\right)} + C
+      | c = substituting for $z$ and simplifying
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of x squared over a squared minus x squared squared}
+Tags: Primitives involving a squared minus x squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {x^2 \rd x} {\paren {a^2 - x^2}^2} = \frac x {2 \paren {a^2 - x^2} } - \frac 1 {4 a} \map \ln {\frac {a + x} {a - x} } + C$
+for $x^2 < a^2$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int \frac {x^2 \rd x} {\paren {a^2 - x^2}^2}
+      | r = \int \frac {x^2 - a^2 + a^2} {\paren {a^2 - x^2}^2} \rd x
+      | c = 
+}}
+{{eqn | r = \int \frac {-\paren {a^2 - x^2} } {\paren {a^2 - x^2}^2} \rd x + a^2 \int \frac {\d x} {\paren {a^2 - x^2}^2}
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = -\int \frac {\d x} {a^2 - x^2} + a^2 \int \frac {\d x} {\paren {a^2 - x^2}^2}
+      | c = simplification
+}}
+{{eqn | r = -\frac 1 {2 a} \map \ln {\frac {a + x} {a - x} } + a^2 \int \frac {\d x} {\paren {x^2 - a^2}^2} + C
+      | c = [[Primitive of Reciprocal of a squared minus x squared/Logarithm Form|Primitive of Reciprocal of $a^2 - x^2$]]
+}}
+{{eqn | r = -\frac 1 {2 a} \map \ln {\frac {a + x} {a - x} } + a^2 \paren {\frac x {2 a^2 \paren {a^2 - x^2} } + \frac 1 {4 a^3} \map \ln {\frac {a + x} {a - x} } } + C
+      | c = [[Primitive of Reciprocal of a squared minus x squared squared|Primitive of Reciprocal of $\paren {a^2 - x^2}^2$]]
+}}
+{{eqn | r = \frac x {2 \paren {a^2 - x^2} } - \frac 1 {4 a} \map \ln {\frac {a + x} {a - x} } + C
+      | c = simplifying
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of x cubed over a squared minus x squared squared}
+Tags: Primitives involving a squared minus x squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {x^3 \rd x} {\paren {a^2 - x^2}^2} = \frac {a^2} {2 \paren {a^2 - x^2} } + \frac 1 2 \map \ln {a^2 - x^2} + C$
+for $x^2 < a^2$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int \frac {x^3 \rd x} {\paren {a^2 - x^2}^2}
+      | r = \int \frac {x \paren {x^2 - a^2 + a^2} } {\paren {a^2 - x^2}^2} \rd x
+      | c = 
+}}
+{{eqn | r = \int \frac {-x \paren {a^2 - x^2} } {\paren {a^2 - x^2}^2} \rd x + a^2 \int \frac {x \rd x} {\paren {a^2 - x^2}^2}
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \int \frac {-x \rd x} {a^2 - x^2} + a^2 \int \frac {x \rd x} {\paren {a^2 - x^2}^2}
+      | c = simplification
+}}
+{{eqn | r = -\frac 1 2 \paren {-\map \ln {a^2 - x^2} } + a^2 \int \frac {x \rd x} {\paren {a^2 - x^2}^2} + C
+      | c = [[Primitive of x over a squared minus x squared|Primitive of $\dfrac x {a^2 - x^2}$]]
+}}
+{{eqn | r = \frac 1 2 \map \ln {a^2 - x^2} + a^2 \paren {\frac 1 {2 \paren {a^2 - x^2} } } + C
+      | c = [[Primitive of x over a squared minus x squared squared|Primitive of Reciprocal of $\dfrac x {\paren {a^2 - x^2}^2}$]]
+}}
+{{eqn | r = \frac {a^2} {2 \paren {a^2 - x^2} } + \frac 1 2 \map \ln {a^2 - x^2} + C
+      | c = simplifying
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Reciprocal of x by a squared minus x squared squared}
+Tags: Primitives involving a squared minus x squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {\d x} {x \paren {a^2 - x^2}^2} = \frac 1 {2 a^2 \paren {a^2 - x^2} } + \frac 1 {2 a^4} \map \ln {\frac {x^2} {a^2 - x^2} } + C$
+for $x^2 < a^2$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int \frac {\d x} {x \paren {a^2 - x^2}^2}
+      | r = \int \paren {\frac 1 {a^4 x} + \frac x {a^4 \paren {a^2 - x^2} } + \frac x {a^2 \paren {a^2 - x^2}^2} } \rd x
+      | c = [[Primitive of Reciprocal of x by a squared minus x squared squared/Partial Fraction Expansion|Partial Fraction Expansion]]
+}}
+{{eqn | r = \frac 1 {a^4} \int \frac {\d x} x + \frac 1 {a^4} \int \frac {x \rd x} {a^2 - x^2} + \frac 1 {a^2} \int \frac {x \rd x} {\paren {a^2 - x^2}^2}
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \frac 1 {a^4} \ln \size x + \frac 1 {a^4} \int \frac {x \rd x} {a^2 - x^2} + \frac 1 {a^2} \int \frac {x \rd x} {\paren {a^2 - x^2}^2} + C
+      | c = [[Primitive of Reciprocal]]
+}}
+{{eqn | r = \frac 1 {a^4} \ln \size x + \frac 1 {a^4} \paren {-\frac 1 2 \map \ln {a^2 - x^2} } + \frac 1 {a^2} \int \frac {x \rd x} {\paren {a^2 - x^2}^2} + C
+      | c = [[Primitive of x over a squared minus x squared|Primitive of $\dfrac x {a^2 - x^2}$]]
+}}
+{{eqn | r = \frac 1 {a^4} \ln \size x - \frac 1 {2 a^4} \map \ln {a^2 - x^2} + \frac 1 {a^2} \paren {\frac 1 {2 \paren {a^2 - x^2} } } + C
+      | c = [[Primitive of x over a squared minus x squared squared|Primitive of $\dfrac x {\paren {a^2 - x^2}^2}$]]
+}}
+{{eqn | r = \frac 1 {2 a^4} \map \ln {x^2} - \frac 1 {2 a^4} \map \ln {a^2 - x^2} + \frac 1 {2 a^2 \paren {a^2 - x^2} } + C
+      | c = [[Logarithm of Power]] and $x^2 > 0$
+}}
+{{eqn | r = \frac 1 {2 a^2 \paren {a^2 - x^2} } + \frac 1 {2 a^4} \map \ln {\frac {x^2} {a^2 - x^2} } + C
+      | c = [[Difference of Logarithms]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Reciprocal of x squared by a squared minus x squared squared}
+Tags: Primitives involving a squared minus x squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {\d x} {x^2 \paren {a^2 - x^2}^2} = \frac {-1} {a^4 x} + \frac x {2 a^4 \paren {a^2 - x^2} } + \frac 3 {4 a^5} \map \ln {\frac {a + x} {a - x} } + C$
+for $x^2 < a^2$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int \frac {\d x} {x^2 \paren {a^2 - x^2}^2}
+      | r = \int \paren {\frac 1 {a^4 x^2} + \frac 3 {4 a^5 \paren {a + x} } + \frac 3 {4 a^5 \paren {a - x} } + \frac 1 {4 a^4 \paren {a + x}^2} + \frac 1 {4 a^4 \paren {a - x}^2} } \rd x
+      | c = [[Primitive of Reciprocal of x squared by a squared minus x squared squared/Partial Fraction Expansion|Partial Fraction Expansion]]
+}}
+{{eqn | r = \frac 1 {a^4} \int \frac {\d x} {x^2} + \frac 3 {4 a^5} \int \frac {\d x} {a + x} + \frac 3 {4 a^5} \int \frac {\d x} {a - x} + \frac 1 {4 a^4} \int \frac {\d x} {\paren {a + x}^2} + \frac 1 {4 a^4} \int \frac {\d x} {\paren {a - x}^2}
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \frac 1 {a^4} \frac {-1} x + \frac 3 {4 a^5} \int \frac {\d x} {a + x} + \frac 3 {4 a^5} \int \frac {\d x} {a - x} + \frac 1 {4 a^4} \int \frac {\d x} {\paren {a + x}^2} + \frac 1 {4 a^4} \int \frac {\d x} {\paren {a - x}^2} + C
+      | c = [[Primitive of Power]]
+}}
+{{eqn | r = \frac {-1} {a^4 x} + \frac 3 {4 a^5} \ln \size {a + x} + \frac 3 {4 a^5} \paren {-\ln \size {a - x} } + \frac 1 {4 a^4} \int \frac {\d x} {\paren {a + x}^2} + \frac 1 {4 a^4} \int \frac {\d x} {\paren {a - x}^2} + C
+      | c = [[Primitive of Reciprocal]]
+}}
+{{eqn | r = \frac {-1} {a^4 x} + \frac 3 {4 a^5} \ln \size {a + x} - \frac 3 {4 a^5} \ln \size {a - x} + \frac 1 {4 a^4} \frac {-1} {a + x} + \frac 1 {4 a^4} \frac 1 {a - x} + C
+      | c = [[Primitive of Power]]
+}}
+{{eqn | r = \frac {-1} {a^4 x} + \frac 3 {4 a^5} \ln \size {\frac {a + x} {a - x} } + \frac {-1} {4 a^4 \paren {a + x} } + \frac 1 {4 a^4 \paren {a - x} } + C
+      | c = [[Difference of Logarithms]]
+}}
+{{eqn | r = \frac {-1} {a^4 x} + \frac 3 {4 a^5} \map \ln {\frac {a + x} {a - x} } + \frac {-1} {4 a^4 \paren {a + x} } + \frac 1 {4 a^4 \paren {a - x} } + C
+      | c = [[Sign of Quotient of Factors of Difference of Squares]]
+}}
+{{eqn | r = \frac {-1} {a^4 x} + \frac x {2 a^4 \paren {a^2 - x^2} } + \frac 3 {4 a^5} \map \ln {\frac {a + x} {a - x} } + C
+      | c = simplification
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Reciprocal of Power of a squared minus x squared}
+Tags: Primitives involving a squared minus x squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {\mathrm d x} {\left({a^2 - x^2}\right)^n} = \frac x {2 \left({n - 1}\right) a^2 \left({a^2 - x^2}\right)^{n - 1} } + \frac {2 n - 3} {\left({2 n - 2}\right) a^2} \int \frac {\mathrm d x} {\left({a^2 - x^2}\right)^{n - 1} }$
+for $x^2 > a^2$.
+\end{theorem}
+
+\begin{proof}
+Aiming for an expression in the form:
+: $\displaystyle \int u \frac {\mathrm d v} {\mathrm d x} \ \mathrm d x = u v - \int v \ \frac {\mathrm d u} {\mathrm d x} \ \mathrm d x$
+in order to use the technique of [[Integration by Parts]], let:
+{{begin-eqn}}
+{{eqn | l = v
+      | r = x
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \frac {\mathrm d v} {\mathrm d x}
+      | r = 1
+      | c = [[Power Rule for Derivatives]]
+}}
+{{end-eqn}}
+Thus:
+{{begin-eqn}}
+{{eqn | l = u
+      | r = \frac 1 {\left({a^2 - x^2}\right)^{n - 1} }
+      | c = 
+}}
+{{eqn | r = \left({a^2 - x^2}\right)^{- \left({n - 1}\right)}
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \frac {\mathrm d u} {\mathrm d x}
+      | r = - \left({n - 1}\right) \left({-2 x}\right) \left({a^2 - x^2}\right)^{- \left({n - 1}\right) - 1}
+      | c = [[Power Rule for Derivatives]] and [[Chain Rule for Derivatives]]
+}}
+{{eqn | r = 2 \left({n - 1}\right) x \left({a^2 - x^2}\right)^{-n}
+      | c = 
+}}
+{{eqn | r = \frac {\left({2 n - 2}\right) x} {\left({a^2 - x^2}\right)^n}
+      | c = 
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = \int \frac {\mathrm d x} {\left({a^2 - x^2}\right)^{n - 1} }
+      | r = \frac 1 {\left({a^2 - x^2}\right)^{n - 1} } x - \int x \frac {\left({2 n - 2}\right) x} {\left({a^2 - x^2}\right)^n} \ \mathrm d x
+      | c = [[Integration by Parts]]
+}}
+{{eqn | r = \frac x {\left({a^2 - x^2}\right)^{n - 1} } - \left({2 n - 2}\right) \int \frac {x^2} {\left({a^2 - x^2}\right)^n} \ \mathrm d x
+      | c = simplifying
+}}
+{{eqn | r = \frac x {\left({a^2 - x^2}\right)^{n - 1} } - \left({2 n - 2}\right) \int \frac {x^2 - a^2 + a^2} {\left({a^2 - x^2}\right)^n} \ \mathrm d x
+      | c = 
+}}
+{{eqn | r = \frac x {\left({a^2 - x^2}\right)^{n - 1} } - \left({2 n - 2}\right) \int \frac {-\left({a^2 - x^2}\right)} {\left({a^2 - x^2}\right)^n} \ \mathrm d x - \left({2 n - 2}\right) a^2 \int \frac {\mathrm d x} {\left({a^2 - x^2}\right)^n}
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \frac x {\left({a^2 - x^2}\right)^{n - 1} } + \left({2 n - 2}\right) \int \frac {\mathrm d x} {\left({a^2 - x^2}\right)^{n - 1} } - \left({2 n - 2}\right) a^2 \int \frac {\mathrm d x} {\left({a^2 - x^2}\right)^n} 
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \left({1 - \left({2 n - 2}\right)}\right) \int \frac {\mathrm d x} {\left({a^2 - x^2}\right)^{n - 1} }
+      | r = \frac x {\left({a^2 - x^2}\right)^{n - 1} } + \left({2 n - 2}\right) a^2 \int \frac {\mathrm d x} {\left({a^2 - x^2}\right)^n}
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \int \frac {\mathrm d x} {\left({a^2 - x^2}\right)^n}
+      | r = \frac x {2 \left({n - 1}\right) a^2 \left({a^2 - x^2}\right)^{n - 1} } + \frac {2 n - 3} {\left({2 n - 2}\right) a^2} \int \frac {\mathrm d x} {\left({a^2 - x^2}\right)^{n - 1} }
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of x over Power of a squared minus x squared}
+Tags: Primitives involving a squared minus x squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {x \rd x} {\paren {a^2 - x^2}^n} = \frac 1 {2 \paren {n - 1} \paren {a^2 - x^2}^{n - 1} }$
+for $x^2 < a^2$.
+\end{theorem}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = z
+      | r = a^2 - x^2
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d z} {\d x}
+      | r = -2 x
+      | c = [[Power Rule for Derivatives]]
+}}
+{{eqn | ll= \leadsto
+      | l = \int \frac {x \rd x} {\paren {a^2 - x^2}^n}
+      | r = \int \frac {\d z} {- 2 z^n}
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \frac {-1} 2 \int z^{-n} \rd z
+      | c = [[Primitive of Constant Multiple of Function]]
+}}
+{{eqn | r = \frac {-1} 2 \frac {z^{-n + 1} } {-n + 1} + C
+      | c = [[Primitive of Power]]
+}}
+{{eqn | r = \frac 1 {2 \paren {n - 1} z^{n - 1} } + C
+      | c = simplifying
+}}
+{{eqn | r = \frac 1 {2 \paren {n - 1} \paren {a^2 - x^2}^{n - 1} } + C
+      | c = substituting for $z$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Reciprocal of x by Power of a squared minus x squared}
+Tags: Primitives involving a squared minus x squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {\mathrm d x} {x \left({a^2 - x^2}\right)^n} = \frac 1 {2 \left({n - 1}\right) a^2 \left({a^2 - x^2}\right)^{n - 1} } + \frac 1 {a^2} \int \frac {\mathrm d x} {x \left({a^2 - x^2}\right)^{n - 1} }$
+for $x^2 < a^2$.
+\end{theorem}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = z
+      | r = a^2 - x^2
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \frac {\mathrm d z} {\mathrm d x}
+      | r = -2 x
+      | c = [[Power Rule for Derivatives]]
+}}
+{{eqn | ll= \implies
+      | l = \int \frac {\mathrm d x} {x \left({a^2 - x^2}\right)^n}
+      | r = \int \frac {\mathrm d z} {-2 x^2 z^n}
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \frac 1 2 \int \frac {\mathrm d z} {\left({z - a^2}\right) z^n}
+      | c = [[Primitive of Constant Multiple of Function]] and substituting for $x^2$
+}}
+{{end-eqn}}
+From [[Reduction Formula for Primitive of Power of x by Power of a x + b/Increment of Power of x|Reduction Formula for Primitive of $x^m \left({a x + b}\right)^n$: Increment of Power of $x$]]:
+:$\displaystyle \int x^m \left({a x + b}\right)^n \ \mathrm d x = \frac {x^{m+1} \left({a x + b}\right)^{n + 1} } {\left({m + 1}\right) b} - \frac {\left({m + n + 2}\right) a} {\left({m + 1}\right) b} \int x^{m + 1} \left({a x + b}\right)^n \ \mathrm d x$
+Let:
+{{begin-eqn}}
+{{eqn | l = a
+      | o = :=
+      | r = 1
+      | c = 
+}}
+{{eqn | l = b
+      | o = :=
+      | r = -a^2
+      | c = 
+}}
+{{eqn | l = x
+      | o = :=
+      | r = z
+      | c = 
+}}
+{{eqn | l = m
+      | o = :=
+      | r = -n
+      | c = 
+}}
+{{eqn | l = n
+      | o = :=
+      | r = -1
+      | c = 
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = \int \frac {\mathrm d x} {x \left({x^2 - a^2}\right)^n}
+      | r = \frac 1 2 \int \frac {\mathrm d z} {\left({z - a^2}\right) z^n}
+      | c = 
+}}
+{{eqn | r = \frac 1 2 \left({\frac {z^{-n+1} \left({z - a^2}\right)^{-1 + 1} } {\left({-n + 1}\right) \left({-a^2}\right)} - \frac {\left({-n + \left({-1}\right) + 2}\right)} {\left({-n + 1}\right) \left({-a^2}\right)} \int z^{-n + 1} \left({z - a^2}\right)^{-1} \ \mathrm d z}\right)
+      | c = 
+}}
+{{eqn | r = \frac 1 {2 \left({n - 1}\right) a^2 z^{n-1} } + \frac 1 {2 a^2} \int \frac {\mathrm d z} {z^{n - 1} \left({z - a^2}\right)}
+      | c = simplification
+}}
+{{eqn | r = \frac 1 {2 \left({n - 1}\right) a^2 \left({a^2 - x^2}\right)^{n - 1} } + \frac 1 {2 a^2} \int \frac {2 x \ \mathrm d x} {x^2 \left({a^2 - x^2}\right)^{n - 1} }
+      | c = substituting for $z$ and $\mathrm d z$
+}}
+{{eqn | r = \frac 1 {2 \left({n - 1}\right) a^2 \left({a^2 - x^2}\right)^{n - 1} } + \frac 1 {a^2} \int \frac {\mathrm d x} {x \left({a^2 - x^2}\right)^{n - 1} }
+      | c = simplifying
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Power of x over Power of a squared minus x squared}
+Tags: Primitives involving a squared minus x squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {x^m \rd x} {\paren {a^2 - x^2}^n} = a^2 \int \frac {x^{m - 2} \rd x} {\paren {a^2 - x^2}^n} - \int \frac {x^{m - 2} \rd x} {\paren {a^2 - x^2}^{n - 1} }$
+for $x^2 < a^2$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int \frac {x^m \rd x} {\paren {a^2 - x^2}^n}
+      | r = \int \frac {x^{m - 2} \paren {x^2} \rd x} {\paren {a^2 - x^2}^n}
+      | c = 
+}}
+{{eqn | r = \int \frac {x^{m - 2} \paren {a^2 + x^2 - a^2} \rd x} {\paren {a^2 - x^2}^n}
+      | c = 
+}}
+{{eqn | r = \int \frac {x^{m - 2} \paren {a^2 - \paren {a^2 - x^2} } \rd x} {\paren {a^2 - x^2}^n}
+      | c = 
+}}
+{{eqn | r = a^2 \int \frac {x^{m - 2} \rd x} {\paren {a^2 - x^2}^n} - \int \frac {x^{m - 2} \paren {a^2 - x^2} \rd x} {\paren {a^2 - x^2}^n}
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = a^2 \int \frac {x^{m - 2} \rd x} {\paren {a^2 - x^2}^n} - \int \frac {x^{m - 2} \rd x} {\paren {a^2 - x^2}^{n - 1} }
+      | c = simplifying
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Reciprocal of Power of x by Power of a squared minus x squared}
+Tags: Primitives involving a squared minus x squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {\mathrm d x} {x^m \left({a^2 - x^2}\right)^n} = \frac 1 {a^2} \int \frac {\mathrm d x} {x^m \left({a^2 - x^2}\right)^{n-1} } + \frac 1 {a^2} \int \frac {\mathrm d x} {x^{m-2} \left({a^2 - x^2}\right)^n}$
+for $x^2 < a^2$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int \frac {\mathrm d x} {x^m \left({a^2 - x^2}\right)^{n - 1} }
+      | r = \int \frac {\left({a^2 - x^2}\right) \ \mathrm d x} {x^m \left({a^2 - x^2}\right)^{n - 1} \left({a^2 - x^2}\right)}
+      | c = 
+}}
+{{eqn | r = \int \frac {\left({a^2 - x^2}\right) \ \mathrm d x} {x^m \left({a^2 - x^2}\right)^{\left({n - 1}\right) + 1} }
+      | c = 
+}}
+{{eqn | r = a^2 \int \frac {\mathrm d x} {x^m \left({a^2 - x^2}\right)^n} - \int \frac {x^2 \ \mathrm d x} {x^m \left({a^2 - x^2}\right)^n}
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = a^2 \int \frac {\mathrm d x} {x^m \left({a^2 - x^2}\right)^n} - \int \frac {\mathrm d x} {x^{m - 2} \left({a^2 - x^2}\right)^n}
+      | c = simplifying
+}}
+{{eqn | ll= \implies
+      | l = a^2 \int \frac {\mathrm d x} {x^m \left({a^2 - x^2}\right)^n} 
+      | r = \int \frac {\mathrm d x} {x^m \left({a^2 - x^2}\right)^{n - 1} } + \int \frac {\mathrm d x} {x^{m - 2} \left({a^2 - x^2}\right)^n}
+      | c = changing sides
+}}
+{{eqn | ll= \implies
+      | l = \int \frac {\mathrm d x} {x^m \left({a^2 - x^2}\right)^n} 
+      | r = \frac 1 {a^2} \int \frac {\mathrm d x} {x^m \left({a^2 - x^2}\right)^{n-1} } + \frac 1 {a^2} \int \frac {\mathrm d x} {x^{m-2} \left({a^2 - x^2}\right)^n}
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of x over Root of x squared plus a squared}
+Tags: Primitives involving Root of x squared plus a squared
+
+\begin{theorem}
+:$\ds \int \frac {x \rd x} {\sqrt {x^2 + a^2} } = \sqrt {x^2 + a^2} + C$
+\end{theorem}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = z^2
+      | r = x^2 + a^2
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = 2 z \frac {\d z} {\d x}
+      | r = 2 x
+      | c = [[Chain Rule for Derivatives]], [[Power Rule for Derivatives]]
+}}
+{{eqn | ll= \leadsto
+      | l = \int \frac {x \rd x} {\sqrt {x^2 + a^2} }
+      | r = \int \frac {z \rd z} {z}
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \int \rd z
+      | c = 
+}}
+{{eqn | r = z + C
+      | c = [[Primitive of Constant]]
+}}
+{{eqn | r = \sqrt {x^2 + a^2} + C
+      | c = substituting for $z$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of x cubed over Root of x squared plus a squared}
+Tags: Primitives involving Root of x squared plus a squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {x^2 \rd x} {\sqrt {x^2 + a^2} } = \frac {\paren {\sqrt {x^2 + a^2} }^3} 3 - a^2 \sqrt {x^2 + a^2} + C$
+\end{theorem}
+
+\begin{proof}
+With a view to expressing the problem in the form:
+:$\displaystyle \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$
+let:
+{{begin-eqn}}
+{{eqn | l = u
+      | r = x^2
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d u} {\d x}
+      | r = 2 x
+      | c = [[Power Rule for Derivatives]]
+}}
+{{end-eqn}}
+and let:
+{{begin-eqn}}
+{{eqn | l = \frac {\d v} {\d x}
+      | r = \frac x {\sqrt {x^2 + a^2} }
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = v
+      | r = \sqrt {x^2 + a^2}
+      | c = [[Primitive of x over Root of x squared plus a squared|Primitive of $\dfrac x {\sqrt {x^2 + a^2} }$]]
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = \int \frac {x^3 \rd x} {\sqrt {x^2 + a^2} }
+      | r = \int x^2 \frac {x \rd x} {\sqrt {x^2 + a^2} }
+      | c = 
+}}
+{{eqn | r = x^2 \sqrt {x^2 + a^2} - \int 2 x \sqrt {x^2 + a^2} \rd x
+      | c = [[Integration by Parts]]
+}}
+{{eqn | r = x^2 \sqrt {x^2 + a^2} - 2 \paren {\frac {\paren {\sqrt {x^2 + a^2} }^3} 3 } + C
+      | c = [[Primitive of x by Root of x squared plus a squared|Primitive of $x \sqrt {x^2 + a^2}$]]
+}}
+{{eqn | r = \paren {x^2 + a^2 - a^2} \sqrt {x^2 + a^2} - 2 \paren {\frac {\paren {\sqrt {x^2 + a^2} }^3} 3 } + C
+      | c = 
+}}
+{{eqn | r = \paren {\sqrt {x^2 + a^2} }^3 - a^2 \sqrt {x^2 + a^2} - 2 \paren {\frac {\paren {\sqrt {x^2 + a^2} }^3} 3 } + C
+      | c = 
+}}
+{{eqn | r = \frac {\paren {\sqrt {x^2 + a^2} }^3} 3 - a^2 \sqrt {x^2 + a^2} + C
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Reciprocal of x squared by Root of x squared plus a squared}
+Tags: Primitives involving Root of x squared plus a squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {\mathrm d x} {x^2 \sqrt {x^2 + a^2} } = \frac {-\sqrt {x^2 + a^2} } {a^2 x} + C$
+\end{theorem}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = z
+      | r = x^2
+}}
+{{eqn | ll= \implies
+      | l = \frac {\mathrm d z} {\mathrm d x}
+      | r = 2 x
+      | c = [[Power Rule for Derivatives]]
+}}
+{{eqn | ll= \implies
+      | l = \int \frac {\mathrm d x} {x^2 \sqrt {x^2 + a^2} }
+      | r = \int \frac {\mathrm d z} {2 z \sqrt z \sqrt {z + a^2} }
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \frac 1 2 \int \frac {\mathrm d z} {z^{3/2} \sqrt {z + a^2} }
+      | c = 
+}}
+{{end-eqn}}
+Using [[Primitive of Reciprocal of Power of x by Root of a x + b|Primitive of $ \dfrac 1{x^m \sqrt{a x + b} }$]]:
+:$\displaystyle \int \frac {\mathrm d x} {x^m \sqrt{a x + b} } = -\frac {\sqrt{a x + b} } {\left({m - 1}\right) b x^{m-1} } - \frac {\left({2 m - 3}\right) a} {\left({2 m - 2}\right) b} \int \frac {\mathrm d x} {x^{m - 1} \sqrt{a x + b} }$
+Setting:
+{{begin-eqn}}
+{{eqn | l = x
+      | o = :=
+      | r = z
+}}
+{{eqn | l = m
+      | o = :=
+      | r = \frac 3 2
+}}
+{{eqn | l = a
+      | o = :=
+      | r = 1
+}}
+{{eqn | l = b
+      | o = :=
+      | r = a^2
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = \frac 1 2 \int \frac {\mathrm d z} {z^{3/2} \sqrt {z + a^2} }
+      | r = \frac {-\sqrt{z + a^2} } {2 \left({\left({\frac 3 2}\right) - 1}\right) a^2 z^{\left({3/2}\right) - 1} } - \frac {2 \left({\frac 3 2}\right) - 3} {2 \left({2 \left({\frac 3 2}\right) - 2}\right) a^2} \int \frac {\mathrm d z} {z^{\left({\frac 3 2}\right) - 1} \sqrt{z + a^2} } + C
+      | c = [[Primitive of Reciprocal of Power of x by Root of a x + b|Primitive of $ \dfrac 1{x^m \sqrt{a x + b} }$]]
+}}
+{{eqn | r = \frac {-\sqrt{z + a^2} } {a^2 z^{1/2} } - 0 + C
+      | c = simplifying
+}}
+{{eqn | r = \frac {-\sqrt {x^2 + a^2} } {a^2 x} + C
+      | c = substituting back for $z$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Reciprocal of x cubed by Root of x squared plus a squared}
+Tags: Primitives involving Root of x squared plus a squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {\mathrm d x} {x^3 \sqrt {x^2 + a^2} } = \frac {- \sqrt {x^2 + a^2} } {2 a^2 x^2} + \frac 1 {2 a^3} \ln \left({\frac {a + \sqrt {x^2 + a^2} } x}\right) + C$
+\end{theorem}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = z
+      | r = x^2
+}}
+{{eqn | ll= \implies
+      | l = \frac {\mathrm d z} {\mathrm d x}
+      | r = 2 x
+      | c = [[Power Rule for Derivatives]]
+}}
+{{eqn | ll= \implies
+      | l = \int \frac {\mathrm d x} {x^3 \sqrt {x^2 + a^2} }
+      | r = \int \frac {\mathrm d z} {2 z^{3/2} \sqrt z \sqrt {z + a^2} }
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \frac 1 2 \int \frac {\mathrm d z} {z^2 \sqrt {z + a^2} }
+      | c = 
+}}
+{{end-eqn}}
+Using [[Primitive of Reciprocal of Power of x by Root of a x + b|Primitive of $ \dfrac 1{x^m \sqrt{a x + b} }$]]:
+:$\displaystyle \int \frac {\mathrm d x} {x^m \sqrt{a x + b} } = -\frac {\sqrt{a x + b} } {\left({m - 1}\right) b x^{m-1} } - \frac {\left({2 m - 3}\right) a} {\left({2 m - 2}\right) b} \int \frac {\mathrm d x} {x^{m - 1} \sqrt{a x + b} }$
+Setting:
+{{begin-eqn}}
+{{eqn | l = x
+      | o = :=
+      | r = z
+}}
+{{eqn | l = m
+      | o = :=
+      | r = 2
+}}
+{{eqn | l = a
+      | o = :=
+      | r = 1
+}}
+{{eqn | l = b
+      | o = :=
+      | r = a^2
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = \frac 1 2 \int \frac {\mathrm d z} {z^2 \sqrt {z + a^2} }
+      | r = \frac {-\sqrt{z + a^2} } {2 \left({2 - 1}\right) a^2 z^{2 - 1} } - \frac {2 \left({2}\right) - 3} {2 \left({2 \left({2}\right) - 2}\right) a^2} \int \frac {\mathrm d z} {z^{2 - 1} \sqrt{z + a^2} } + C
+      | c = [[Primitive of Reciprocal of Power of x by Root of a x + b|Primitive of $ \dfrac 1{x^m \sqrt{a x + b} }$]]
+}}
+{{eqn | r = \frac {-\sqrt{z + a^2} } {2 a^2 z} - \frac 1 {4 a^2} \int \frac {\mathrm d z} {z \sqrt{z + a^2} } + C
+      | c = simplifying
+}}
+{{eqn | r = \frac {-\sqrt{x^2 + a^2} } {2 a^2 x^2} - \frac 1 {4 a^2} \left({\int \frac {2 x \ \mathrm d x} {x^2 \sqrt{x^2 + a^2} } }\right) + C
+      | c = substituting for $z$
+}}
+{{eqn | r = \frac {-\sqrt{x^2 + a^2} } {2 a^2 x^2} - \frac 1 {2 a^2} \left({\int \frac {\mathrm d x} {x \sqrt{x^2 + a^2} } }\right) + C
+      | c = simplification
+}}
+{{eqn | r =\frac {- \sqrt {x^2 + a^2} } {2 a^2 x^2} + \frac 1 {2 a^3} \ln \left({\frac {a + \sqrt {x^2 + a^2} } x}\right) + C
+      | c = [[Primitive of Reciprocal of x by Root of x squared plus a squared|Primitive of $\dfrac 1 {x \sqrt{x^2 + a^2} }$]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of x by Root of x squared plus a squared}
+Tags: Primitives involving Root of x squared plus a squared
+
+\begin{theorem}
+:$\displaystyle \int x \sqrt {x^2 + a^2} \rd x = \frac {\paren {\sqrt {x^2 + a^2} }^3} 3 + C$
+\end{theorem}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = z
+      | r = x^2
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d z} {\d x}
+      | r = 2 x
+      | c = [[Power Rule for Derivatives]]
+}}
+{{eqn | ll= \leadsto
+      | l = \int x \sqrt {x^2 + a^2} \rd x
+      | r = \int \frac {\sqrt z \sqrt {z + a^2} \rd z} {2 \sqrt z}
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \frac 1 2 \int \sqrt {z + a^2} \rd z
+      | c = [[Primitive of Constant Multiple of Function]]
+}}
+{{eqn | r = \frac 1 2 \frac {2 \paren {\sqrt {z + a^2} }^3} 3 + C
+      | c = [[Primitive of Root of a x + b|Primitive of $\sqrt {a x + b}$]]
+}}
+{{eqn | r = \frac {\paren {\sqrt {x^2 + a^2} }^3} 3 + C
+      | c = substituting for $z$ and simplifying
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of x squared by Root of x squared plus a squared}
+Tags: Primitives involving Root of x squared plus a squared
+
+\begin{theorem}
+:$\displaystyle \int x^2 \sqrt {x^2 + a^2} \rd x = \frac {x \paren {\sqrt {x^2 + a^2} }^3} 4 - \frac {a^2 x \sqrt {x^2 + a^2} } 8 - \frac {a^4} 8 \map \ln {x + \sqrt {x^2 + a^2} } + C$
+\end{theorem}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = z
+      | r = x^2
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d z} {\d x}
+      | r = 2 x
+      | c = [[Power Rule for Derivatives]]
+}}
+{{eqn | ll= \leadsto
+      | l = \int x^2 \sqrt {x^2 + a^2} \rd x
+      | r = \int \frac {z \sqrt {z + a^2} \rd z} {2 \sqrt z}
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \frac 1 2 \int \sqrt z \sqrt {z + a^2} \rd z
+      | c = [[Primitive of Constant Multiple of Function]]
+}}
+{{eqn | r = \frac 1 2 \paren {\frac {2 z + a^2} 4 \sqrt z \sqrt {z + a^2} - \frac {a^4} 8 \int \frac {\d z} {\sqrt z \sqrt {z + a^2} } } + C
+      | c = [[Primitive of Root of a x + b by Root of p x + q|Primitive of $\sqrt {\paren {a x + b} \paren {p x + q} }$]]
+}}
+{{eqn | r = \frac {2 z + a^2} 8 \sqrt z \sqrt {z + a^2} - \frac {a^4} {16} \paren {2 \map \ln {\sqrt {z + a^2} + \sqrt z} } + C
+      | c = [[Primitive of Reciprocal of Root of a x + b by Root of p x + q|Primitive of $\dfrac 1 {\sqrt {\paren {a x + b} \paren {p x + q} } }$]]
+}}
+{{eqn | r = \frac {2 z + 2 a^2} 8 \sqrt z \sqrt {z + a^2} - \frac {a^2} 8 \sqrt z \sqrt {z + a^2} - \frac {a^4} 8 \map \ln {\sqrt {z + a^2} + \sqrt z} + C
+      | c = 
+}}
+{{eqn | r = \frac {z + a^2} 4 \sqrt z \sqrt {z + a^2} - \frac {a^2} 8 \sqrt z \sqrt {z + a^2} - \frac {a^4} 8 \map \ln {\sqrt {z + a^2} + \sqrt z} + C
+      | c = 
+}}
+{{eqn | r = \frac {x \paren {\sqrt {x^2 + a^2} }^3} 4 - \frac {a^2 x \sqrt {x^2 + a^2} } 8 - \frac {a^4} 8 \map \ln {x + \sqrt {x^2 + a^2} } + C
+      | c = substituting for $z$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of x cubed by Root of x squared plus a squared}
+Tags: Primitives involving Root of x squared plus a squared
+
+\begin{theorem}
+:$\displaystyle \int x^3 \sqrt {x^2 + a^2} \rd x = \frac {\paren {\sqrt {x^2 + a^2} }^5} 5 - \frac {a^2 \paren {\sqrt {x^2 + a^2} }^3} 3 + C$
+\end{theorem}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = z
+      | r = x^2
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d z} {\d x}
+      | r = 2 x
+      | c = [[Power Rule for Derivatives]]
+}}
+{{eqn | ll= \leadsto
+      | l = \int x^3 \sqrt {x^2 + a^2} \rd x
+      | r = \int \frac {z^{3/2} \sqrt {z + a^2} \rd z} {2 \sqrt z}
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \frac 1 2 \int z \sqrt {z + a^2} \rd z
+      | c = [[Primitive of Constant Multiple of Function]]
+}}
+{{eqn | r = \frac 1 2 \paren {\frac {2 \paren {3 z - 2 a^2} } {15} \paren {\sqrt {z + a^2} }^3} + C
+      | c = [[Primitive of x by Root of a x + b|Primitive of $x \sqrt {\paren {a x + b} }$]]
+}}
+{{eqn | r = \frac {3 z - 2 a^2} {15} \paren {\sqrt {z + a^2} }^3 + C
+      | c = simplification
+}}
+{{eqn | r = \frac {3 x^2 - 2 a^2} {15} \paren {\sqrt {x^2 + a^2} }^3 + C
+      | c = substituting for $z$
+}}
+{{eqn | r = \frac {3 x^2 + 3 a^2 - 5 a^2} {15} \paren {\sqrt {x^2 + a^2} }^3 + C
+      | c = 
+}}
+{{eqn | r = \frac {3 \paren {x^2 + a^2} \paren {\sqrt {a^2 - x^2} }^3 - 5 a^2 \paren {\sqrt {a^2 - x^2} }^3} {15} + C
+      | c = 
+}}
+{{eqn | r = \frac {\paren {\sqrt {x^2 + a^2} }^5} 5 - \frac {a^2 \paren {\sqrt {x^2 + a^2} }^3} 3 + C
+      | c = simplification
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Root of x squared plus a squared over x}
+Tags: Primitives involving Root of x squared plus a squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {\sqrt {x^2 + a^2} } x \ \mathrm d x = \sqrt {x^2 + a^2} - a \ln \left({\frac {a + \sqrt {x^2 + a^2} } a}\right) + C$
+\end{theorem}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = z
+      | r = x^2
+}}
+{{eqn | ll= \implies
+      | l = \frac {\mathrm d z} {\mathrm d x}
+      | r = 2 x
+      | c = [[Power Rule for Derivatives]]
+}}
+{{eqn | ll= \implies
+      | l = \int \frac {\sqrt {x^2 + a^2} } x \ \mathrm d x
+      | r = \int \frac {\sqrt {z + a^2} \ \mathrm d z} {2 \sqrt z \sqrt z}
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \frac 1 2 \int \frac {\sqrt {z + a^2} \ \mathrm d z} z
+      | c = [[Primitive of Constant Multiple of Function]]
+}}
+{{eqn | r = \frac 1 2 \left({2 \sqrt {z + a^2} + a^2 \int \frac {\mathrm d z} {z \sqrt {z + a^2} } }\right) + C
+      | c = [[Primitive of Root of a x + b over x|Primitive of $\dfrac {\sqrt {a x + b} } x$]]
+}}
+{{eqn | r = \sqrt {x^2 + a^2} + \frac {a^2} 2 \int \frac {2 x \ \mathrm d x} {x^2 \sqrt {x^2 + a^2} } + C
+      | c = substituting for $z$
+}}
+{{eqn | r = \sqrt {x^2 + a^2} + a^2 \int \frac {\mathrm d x} {x \sqrt {x^2 + a^2} } + C
+      | c = simplifying
+}}
+{{eqn | r = \sqrt {x^2 + a^2} + a^2 \left({-\frac 1 a \ln \left({\frac {a + \sqrt {x^2 + a^2} } x}\right)}\right) + C
+      | c = [[Primitive of Reciprocal of x by Root of x squared plus a squared/Logarithm Form|Primitive of $\dfrac 1 {x \sqrt {x^2 + a^2} }$]]
+}}
+{{eqn | r = \sqrt {x^2 + a^2} - a \ln \left({\frac {a + \sqrt {x^2 + a^2} } a}\right) + C
+      | c = simplification
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Root of x squared plus a squared over x squared}
+Tags: Primitives involving Root of x squared plus a squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {\sqrt {x^2 + a^2} } {x^2} \rd x = \frac {-\sqrt {x^2 + a^2} } x + \map \ln {x + \sqrt {x^2 + a^2} } + C$
+\end{theorem}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = z
+      | r = x^2
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d z} {\d x}
+      | r = 2 x
+      | c = [[Power Rule for Derivatives]]
+}}
+{{eqn | ll= \leadsto
+      | l = \int \frac {\sqrt {x^2 + a^2} } {x^2} \rd x
+      | r = \int \frac {\sqrt {z + a^2} \rd z} {2 z \sqrt z}
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \frac 1 2 \int \frac {\sqrt {z + a^2} \rd z} {z^{3/2} }
+      | c = [[Primitive of Constant Multiple of Function]]
+}}
+{{eqn | r = \frac 1 2 \paren {\frac {-\sqrt {z + a^2} } {\frac 1 2 \sqrt z} + \frac 1 2 \int \frac {\d z} {\sqrt z \sqrt {z + a^2} } } + C
+      | c = [[Primitive of Root of a x + b over Power of x/Formulation 1|Primitive of $\dfrac {\sqrt {a x + b} } {x^m}$]]
+}}
+{{eqn | r = \frac {-\sqrt {x^2 + a^2} } x + \frac 1 2 \int \frac {2 x \rd x} {x \sqrt {x^2 + a^2} } + C
+      | c = substituting for $z$
+}}
+{{eqn | r = \frac {-\sqrt {x^2 + a^2} } x + \int \frac {\d x} {\sqrt {x^2 + a^2} } + C
+      | c = simplifying
+}}
+{{eqn | r = \frac {-\sqrt {x^2 + a^2} } x + \map \ln {x + \sqrt {x^2 + a^2} } + C
+      | c = [[Primitive of Reciprocal of Root of x squared plus a squared/Logarithm Form|Primitive of $\dfrac 1 {\sqrt {x^2 + a^2} }$]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Root of x squared plus a squared over x cubed}
+Tags: Primitives involving Root of x squared plus a squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {\sqrt {x^2 + a^2} } {x^3} \ \mathrm d x = \frac {-\sqrt {x^2 + a^2} } {2 x^2} - \frac 1 {2 a} \ln \left({\frac {a + \sqrt {x^2 + a^2} } x}\right) + C$
+\end{theorem}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = z
+      | r = x^2
+}}
+{{eqn | ll= \implies
+      | l = \frac {\mathrm d z} {\mathrm d x}
+      | r = 2 x
+      | c = [[Power Rule for Derivatives]]
+}}
+{{eqn | ll= \implies
+      | l = \int \frac {\sqrt {x^2 + a^2} } {x^3} \ \mathrm d x
+      | r = \int \frac {\sqrt {z + a^2} \ \mathrm d z} {2 z^{3/2} \sqrt z}
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \frac 1 2 \int \frac {\sqrt {z + a^2} \ \mathrm d z} {z^2}
+      | c = [[Primitive of Constant Multiple of Function]]
+}}
+{{eqn | r = \frac 1 2 \left({\frac {-\sqrt {z + a^2} } z + \frac 1 2 \int \frac {\mathrm d z} {z \sqrt {z + a^2} } }\right) + C
+      | c = [[Primitive of Root of a x + b over Power of x/Formulation 1|Primitive of $\dfrac {\sqrt {a x + b} } {x^m}$]]
+}}
+{{eqn | r = \frac {-\sqrt {x^2 + a^2} } {2 x^2} + \frac 1 4 \int \frac {2 x \ \mathrm d x} {x^2 \sqrt {x^2 + a^2} } + C
+      | c = substituting for $z$
+}}
+{{eqn | r = \frac {-\sqrt {x^2 + a^2} } {2 x^2} + \frac 1 2 \int \frac {\mathrm d x} {x \sqrt {x^2 + a^2} } + C
+      | c = simplifying
+}}
+{{eqn | r = \frac {-\sqrt {x^2 + a^2} } {2 x^2} + \frac 1 2 \left({-\frac 1 a \ln \left({\frac {a + \sqrt {x^2 + a^2} } x}\right) }\right) + C
+      | c = [[Primitive of Reciprocal of x by Root of x squared plus a squared/Logarithm Form|Primitive of $\dfrac 1 {x \sqrt {x^2 + a^2} }$]]
+}}
+{{eqn | r = \frac {-\sqrt {x^2 + a^2} } {2 x^2} - \frac 1 {2 a} \ln \left({\frac {a + \sqrt {x^2 + a^2} } x}\right) + C
+      | c = simplifying
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Reciprocal of Root of x squared plus a squared cubed}
+Tags: Primitives involving Root of x squared plus a squared
+
+\begin{theorem}
+:$\ds \int \frac {\d x} {\paren {\sqrt {x^2 + a^2} }^3} = \frac x {a^2 \sqrt {x^2 + a^2} } + C$
+\end{theorem}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = z
+      | r = x^2
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d z} {\d x}
+      | r = 2 x
+      | c = [[Power Rule for Derivatives]]
+}}
+{{eqn | ll= \leadsto
+      | l = \int \frac {\d x} {\paren {\sqrt {x^2 + a^2} }^3}
+      | r = \int \frac {\d z} {2 \sqrt z \paren {\sqrt {z + a^2} }^3}
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \frac 1 2 \int \frac {\d z} {\paren {z + a^2} \sqrt z \sqrt {z + a^2} }
+      | c = [[Primitive of Constant Multiple of Function]]
+}}
+{{eqn | r = \frac 1 2 \paren {\frac {2 \sqrt z} {a^2 \sqrt {z + a^2} } } + C
+      | c = [[Primitive of Reciprocal of p x + q by Root of a x + b by Root of p x + q|Primitive of $\dfrac 1 {\paren {p x + q} \sqrt {\paren {a x + b} \paren {p x + q} } }$]]
+}}
+{{eqn | r = \frac 1 2 \paren {\frac {2 x} {a^2 \sqrt {x^2 + a^2} } } + C
+      | c = substituting for $z$
+}}
+{{eqn | r = \frac x {a^2 \sqrt {x^2 + a^2} } + C
+      | c = simplifying
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of x over Root of x squared plus a squared cubed}
+Tags: Primitives involving Root of x squared plus a squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {x \rd x} {\paren {\sqrt {x^2 + a^2} }^3} = \frac {-1} {\sqrt {x^2 + a^2} } + C$
+\end{theorem}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = z
+      | r = x^2 + a^2
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d z} {\d x}
+      | r = 2 x
+      | c = [[Power Rule for Derivatives]]
+}}
+{{eqn | ll= \leadsto
+      | l = \int \frac {x \rd x} {\paren {\sqrt {x^2 + a^2} }^3}
+      | r = \int \frac {x \rd z} {2 x z^{3/2} }
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \frac 1 2 \int z^{-3/2} \rd z
+      | c = [[Primitive of Constant Multiple of Function]]
+}}
+{{eqn | r = \frac 1 2 \paren {\frac {-z^{1/2} } {\frac 1 2} } + C
+      | c = [[Primitive of Power]]
+}}
+{{eqn | r = \frac {-1} {\sqrt z} + C
+      | c = simplifying
+}}
+{{eqn | r = \frac {-1} {\sqrt {x^2 + a^2} } + C
+      | c = substituting for $z$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of x squared over Root of x squared plus a squared cubed}
+Tags: Primitives involving Root of x squared plus a squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {x^2 \ \mathrm d x} {\left({\sqrt {x^2 + a^2} }\right)^3} = \frac {-x} {\sqrt {x^2 + a^2} } + \ln \left({x + \sqrt {x^2 + a^2} }\right) + C$
+\end{theorem}
+
+\begin{proof}
+With a view to expressing the problem in the form:
+:$\displaystyle \int u \frac {\mathrm d v}{\mathrm d x} \mathrm d x = u v - \int v \frac {\mathrm d u}{\mathrm d x} \mathrm d x$
+let:
+{{begin-eqn}}
+{{eqn | l = u
+      | r = x
+}}
+{{eqn | ll= \implies
+      | l = \frac {\mathrm d u} {\mathrm d x}
+      | r = 1
+      | c = [[Power Rule for Derivatives]]
+}}
+{{end-eqn}}
+and let:
+{{begin-eqn}}
+{{eqn | l = \frac {\mathrm d v} {\mathrm d x}
+      | r = \frac x {\left({\sqrt {x^2 + a^2} }\right)^3}
+}}
+{{eqn | ll= \implies
+      | l = v
+      | r = \frac {-1} {\sqrt {x^2 + a^2} }
+      | c = [[Primitive of x over Root of x squared plus a squared cubed|Primitive of $\dfrac x {\left({\sqrt {x^2 + a^2} }\right)^3}$]]
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = \int \frac {x^2 \ \mathrm d x} {\left({\sqrt {x^2 + a^2} }\right)^3}
+      | r = \int x \frac {x \ \mathrm d x} {\left({\sqrt {x^2 + a^2} }\right)^3}
+      | c = 
+}}
+{{eqn | r = x \frac {-1} {\sqrt {x^2 + a^2} } - \int \frac {-1} {\sqrt {x^2 + a^2} } 1 + C
+      | c = [[Integration by Parts]]
+}}
+{{eqn | r = \frac {-x} {\sqrt {x^2 + a^2} } + \int \frac 1 {\sqrt {x^2 + a^2} } + C
+      | c = simplifying
+}}
+{{eqn | r = \frac {-x} {\sqrt {x^2 + a^2} } + \ln \left({x + \sqrt {x^2 + a^2} }\right) + C
+      | c = [[Primitive of Reciprocal of Root of x squared plus a squared/Logarithm Form|Primitive of $\dfrac 1 {\sqrt {x^2 + a^2} }$]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of x cubed over Root of x squared plus a squared cubed}
+Tags: Primitives involving Root of x squared plus a squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {x^3 \ \mathrm d x} {\left({\sqrt {x^2 + a^2} }\right)^3} = \sqrt {x^2 + a^2} + \frac {a^2} {\sqrt {x^2 + a^2} } + C$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int \frac {x^3 \ \mathrm d x} {\left({\sqrt {x^2 + a^2} }\right)^3}
+      | r = \int \frac {x \left({x^2}\right) \ \mathrm d x} {\left({\sqrt {x^2 + a^2} }\right)^3}
+      | c = 
+}}
+{{eqn | r = \int \frac {x \left({x^2 + a^2 - a^2}\right) \ \mathrm d x} {\left({\sqrt {x^2 + a^2} }\right)^3}
+      | c = 
+}}
+{{eqn | r = \int \frac {x \left({x^2 + a^2}\right) \ \mathrm d x} {\left({\sqrt {x^2 + a^2} }\right)^3} - a^2 \int \frac {x \ \mathrm d x} {\left({\sqrt {x^2 + a^2} }\right)^3}
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \int \frac {x \ \mathrm d x} {\sqrt {x^2 + a^2} } - a^2 \int \frac {x \ \mathrm d x} {\left({\sqrt {x^2 + a^2} }\right)^3}
+      | c = simplifying
+}}
+{{eqn | r = \sqrt {x^2 + a^2} - a^2 \int \frac {x \ \mathrm d x} {\left({\sqrt {x^2 + a^2} }\right)^3} + C
+      | c = [[Primitive of x over Root of x squared plus a squared|Primitive of $\dfrac x {\sqrt {x^2 + a^2} }$]]
+}}
+{{eqn | r = \sqrt {x^2 + a^2} - a^2 \frac {-1} {\sqrt {x^2 + a^2} } + C
+      | c = [[Primitive of x over Root of x squared plus a squared cubed|Primitive of $\dfrac x {\left({\sqrt {x^2 + a^2} }\right)^3}$]]
+}}
+{{eqn | r = \sqrt {x^2 + a^2} + \frac {a^2} {\sqrt {x^2 + a^2} } + C
+      | c = simplifying
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Reciprocal of x by Root of x squared plus a squared cubed}
+Tags: Primitives involving Root of x squared plus a squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {\d x} {x \paren {\sqrt {x^2 + a^2} }^3} = \frac 1 {a^2 \sqrt {x^2 + a^2} } - \frac 1 {a^3} \map \ln {\frac {a + \sqrt {x^2 + a^2} } x} + C$
+\end{theorem}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = z
+      | r = x^2
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d z} {\d x}
+      | r = 2 x
+      | c = [[Power Rule for Derivatives]]
+}}
+{{eqn | ll= \leadsto
+      | l = \int \frac {\d x} {x \paren {\sqrt {x^2 + a^2} }^3}
+      | r = \int \frac {\d z} {2 \sqrt z \sqrt z \paren {\sqrt {z + a^2} }^3}
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \frac 1 2 \int \frac {\d z} {z \paren {\sqrt {z + a^2} }^3}
+      | c = [[Primitive of Constant Multiple of Function]]
+}}
+{{eqn | r = \frac 1 2 \paren {\frac 2 {a^2 \sqrt {z + a^2} } + \frac 1 {a^2} \int \frac {\d z} {z \sqrt {z + a^2} } } + C
+      | c = [[Primitive of Reciprocal of x by Power of Root of a x + b|Primitive of $\dfrac 1 {x \paren {\sqrt{a x + b} }^m }$]]
+}}
+{{eqn | r = \frac 1 {a^2 \sqrt {z + a^2} } + \frac 1 {2 a^2} \int \frac {\d z} {z \sqrt {z + a^2} } + C
+      | c = simplifying
+}}
+{{eqn | r = \frac 1 {a^2 \sqrt {x^2 + a^2} } + \frac 1 {2 a^2} \int \frac {2 x \rd x} {x^2 \sqrt {x^2 + a^2} } + C
+      | c = substituting for $z$
+}}
+{{eqn | r = \frac 1 {a^2 \sqrt {x^2 + a^2} } + \frac 1 {a^2} \int \frac {\d x} {x \sqrt {x^2 + a^2} } + C
+      | c = simplifying
+}}
+{{eqn | r = \frac 1 {a^2 \sqrt {x^2 + a^2} } - \frac 1 {a^3} \map \ln {\frac {a + \sqrt {x^2 + a^2} } x} + C
+      | c = [[Primitive of Reciprocal of x by Root of x squared plus a squared/Logarithm Form|Primitive of $\dfrac 1 {x \sqrt {x^2 + a^2} }$]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Reciprocal of x squared by Root of x squared plus a squared cubed}
+Tags: Primitives involving Root of x squared plus a squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {\d x} {x^2 \paren {\sqrt {x^2 + a^2} }^3} = \frac {-\sqrt {x^2 + a^2} } {a^4 x} - \frac x {a^4 \sqrt {x^2 + a^2} } + C$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int \frac {\d x} {x^2 \paren {\sqrt {x^2 + a^2} }^3}
+      | r = \int \frac {a^2 \rd x} {a^2 x^2 \paren {\sqrt {x^2 + a^2} }^3}
+      | c = 
+}}
+{{eqn | r = \int \frac {\paren {x^2 + a^2 - x^2} \rd x} {a^2 x^2 \paren {\sqrt {x^2 + a^2} }^3}
+      | c = 
+}}
+{{eqn | r = \frac 1 {a^2} \int \frac {\paren {x^2 + a^2} \rd x} {a^2 x^2 \paren {\sqrt {x^2 + a^2} }^3} - \frac 1 {a^2} \int \frac {x^2 \rd x} {x^2 \paren {\sqrt {x^2 + a^2} }^3}
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \frac 1 {a^2} \int \frac {\d x} {x^2 \sqrt {x^2 + a^2} } - \frac 1 {a^2} \int \frac {\d x} {\paren {\sqrt {x^2 + a^2} }^3}
+      | c = simplifying
+}}
+{{eqn | r = \frac 1 {a^2} \frac {-\sqrt {x^2 + a^2} } {a^2 x} - \frac 1 {a^2} \int \frac {\d x} {\paren {\sqrt {x^2 + a^2} }^3} + C
+      | c = [[Primitive of Reciprocal of x squared by Root of x squared plus a squared|Primitive of $\dfrac 1 {x^2 \sqrt {x^2 + a^2} }$]]
+}}
+{{eqn | r = \frac {-\sqrt {x^2 + a^2} } {a^4 x} - \frac 1 {a^2} \frac x {a^2 \sqrt {x^2 + a^2} } + C
+      | c = [[Primitive of Reciprocal of Root of x squared plus a squared cubed|Primitive of $\dfrac 1 {\paren {\sqrt {x^2 + a^2} }^3}$]]
+}}
+{{eqn | r = \frac {-\sqrt {x^2 + a^2} } {a^4 x} - \frac x {a^4 \sqrt {x^2 + a^2} } + C
+      | c = simplification
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Reciprocal of x cubed by Root of x squared plus a squared cubed}
+Tags: Primitives involving Root of x squared plus a squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {\mathrm d x} {x^3 \left({\sqrt {x^2 + a^2} }\right)^3} = \frac {-1} {2 a^2 x^2 \sqrt {x^2 + a^2} } - \frac 3 {2 a^4 \sqrt {x^2 + a^2} } + \frac 3 {2 a^5} \ln \left({\frac {a + \sqrt {x^2 + a^2} } x}\right) + C$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int \frac {\mathrm d x} {x^3 \left({\sqrt {x^2 + a^2} }\right)^3}
+      | r = \int \frac {a^2 \ \mathrm d x} {a^2 x^3 \left({\sqrt {x^2 + a^2} }\right)^3}
+      | c = 
+}}
+{{eqn | r = \int \frac {\left({x^2 + a^2 - x^2}\right) \ \mathrm d x} {a^2 x^3 \left({\sqrt {x^2 + a^2} }\right)^3}
+      | c = 
+}}
+{{eqn | r = \frac 1 {a^2} \int \frac {\left({x^2 + a^2}\right) \ \mathrm d x} {x^3 \left({\sqrt {x^2 + a^2} }\right)^3} - \frac 1 {a^2} \int \frac {x^2 \ \mathrm d x} {x^3 \left({\sqrt {x^2 + a^2} }\right)^3}
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \frac 1 {a^2} \int \frac {\mathrm d x} {x^3 \sqrt {x^2 + a^2} } - \frac 1 {a^2} \int \frac {\mathrm d x} {x \left({\sqrt {x^2 + a^2} }\right)^3}
+      | c = simplifying
+}}
+{{eqn | r = \frac 1 {a^2} \left({\frac {- \sqrt {x^2 + a^2} } {2 a^2 x^2} + \frac 1 {2 a^3} \ln \left({\frac {a + \sqrt {x^2 + a^2} } x}\right)}\right) - \frac 1 {a^2} \int \frac {\mathrm d x} {x \left({\sqrt {x^2 + a^2} }\right)^3} + C
+      | c = [[Primitive of Reciprocal of x cubed by Root of x squared plus a squared|Primitive of $\dfrac 1 {x^3 \sqrt {x^2 + a^2} }$]]
+}}
+{{eqn | r = \frac 1 {a^2} \left({\frac {- \sqrt {x^2 + a^2} } {2 a^2 x^2} + \frac 1 {2 a^3} \ln \left({\frac {a + \sqrt {x^2 + a^2} } x}\right)}\right)
+      | c = 
+}}
+{{eqn | o = 
+      | ro= -
+      | r = \frac 1 {a^2} \left({\frac 1 {a^2 \sqrt {x^2 + a^2} } - \frac 1 {a^3} \ln \left({\frac {a + \sqrt {x^2 + a^2} } x}\right) }\right) + C
+      | c = [[Primitive of Reciprocal of x by Root of x squared plus a squared cubed|Primitive of $\dfrac 1 {x \left({\sqrt {x^2 + a^2} }\right)^3}$]]
+}}
+{{eqn | r = \frac {-1} {2 a^2 x^2 \sqrt {x^2 + a^2} } - \frac 3 {2 a^4 \sqrt {x^2 + a^2} } + \frac 3 {2 a^5} \ln \left({\frac {a + \sqrt {x^2 + a^2} } x}\right) + C
+      | c = simplification
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Root of x squared plus a squared cubed}
+Tags: Primitives involving Root of x squared plus a squared
+
+\begin{theorem}
+:$\ds \int \paren {\sqrt {x^2 + a^2} }^3 \rd x = \frac {x \paren {\sqrt {x^2 + a^2} }^3} 4 + \frac {3 a^2 x \sqrt {x^2 + a^2} } 8 + \frac {3 a^4} 8 \map \ln {x + \sqrt {x^2 + a^2} } + C$
+\end{theorem}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = z
+      | r = x^2
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d z} {\d x}
+      | r = 2 x
+      | c = [[Power Rule for Derivatives]]
+}}
+{{eqn | ll= \leadsto
+      | l = \int \paren {\sqrt {x^2 + a^2} }^3 \rd x
+      | r = \int \frac {\paren {\sqrt {z + a^2} }^3} {2 \sqrt z} \rd x
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \frac {\sqrt z \paren {\sqrt {z + a^2} }^3} 4 + \frac {3 a^2} 8 \int \frac {\sqrt {z + a^2} } {\sqrt z} \rd x + C
+      | c = [[Primitive of Power of p x + q over Root of a x + b|Primitive of $\dfrac {\paren {p x + q}^n} {\sqrt{a x + b} }$]]
+}}
+{{eqn | r = \frac {\sqrt z \paren {\sqrt {z + a^2} }^3} 4 + \frac {3 a^2} 8 \paren {\sqrt z \sqrt {z + a^2} + \frac {a^2} 2 \int \frac {\d z} {\sqrt z \sqrt {z + a^2} } } + C
+      | c = [[Primitive of Root of p x + q over Root of a x + b|Primitive of $\dfrac {\sqrt{p x + q} } {\sqrt{a x + b} }$]]
+}}
+{{eqn | r = \frac {x \paren {\sqrt {x^2 + a^2} }^3} 4 + \frac {3 a^2} 8 x \sqrt {x^2 + a^2} + \frac {3 a^4} 8 \int \frac {\d x} {\sqrt {x^2 + a^2} } + C
+      | c = substituting for $z$ and simplifying
+}}
+{{eqn | r = \frac {x \paren {\sqrt {x^2 + a^2} }^3} 4 + \frac {3 a^2 x \sqrt {x^2 + a^2} } 8 + \frac {3 a^4} 8 \map \ln {x + \sqrt {x^2 + a^2} } + C
+      | c = [[Primitive of Reciprocal of Root of x squared plus a squared/Logarithm Form|Primitive of $\dfrac 1 {\sqrt {x^2 + a^2} }$]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of x by Root of x squared plus a squared cubed}
+Tags: Primitives involving Root of x squared plus a squared
+
+\begin{theorem}
+:$\displaystyle \int x \left({\sqrt {x^2 + a^2} }\right)^3 \ \mathrm d x = \frac {\left({\sqrt {x^2 + a^2} }\right)^5} 5 + C$
+\end{theorem}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = z
+      | r = x^2 + a^2
+}}
+{{eqn | ll= \implies
+      | l = \frac {\mathrm d z} {\mathrm d x}
+      | r = 2 x
+      | c = [[Power Rule for Derivatives]]
+}}
+{{eqn | ll= \implies
+      | l = \int x \left({\sqrt {x^2 + a^2} }\right)^3 \ \mathrm d x
+      | r = \int \frac {z^{3/2} } 2 \ \mathrm d z
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \frac 1 2 \frac {z^{5/2} } {\frac 5 2} + C
+      | c = [[Primitive of Power]]
+}}
+{{eqn | r = \frac {z^{5/2} } 5 + C
+      | c = simplifying
+}}
+{{eqn | r = \frac {\left({\sqrt {x^2 + a^2} }\right)^5} 5 + C
+      | c = substituting for $z$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of x squared by Root of x squared plus a squared cubed}
+Tags: Primitives involving Root of x squared plus a squared
+
+\begin{theorem}
+:$\displaystyle \int x^2 \left({\sqrt {x^2 + a^2} }\right)^3 \ \mathrm d x = \frac {x \left({\sqrt {x^2 + a^2} }\right)^5} 6 - \frac {a^2 x \left({\sqrt {x^2 + a^2} }\right)^3} {24} - \frac {a^4 x \sqrt {x^2 + a^2} } {16} - \frac {a^6} {16} \ln \left({x + \sqrt {x^2 + a^2} }\right) + C$
+\end{theorem}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = z
+      | r = x^2
+}}
+{{eqn | ll= \implies
+      | l = \frac {\mathrm d z} {\mathrm d x}
+      | r = 2 x
+      | c = [[Power Rule for Derivatives]]
+}}
+{{eqn | ll= \implies
+      | l = \int x^2 \left({\sqrt {x^2 + a^2} }\right)^3 \ \mathrm d x
+      | r = \int \frac {\sqrt z \left({z + a^2}\right)^{3/2} } 2 \ \mathrm d z
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \frac {\sqrt z \left({z + a^2}\right)^{5/2} } 6 - \frac {a^2} {12} \int \frac {\left({z + a^2}\right)^{3/2} } {\sqrt z} \ \mathrm d z + C
+      | c = [[Primitive of Power of p x + q by Root of a x + b|Primitive of $\left({p x + q}\right)^n \sqrt{a x + b}$]]
+}}
+{{eqn | r = \frac {x \left({x^2 + a^2}\right)^{5/2} } 6 - \frac {a^2} 6 \int \left({x^2 + a^2}\right)^{3/2} \ \mathrm d x + C
+      | c = substituting for $z$ and simplifying
+}}
+{{eqn | r = \frac {x \left({x^2 + a^2}\right)^{5/2} } 6
+}}
+{{eqn | o =
+      | ro= -
+      | r = \frac {a^2} 6 \left({\frac {x \left({\sqrt {x^2 + a^2} }\right)^3} 4 + \frac {3 a^2 x \sqrt {x^2 + a^2} } 8 + \frac {3 a^4} 8 \ln \left({x + \sqrt {x^2 + a^2} }\right)}\right) + C
+      | c = [[Primitive of Root of x squared plus a squared cubed|Primitive of $\left({\sqrt {x^2 + a^2} }\right)^3$]]
+}}
+{{eqn | r = \frac {x \left({x^2 + a^2}\right)^{5/2} } 6 - \frac {a^2 x \left({\sqrt {x^2 + a^2} }\right)^3} {24}
+}}
+{{eqn | o =
+      | ro= -
+      | r = \frac {a^4 x \sqrt {x^2 + a^2} } {16} - \frac {a^6} {16} \ln \left({x + \sqrt {x^2 + a^2} }\right) + C
+      | c = simplifying
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of x cubed by Root of x squared plus a squared cubed}
+Tags: Primitives involving Root of x squared plus a squared
+
+\begin{theorem}
+:$\displaystyle \int x^3 \paren {\sqrt {x^2 + a^2} }^3 \rd x = \frac {\paren {\sqrt {x^2 + a^2} }^7} 7 - \frac {a^2 \paren {\sqrt {x^2 + a^2} }^5} 5 + C$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int x^3 \paren {\sqrt {x^2 + a^2} }^3 \rd x
+      | r = \int x \paren {x^2} \paren {\sqrt {x^2 + a^2} }^3 \rd x
+      | c = 
+}}
+{{eqn | r = \int x \paren {x^2 + a^2 - a^2} \paren {\sqrt {x^2 + a^2} }^3 \rd x
+      | c = [[Primitive of Power]]
+}}
+{{eqn | r = \int x \paren {x^2 + a^2} \paren {\sqrt {x^2 + a^2} }^3 \rd x - a^2 \int x \paren {\sqrt {x^2 + a^2} }^3 \rd x
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \int x \paren {\sqrt {x^2 + a^2} }^5 \rd x - a^2 \int x \paren {\sqrt {x^2 + a^2} }^3 \rd x
+      | c = simplifying
+}}
+{{end-eqn}}
+Let:
+{{begin-eqn}}
+{{eqn | l = z
+      | r = x^2 + a^2
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d z} {\d x}
+      | r = 2 x
+      | c = [[Power Rule for Derivatives]]
+}}
+{{eqn | ll= \leadsto
+      | o = 
+      | r = \int x \paren {\sqrt {x^2 + a^2} }^5 \rd x - a^2 \int x \paren {\sqrt {x^2 + a^2} }^3 \rd x
+      | c = 
+}}
+{{eqn | r = \int \frac {\sqrt z z^{5/2} } {2 \sqrt z} \rd z - a^2 \int \frac {\sqrt z z^{3/2} } {2 \sqrt z} \rd z
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r =  \frac 1 2 \int z^{5/2} \rd z - \frac {a^2} 2 \int z^{3/2} \rd z
+      | c = simplifying
+}}
+{{eqn | r = \frac 1 2 \frac {z^{7/2} } {7/2} - \frac {a^2} 2 \frac {z^{5/2} } {5/2} + C
+      | c = [[Primitive of Power]]
+}}
+{{eqn | r = \frac {z^{7/2} } 7 - a^2 \frac {z^{5/2} } 5 + C
+      | c = simplifying
+}}
+{{eqn | r = \frac {\paren {\sqrt {x^2 + a^2} }^7} 7 - \frac {a^2 \paren {\sqrt {x^2 + a^2} }^5} 5 + C
+      | c = substituting for $z$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Root of x squared plus a squared cubed over x}
+Tags: Primitives involving Root of x squared plus a squared
+
+\begin{theorem}
+:$\ds \int \frac {\paren {\sqrt {x^2 + a^2} }^3} x \rd x = \frac {\paren {\sqrt {x^2 + a^2} }^3} 3 + a^2 \sqrt {x^2 + a^2} - a^3 \map \ln {\frac {a + \sqrt {x^2 + a^2} } x} + C$
+\end{theorem}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = z
+      | r = x^2
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d z} {\d x}
+      | r = 2 x
+      | c = [[Power Rule for Derivatives]]
+}}
+{{eqn | ll= \leadsto
+      | l = \int \frac {\paren {\sqrt {x^2 + a^2} }^3} x \rd x
+      | r = \int \frac {\paren {\sqrt {z + a^2} }^3} {2 z} \rd z
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \frac {\paren {\sqrt {z + a^2} }^3} 3 + \frac {a^2} 2 \int \frac {\sqrt {z + a^2} } z \rd z
+      | c = [[Primitive of Power of Root of a x + b over x|Primitive of $\dfrac {\paren {\sqrt {a x + b} }^m} x$]]
+}}
+{{eqn | r = \frac {\paren {\sqrt {z + a^2} }^3} 3 + \frac {a^2} 2 \paren {2 \sqrt {z + a^2} + a^2 \int \frac {\d z} {z \sqrt {z + a^2} } }
+      | c = [[Primitive of Root of a x + b over x|Primitive of $\dfrac {\sqrt {a x + b} } x$]]
+}}
+{{eqn | r = \frac {\paren {\sqrt {x^2 + a^2} }^3} 3 + a^2 \sqrt {x^2 + a^2} + a^4 \int \frac {\d x} {x \sqrt {x^2 + a^2} }
+      | c = substituting for $z$ and simplifying
+}}
+{{eqn | r = \frac {\paren {\sqrt {x^2 + a^2} }^3} 3 + a^2 \sqrt {x^2 + a^2} - a^3 \map \ln {\frac {a + \sqrt {x^2 + a^2} } x} + C
+      | c = [[Primitive of Reciprocal of x by Root of x squared plus a squared|Primitive of $\dfrac 1 {x \sqrt {x^2 + a^2} }$]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Root of x squared plus a squared cubed over x squared}
+Tags: Primitives involving Root of x squared plus a squared
+
+\begin{theorem}
+:$\displaystyle \int \frac{\left({\sqrt {x^2 + a^2} }\right)^3} {x^2} \ \mathrm d x = \frac {-\left({\sqrt {x^2 + a^2} }\right)^3} x + \frac{3 x \sqrt {x^2 + a^2} } 2 + \frac {3 a^2} 2 \ln \left({x + \sqrt {x^2 + a^2} }\right) + C$
+\end{theorem}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = z
+      | r = x^2
+}}
+{{eqn | ll= \implies
+      | l = \frac {\mathrm d z} {\mathrm d x}
+      | r = 2 x
+      | c = [[Power Rule for Derivatives]]
+}}
+{{eqn | ll= \implies
+      | l = \int \frac{\left({\sqrt {x^2 + a^2} }\right)^3} {x^2} \ \mathrm d x
+      | r = \int \frac{\left({\sqrt {z + a^2} }\right)^3} {2 z \sqrt z} \ \mathrm d z
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \frac {-\left({\sqrt {z + a^2} }\right)^3} {\sqrt z} + \frac 3 2 \int \frac{\sqrt {z + a^2} } {\sqrt z} \ \mathrm d z
+      | c = [[Primitive of Power of a x + b over Power of p x + q/Formulation 3|Primitive of $\dfrac {\left({a x + b}\right)^m} {\left({p x + q}\right)^n}$: Formulation 3]]
+}}
+{{eqn | r = \frac {-\left({\sqrt {x^2 + a^2} }\right)^3} x + 3 \int \sqrt {x^2 + a^2} \ \mathrm d x
+      | c = substituting for $z$ and simplifying
+}}
+{{eqn | r = \frac {-\left({\sqrt {x^2 + a^2} }\right)^3} x + \frac{3 x \sqrt {x^2 + a^2} } 2 + \frac {3 a^2} 2 \ln \left({x + \sqrt {x^2 + a^2} }\right) + C
+      | c = [[Primitive of Root of x squared plus a squared|Primitive of $\sqrt {x^2 + a^2}$]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Root of x squared plus a squared cubed over x cubed}
+Tags: Primitives involving Root of x squared plus a squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {\paren {\sqrt {x^2 + a^2} }^3} {x^3} \rd x = \frac {-\paren {\sqrt {x^2 + a^2} }^3} {2 x^2} + \frac {3 \sqrt {x^2 + a^2} } 2 - \frac {3 a} 2 \map \ln {\frac {a + \sqrt {x^2 + a^2} } x} + C$
+\end{theorem}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = z
+      | r = x^2
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d z} {\d x}
+      | r = 2 x
+      | c = [[Power Rule for Derivatives]]
+}}
+{{eqn | ll= \leadsto
+      | l = \int \frac {\paren {\sqrt {x^2 + a^2} }^3} {x^3} \rd x
+      | r = \int \frac {\paren {\sqrt {z + a^2} }^3} {2 z^2} \rd z
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \frac 1 2 \paren {\frac {-\paren {\sqrt {z + a^2} }^3} z + \frac 3 2 \int \frac {\sqrt {z + a^2} } z \rd z}
+      | c = [[Primitive of Power of a x + b over Power of p x + q/Formulation 3|Primitive of $\dfrac {\paren {a x + b}^m} {\paren {p x + q}^n}$: Formulation 3]]
+}}
+{{eqn | r = \frac {-\paren {\sqrt {x^2 + a^2} }^3} {2 x^2} + \frac 3 2 \int \frac {\sqrt {x^2 + a^2} } x \rd x
+      | c = substituting for $z$ and simplifying
+}}
+{{eqn | r = \frac {-\paren {\sqrt {x^2 + a^2} }^3} {2 x^2} + \frac 3 2 \paren {\sqrt {x^2 + a^2} - a \map \ln {\frac {a + \sqrt {x^2 + a^2} } a} } + C
+      | c = [[Primitive of Root of x squared plus a squared over x|Primitive of $\dfrac {\sqrt {x^2 + a^2} } x$]]
+}}
+{{eqn | r = \frac {-\paren {\sqrt {x^2 + a^2} }^3} {2 x^2} + \frac {3 \sqrt {x^2 + a^2} } 2 - \frac {3 a} 2 \map \ln {\frac {a + \sqrt {x^2 + a^2} } x} + C
+      | c = simplifying
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of x over Root of x squared minus a squared}
+Tags: Primitives involving Root of x squared minus a squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {x \rd x} {\sqrt {x^2 - a^2} } = \sqrt {x^2 - a^2} + C$
+\end{theorem}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = z^2
+      | r = x^2 - a^2
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = 2 z \frac {\d z} {\d x}
+      | r = 2 x
+      | c = [[Chain Rule for Derivatives]], [[Power Rule for Derivatives]]
+}}
+{{eqn | ll= \leadsto
+      | l = \int \frac {x \rd x} {\sqrt {x^2 - a^2} }
+      | r = \int \frac {z \rd z} z
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \int \rd z
+      | c = 
+}}
+{{eqn | r = z + C
+      | c = [[Primitive of Constant]]
+}}
+{{eqn | r = \sqrt {x^2 - a^2} + C
+      | c = substituting for $z$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of x squared over Root of x squared minus a squared}
+Tags: Primitives involving Root of x squared minus a squared, Primitive of x squared over Root of x squared minus a squared
+
+\begin{theorem}
+{{:Primitive of x squared over Root of x squared minus a squared/Inverse Hyperbolic Cosine Form}}
+\end{theorem}<|endoftext|>
+\section{Primitive of x cubed over Root of x squared minus a squared}
+Tags: Primitives involving Root of x squared minus a squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {x^3 \rd x} {\sqrt {x^2 - a^2} } = \frac {\paren {\sqrt {x^2 - a^2} }^3} 3 + a^2 \sqrt {x^2 - a^2} + C$
+\end{theorem}
+
+\begin{proof}
+With a view to expressing the problem in the form:
+:$\displaystyle \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$
+let:
+{{begin-eqn}}
+{{eqn | l = u
+      | r = x^2
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d u} {\d x}
+      | r = 2 x
+      | c = [[Power Rule for Derivatives]]
+}}
+{{end-eqn}}
+and let:
+{{begin-eqn}}
+{{eqn | l = \frac {\d v} {\d x}
+      | r = \frac x {\sqrt {x^2 - a^2} }
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = v
+      | r = \sqrt {x^2 - a^2}
+      | c = [[Primitive of x over Root of x squared minus a squared|Primitive of $\dfrac x {\sqrt {x^2 - a^2} }$]]
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = \int \frac {x^3 \rd x} {\sqrt {x^2 - a^2} }
+      | r = \int x^2 \frac {x \rd x} {\sqrt {x^2 - a^2} }
+      | c = 
+}}
+{{eqn | r = x^2 \sqrt {x^2 - a^2} - \int 2 x \sqrt {x^2 - a^2} \rd x
+      | c = [[Integration by Parts]]
+}}
+{{eqn | r = x^2 \sqrt {x^2 - a^2} - 2 \paren {\frac {\paren {\sqrt {x^2 - a^2} }^3} 3} + C
+      | c = [[Primitive of x by Root of x squared minus a squared|Primitive of $x \sqrt {x^2 - a^2}$]]
+}}
+{{eqn | r = \paren {x^2 - a^2 + a^2} \sqrt {x^2 - a^2} - 2 \paren {\frac {\paren {\sqrt {x^2 - a^2} }^3} 3} + C
+      | c = 
+}}
+{{eqn | r = \paren {\sqrt {x^2 - a^2} }^3 + a^2 \sqrt {x^2 - a^2} - 2 \paren {\frac {\paren {\sqrt {x^2 - a^2} }^3} 3} + C
+      | c = 
+}}
+{{eqn | r = \frac {\paren {\sqrt {x^2 - a^2} }^3} 3 + a^2 \sqrt {x^2 - a^2} + C
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Reciprocal of x squared by Root of x squared minus a squared}
+Tags: Primitives involving Root of x squared minus a squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {\d x} {x^2 \sqrt {x^2 - a^2} } = \frac {\sqrt {x^2 - a^2} } {a^2 x} + C$
+\end{theorem}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = z
+      | r = x^2
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d z} {\d x}
+      | r = 2 x
+      | c = [[Power Rule for Derivatives]]
+}}
+{{eqn | ll= \leadsto
+      | l = \int \frac {\d x} {x^2 \sqrt {x^2 - a^2} }
+      | r = \int \frac {\d z} {2 z \sqrt z \sqrt {z - a^2} }
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \frac 1 2 \int \frac {\d z} {z^{3/2} \sqrt {z - a^2} }
+      | c = 
+}}
+{{end-eqn}}
+Using [[Primitive of Reciprocal of Power of x by Root of a x + b|Primitive of $\dfrac 1 {u^m \sqrt {a u + b} }$]]:
+:$\displaystyle \int \frac {\d u} {u^m \sqrt {a u + b} } = -\frac {\sqrt {a u + b} } {\paren {m - 1} b u^{m - 1} } - \frac {\paren {2 m - 3} a} {\paren {2 m - 2} b} \int \frac {\d u} {u^{m - 1} \sqrt {a u + b} }$
+Setting:
+{{begin-eqn}}
+{{eqn | l = u
+      | o = :=
+      | r = z
+}}
+{{eqn | l = m
+      | o = :=
+      | r = \frac 3 2
+}}
+{{eqn | l = a
+      | o = :=
+      | r = 1
+}}
+{{eqn | l = b
+      | o = :=
+      | r = -a^2
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = \frac 1 2 \int \frac {\d z} {z^{3/2} \sqrt {z - a^2} }
+      | r = \frac {-\sqrt {z - a^2} } {2 \paren {\paren {\frac 3 2} - 1} \paren {-a^2} z^{\paren {3/2} - 1} } - \frac {2 \paren {\frac 3 2} - 3} {2 \paren {2 \paren {\frac 3 2} - 2} \paren {-a^2} } \int \frac {\d z} {z^{\paren {\frac 3 2} - 1} \sqrt{z - a^2} } + C
+      | c = [[Primitive of Reciprocal of Power of x by Root of a x + b|Primitive of $ \dfrac 1 {x^m \sqrt {a x + b} }$]]
+}}
+{{eqn | r = \frac {-\sqrt {z - a^2} } {-a^2 z^{1/2} } - 0 + C
+      | c = simplifying
+}}
+{{eqn | r = \frac {\sqrt {x^2 - a^2} } {a^2 x} + C
+      | c = substituting $x$ back for $z$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Reciprocal of x cubed by Root of x squared minus a squared}
+Tags: Primitives involving Root of x squared minus a squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {\d x} {x^3 \sqrt {x^2 - a^2} } = \frac {\sqrt {x^2 - a^2} } {2 a^2 x^2} + \frac 1 {2 a^3} \arcsec \size {\frac x a} + C$
+\end{theorem}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = z
+      | r = x^2
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d z} {\d x}
+      | r = 2 x
+      | c = [[Power Rule for Derivatives]]
+}}
+{{eqn | ll= \leadsto
+      | l = \int \frac {\d x} {x^3 \sqrt {x^2 - a^2} }
+      | r = \int \frac {\d z} {2 z^{3/2} \sqrt z \sqrt {z - a^2} }
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \frac 1 2 \int \frac {\d z} {z^2 \sqrt {z - a^2} }
+      | c = 
+}}
+{{end-eqn}}
+Using [[Primitive of Reciprocal of Power of x by Root of a x + b|Primitive of $ \dfrac 1 {x^m \sqrt {a x + b} }$]]:
+:$\displaystyle \int \frac {\d x} {x^m \sqrt {a x + b} } = -\frac {\sqrt {a x + b} } {\paren {m - 1} b x^{m-1} } - \frac {\paren {2 m - 3} a} {\paren {2 m - 2} b} \int \frac {\d x} {x^{m - 1} \sqrt {a x + b} }$
+Setting:
+{{begin-eqn}}
+{{eqn | l = x
+      | o = :=
+      | r = z
+}}
+{{eqn | l = m
+      | o = :=
+      | r = 2
+}}
+{{eqn | l = a
+      | o = :=
+      | r = 1
+}}
+{{eqn | l = b
+      | o = :=
+      | r = -a^2
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = \frac 1 2 \int \frac {\d z} {z^2 \sqrt {z - a^2} }
+      | r = \frac {-\sqrt {z - a^2} } {2 \paren {2 - 1} \paren {-a^2} z^{2 - 1} } - \frac {2 \paren 2 - 3} {2 \paren {2 \paren 2 - 2} \paren {-a^2} } \int \frac {\d z} {z^{2 - 1} \sqrt {z - a^2} } + C
+      | c = [[Primitive of Reciprocal of Power of x by Root of a x + b|Primitive of $ \dfrac 1 {x^m \sqrt{a x + b} }$]]
+}}
+{{eqn | r = \frac {\sqrt {z - a^2} } {2 a^2 z} + \frac 1 {4 a^2} \int \frac {\d z} {z \sqrt{z - a^2} } + C
+      | c = simplifying
+}}
+{{eqn | r = \frac {\sqrt {x^2 - a^2} } {2 a^2 x^2} + \frac 1 {4 a^2} \paren {\int \frac {2 x \rd x} {x^2 \sqrt {x^2 - a^2} } } + C
+      | c = substituting for $z$
+}}
+{{eqn | r = \frac {\sqrt {x^2 - a^2} } {2 a^2 x^2} + \frac 1 {2 a^2} \paren {\int \frac {\d x} {x \sqrt {x^2 - a^2} } } + C
+      | c = simplification
+}}
+{{eqn | r = \frac {\sqrt {x^2 - a^2} } {2 a^2 x^2} + \frac 1 {2 a^3} \arcsec \size {\frac x a} + C
+      | c = [[Primitive of Reciprocal of x by Root of x squared minus a squared|Primitive of $\dfrac 1 {x \sqrt {x^2 - a^2} }$]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Root of x squared minus a squared}
+Tags: Primitives involving Root of x squared minus a squared, Primitive of Root of x squared minus a squared
+
+\begin{theorem}
+{{:Primitive of Root of x squared minus a squared/Inverse Hyperbolic Cosine Form}}
+\end{theorem}<|endoftext|>
+\section{Primitive of x by Root of x squared minus a squared}
+Tags: Primitives involving Root of x squared minus a squared
+
+\begin{theorem}
+:$\displaystyle \int x \sqrt {x^2 - a^2} \rd x = \frac {\paren {\sqrt {x^2 - a^2} }^3} 3 + C$
+\end{theorem}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = z
+      | r = x^2
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d z} {\d x}
+      | r = 2 x
+      | c = [[Power Rule for Derivatives]]
+}}
+{{eqn | ll= \leadsto
+      | l = \int x \sqrt {x^2 - a^2} \rd x
+      | r = \int \frac {\sqrt z \sqrt {z - a^2} \rd z} {2 \sqrt z}
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \frac 1 2 \int \sqrt {z - a^2} \rd z
+      | c = [[Primitive of Constant Multiple of Function]]
+}}
+{{eqn | r = \frac 1 2 \frac {2 \paren {\sqrt {z - a^2} }^3} 3 + C
+      | c = [[Primitive of Root of a x + b|Primitive of $\sqrt {a x + b}$]]
+}}
+{{eqn | r = \frac {\paren {\sqrt {x^2 - a^2} }^3} 3 + C
+      | c = substituting for $z$ and simplifying
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of x squared by Root of x squared minus a squared}
+Tags: Primitives involving Root of x squared minus a squared
+
+\begin{theorem}
+:$\displaystyle \int x^2 \sqrt {x^2 - a^2} \rd x = \frac {x \paren {\sqrt {x^2 - a^2} }^3} 4 + \frac {a^2 x \sqrt {x^2 - a^2} } 8 - \frac {a^4} 8 \ln \size {x + \sqrt {x^2 - a^2} } + C$
+\end{theorem}
+
+\begin{proof}
+We have that $\sqrt {x^2 - a^2}$ is defined only when $x^2 \ge a^2$, that is, either:
+:$x \ge a$
+or:
+:$x \le -a$
+where it is assumed that $a > 0$.
+First let $x \ge a$.
+Let:
+{{begin-eqn}}
+{{eqn | l = z
+      | r = x^2
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d z} {\d x}
+      | r = 2 x
+      | c = [[Power Rule for Derivatives]]
+}}
+{{eqn | ll= \leadsto
+      | l = \int x^2 \sqrt {x^2 - a^2} \rd x
+      | r = \int \frac {z \sqrt {z - a^2} \rd z} {2 \sqrt z}
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \frac 1 2 \int \sqrt z \sqrt {z - a^2} \rd z
+      | c = [[Primitive of Constant Multiple of Function]]
+}}
+{{eqn | r = \frac 1 2 \paren {\frac {2 z - a^2} 4 \sqrt z \sqrt {z - a^2} - \frac {a^4} 8 \int \frac {\d z} {\sqrt z \sqrt {z - a^2} } } + C
+      | c = [[Primitive of Root of a x + b by Root of p x + q|Primitive of $\sqrt {\paren {a x + b} \paren {p x + q} }$]]
+}}
+{{eqn | r = \frac {2 z - a^2} 8 \sqrt z \sqrt {z - a^2} - \frac {a^4} {16} \paren {2 \map \ln {\sqrt {z - a^2} + \sqrt z} } + C
+      | c = [[Primitive of Reciprocal of Root of a x + b by Root of p x + q|Primitive of $\dfrac 1 {\sqrt {\paren {a x + b} \paren {p x + q} } }$]]
+}}
+{{eqn | r = \frac {2 z - 2 a^2} 8 \sqrt z \sqrt {z - a^2} + \frac {a^2} 8 \sqrt z \sqrt {z - a^2} - \frac {a^4} 8 \ln \paren {\sqrt {z - a^2} + \sqrt z} + C
+      | c = 
+}}
+{{eqn | r = \frac {z - a^2} 4 \sqrt z \sqrt {z - a^2} + \frac {a^2} 8 \sqrt z \sqrt {z - a^2} - \frac {a^4} 8 \ln \paren {\sqrt {z - a^2} + \sqrt z} + C
+      | c = 
+}}
+{{eqn | r = \frac {x \paren {\sqrt {x^2 - a^2} }^3} 4 + \frac {a^2 x \sqrt {x^2 - a^2} } 8 - \frac {a^4} 8 \ln \paren {x + \sqrt {x^2 - a^2} } + C
+      | c = substituting for $z$
+}}
+{{eqn | r = \frac {x \paren {\sqrt {x^2 - a^2} }^3} 4 + \frac {a^2 x \sqrt {x^2 - a^2} } 8 - \frac {a^4} 8 \ln \size {x + \sqrt {x^2 - a^2} } + C
+      | c = as $x + \sqrt {x^2 - a^2} > 0$
+}}
+{{end-eqn}}
+Now suppose $x \le -a$.
+Let $z = -x$.
+Then:
+:$\d x = -\d z$
+and we then have:
+{{begin-eqn}}
+{{eqn | l = \int x^2 \sqrt {x^2 - a^2} \rd x
+      | r = -\int \paren {-z}^2 \sqrt {\paren {-z}^2 - a^2} \rd z
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = -\int z^2 \sqrt {z^2 - a^2} \rd z
+      | c = simplifying
+}}
+{{eqn | r = -\paren {\frac {z \paren {\sqrt {z^2 - a^2} }^3} 4 + \frac {a^2 z \sqrt {z^2 - a^2} } 8 - \frac {a^4} 8 \map \ln {z + \sqrt {z^2 - a^2} } } + C
+      | c = from above
+}}
+{{eqn | r = -\frac {z \paren {\sqrt {z^2 - a^2} }^3} 4 - \frac {a^2 z \sqrt {z^2 - a^2} } 8 + \frac {a^4} 8 \map \ln {z + \sqrt {z^2 - a^2} } + C
+      | c = simplifying
+}}
+{{eqn | r = -\frac {z \paren {\sqrt {z^2 - a^2} }^3} 4 - \frac {a^2 z \sqrt {z^2 - a^2} } 8 - \frac {a^4} 8 \paren {\map \ln {z - \sqrt {z^2 - a^2} } - \map \ln {a^2} } + C
+      | c = [[Negative of Logarithm of x plus Root x squared minus a squared|Negative of $\map \ln {z + \sqrt {z^2 - a^2} }$]]
+}}
+{{eqn | r = -\frac {z \paren {\sqrt {z^2 - a^2} }^3} 4 - \frac {a^2 z \sqrt {z^2 - a^2} } 8 - \frac {a^4} 8 \paren {\map \ln {z - \sqrt {z^2 - a^2} } } + C
+      | c = subsuming $\dfrac {a^4 \map \ln {-a^2} } 8$ into [[Definition:Arbitrary Constant (Calculus)|constant]]
+}}
+{{eqn | r = -\frac {\paren {-x} \paren {\sqrt {\paren {-x}^2 - a^2} }^3} 4 - \frac {a^2 \paren {-x} \sqrt {\paren {-x}^2 - a^2} } 8 - \frac {a^4} 8 \paren {\map \ln {\paren {-x} - \sqrt {\paren {-x}^2 - a^2} } } + C
+      | c = substituting back for $x$
+}}
+{{eqn | r = \frac {x \paren {\sqrt {x^2 - a^2} }^3} 4 + \frac {a^2 x \sqrt {x^2 - a^2} } 8 - \frac {a^4} 8 \map \ln {-\paren {x + \sqrt {x^2 - a^2} } } + C
+      | c = simplifying
+}}
+{{eqn | r = \frac {x \paren {\sqrt {x^2 - a^2} }^3} 4 + \frac {a^2 x \sqrt {x^2 - a^2} } 8 - \frac {a^4} 8 \ln \size {x + \sqrt {x^2 - a^2} } + C
+      | c = as $-\paren {x + \sqrt {x^2 - a^2} } > 0$: {{Defof |Absolute Value}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of x cubed by Root of x squared minus a squared}
+Tags: Primitives involving Root of x squared minus a squared
+
+\begin{theorem}
+:$\displaystyle \int x^3 \sqrt {x^2 - a^2} \rd x = \frac {\paren {\sqrt {x^2 - a^2} }^5} 5 + \frac {a^2 \paren {\sqrt {x^2 - a^2} }^3} 3 + C$
+\end{theorem}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = z
+      | r = x^2
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d z} {\d x}
+      | r = 2 x
+      | c = [[Power Rule for Derivatives]]
+}}
+{{eqn | ll= \leadsto
+      | l = \int x^3 \sqrt {x^2 - a^2} \rd x
+      | r = \int \frac {z^{3/2} \sqrt {z - a^2} \rd z} {2 \sqrt z}
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \frac 1 2 \int z \sqrt {z - a^2} \rd z
+      | c = [[Primitive of Constant Multiple of Function]]
+}}
+{{eqn | r = \frac 1 2 \paren {\frac {2 \paren {3 z + 2 a^2} } {15} \paren {\sqrt {z - a^2} }^3 } + C
+      | c = [[Primitive of x by Root of a x + b|Primitive of $x \sqrt {\paren {a x + b} }$]]
+}}
+{{eqn | r = \frac {3 z + 2 a^2} {15} \paren {\sqrt {z - a^2} }^3 + C
+      | c = simplification
+}}
+{{eqn | r = \frac {3 x^2 + 2 a^2} {15} \paren {\sqrt {x^2 - a^2} }^3 + C
+      | c = substituting for $z$
+}}
+{{eqn | r = \frac {3 x^2 - 3 a^2 + 5 a^2} {15} \paren {\sqrt {x^2 - a^2} }^3 + C
+      | c = 
+}}
+{{eqn | r = \frac {3 \paren {x^2 - a^2} \paren {\sqrt {a^2 - x^2} }^3 + 5 a^2 \paren {\sqrt {a^2 - x^2} }^3} {15} + C
+      | c = 
+}}
+{{eqn | r = \frac {\paren {\sqrt {x^2 - a^2} }^5} 5 + \frac {a^2 \paren {\sqrt {x^2 - a^2} }^3} 3 + C
+      | c = simplification
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Root of x squared minus a squared over x}
+Tags: Primitives involving Root of x squared minus a squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {\sqrt {x^2 - a^2} } x \rd x = \sqrt {x^2 - a^2} - \frac 1 {2 a} \arcsec \size {\frac x a} + C$
+\end{theorem}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = z
+      | r = x^2
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d z} {\d x}
+      | r = 2 x
+      | c = [[Power Rule for Derivatives]]
+}}
+{{eqn | ll= \leadsto
+      | l = \int \frac {\sqrt {x^2 - a^2} } x \rd x
+      | r = \int \frac {\sqrt {z - a^2} \rd z} {2 \sqrt z \sqrt z}
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \frac 1 2 \int \frac {\sqrt {z - a^2} \rd z} z
+      | c = [[Primitive of Constant Multiple of Function]]
+}}
+{{eqn | r = \frac 1 2 \paren {2 \sqrt {z - a^2} - a^2 \int \frac {\d z} {z \sqrt {z - a^2} } } + C
+      | c = [[Primitive of Root of a x + b over x|Primitive of $\dfrac {\sqrt {a x + b} } x$]]
+}}
+{{eqn | r = \sqrt {x^2 - a^2} - \frac {a^2} 2 \int \frac {2 x \rd x} {x^2 \sqrt {x^2 - a^2} } + C
+      | c = substituting for $z$
+}}
+{{eqn | r = \sqrt {x^2 - a^2} - a^2 \int \frac {\d x} {x \sqrt {x^2 - a^2} } + C
+      | c = simplifying
+}}
+{{eqn | r = \sqrt {x^2 - a^2} - a^2 \paren {\frac 1 a \arcsec \size {\frac x a} } + C
+      | c = [[Primitive of Reciprocal of x by Root of x squared minus a squared|Primitive of $\dfrac 1 {x \sqrt {x^2 - a^2} }$]]
+}}
+{{eqn | r = \sqrt {x^2 - a^2} - \frac 1 {2 a} \arcsec \size {\frac x a} + C
+      | c = simplification
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Root of x squared minus a squared over x squared}
+Tags: Primitives involving Root of x squared minus a squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {\sqrt {x^2 - a^2} } {x^2} \rd x = \frac {-\sqrt {x^2 - a^2} } x + \ln \size {x + \sqrt {x^2 - a^2} } + C$
+\end{theorem}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = z
+      | r = x^2
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d z} {\d x}
+      | r = 2 x
+      | c = [[Power Rule for Derivatives]]
+}}
+{{eqn | ll= \leadsto
+      | l = \int \frac {\sqrt {x^2 - a^2} } {x^2} \rd x
+      | r = \int \frac {\sqrt {z - a^2} \rd z} {2 z \sqrt z}
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \frac 1 2 \int \frac {\sqrt {z - a^2} \rd z} {z^{3/2} }
+      | c = [[Primitive of Constant Multiple of Function]]
+}}
+{{eqn | r = \frac 1 2 \paren {\frac {-\sqrt {z - a^2} } {\frac 1 2 \sqrt z} + \frac 1 2 \int \frac {\d z} {\sqrt z \sqrt {z - a^2} } } + C
+      | c = [[Primitive of Root of a x + b over Power of x/Formulation 1|Primitive of $\dfrac {\sqrt {a x + b} } {x^m}$]]
+}}
+{{eqn | r = \frac {-\sqrt {x^2 - a^2} } x + \frac 1 2 \int \frac {2 x \rd x} {x \sqrt {x^2 - a^2} } + C
+      | c = substituting for $z$
+}}
+{{eqn | r = \frac {-\sqrt {x^2 - a^2} } x + \int \frac {\d x} {\sqrt {x^2 - a^2} } + C
+      | c = simplifying
+}}
+{{eqn | r = \frac {-\sqrt {x^2 - a^2} } x + \ln \size {x + \sqrt {x^2 - a^2} } + C
+      | c = [[Primitive of Reciprocal of Root of x squared minus a squared/Logarithm Form|Primitive of $\dfrac 1 {\sqrt {x^2 - a^2} }$]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Root of x squared minus a squared over x cubed}
+Tags: Primitives involving Root of x squared minus a squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {\sqrt {x^2 - a^2} } {x^3} \rd x = \frac {-\sqrt {x^2 - a^2} } {2 x^2} + \frac 1 {2 a} \arcsec \size {\frac x a} + C$
+\end{theorem}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = z
+      | r = x^2
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d z} {\d x}
+      | r = 2 x
+      | c = [[Power Rule for Derivatives]]
+}}
+{{eqn | ll= \leadsto
+      | l = \int \frac {\sqrt {x^2 - a^2} } {x^3} \rd x
+      | r = \int \frac {\sqrt {z - a^2} \rd z} {2 z^{3/2} \sqrt z}
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \frac 1 2 \int \frac {\sqrt {z - a^2} \rd z} {z^2}
+      | c = [[Primitive of Constant Multiple of Function]]
+}}
+{{eqn | r = \frac 1 2 \paren {\frac {-\sqrt {z - a^2} } z + \frac 1 2 \int \frac {\d z} {z \sqrt {z - a^2} } } + C
+      | c = [[Primitive of Root of a x + b over Power of x/Formulation 1|Primitive of $\dfrac {\sqrt {a x + b} } {x^m}$]]
+}}
+{{eqn | r = \frac {-\sqrt {x^2 - a^2} } {2 x^2} + \frac 1 4 \int \frac {2 x \rd x} {x^2 \sqrt {x^2 - a^2} } + C
+      | c = substituting for $z$
+}}
+{{eqn | r = \frac {-\sqrt {x^2 - a^2} } {2 x^2} + \frac 1 2 \int \frac {\d x} {x \sqrt {x^2 - a^2} } + C
+      | c = simplifying
+}}
+{{eqn | r = \frac {-\sqrt {x^2 - a^2} } {2 x^2} + \frac 1 2 \paren {\frac 1 a \arcsec \size {\frac x a} } + C
+      | c = [[Primitive of Reciprocal of x by Root of x squared minus a squared|Primitive of $\dfrac 1 {x \sqrt {x^2 - a^2} }$]]
+}}
+{{eqn | r = \frac {-\sqrt {x^2 - a^2} } {2 x^2} + \frac 1 {2 a} \arcsec \size {\frac x a} + C
+      | c = simplifying
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Reciprocal of Root of x squared minus a squared cubed}
+Tags: Primitives involving Root of x squared minus a squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {\d x} {\paren {\sqrt {x^2 - a^2} }^3} = \frac {-x} {a^2 \sqrt {x^2 - a^2} } + C$
+\end{theorem}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = z
+      | r = x^2
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d z} {\d x}
+      | r = 2 x
+      | c = [[Power Rule for Derivatives]]
+}}
+{{eqn | ll= \leadsto
+      | l = \int \frac {\d x} {\paren {\sqrt {x^2 - a^2} }^3}
+      | r = \int \frac {\d z} {2 \sqrt z \paren {\sqrt {z - a^2} }^3}
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \frac 1 2 \int \frac {\mathrm d z} {\paren {z - a^2} \sqrt z \sqrt {z - a^2} }
+      | c = [[Primitive of Constant Multiple of Function]]
+}}
+{{eqn | r = \frac 1 2 \paren {\frac {2 \sqrt z} {\paren {-a^2} \sqrt {z - a^2} } } + C
+      | c = [[Primitive of Reciprocal of p x + q by Root of a x + b by Root of p x + q|Primitive of $\dfrac 1 {\paren {p x + q} \sqrt {\paren {a x + b} \paren {p x + q} } }$]]
+}}
+{{eqn | r = \frac 1 2 \paren {\frac {2 x} {\paren {-a^2} \sqrt {x^2 - a^2} } } + C
+      | c = substituting for $z$
+}}
+{{eqn | r = \frac {-x} {a^2 \sqrt {x^2 - a^2} } + C
+      | c = simplifying
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of x over Root of x squared minus a squared cubed}
+Tags: Primitives involving Root of x squared minus a squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {x \rd x} {\paren {\sqrt {x^2 - a^2} }^3} = \frac {-1} {\sqrt {x^2 - a^2} } + C$
+\end{theorem}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = z
+      | r = x^2 - a^2
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d z} {\d x}
+      | r = 2 x
+      | c = [[Power Rule for Derivatives]]
+}}
+{{eqn | ll= \leadsto
+      | l = \int \frac {x \rd x} {\paren {\sqrt {x^2 - a^2} }^3}
+      | r = \int \frac {x \rd z} {2 x z^{3/2} }
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \frac 1 2 \int z^{-3/2} \rd z
+      | c = [[Primitive of Constant Multiple of Function]]
+}}
+{{eqn | r = \frac 1 2 \paren {\frac {-z^{1/2} } {\frac 1 2} } + C
+      | c = [[Primitive of Power]]
+}}
+{{eqn | r = \frac {-1} {\sqrt z} + C
+      | c = simplifying
+}}
+{{eqn | r = \frac {-1} {\sqrt {x^2 - a^2} } + C
+      | c = substituting for $z$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of x squared over Root of x squared minus a squared cubed}
+Tags: Primitives involving Root of x squared minus a squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {x^2 \rd x} {\paren {\sqrt {x^2 - a^2} }^3} = \frac {-x} {\sqrt {x^2 - a^2} } + \ln \size {x + \sqrt {x^2 - a^2} } + C$
+\end{theorem}
+
+\begin{proof}
+With a view to expressing the problem in the form:
+:$\displaystyle \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$
+let:
+{{begin-eqn}}
+{{eqn | l = u
+      | r = x
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d u} {\d x}
+      | r = 1
+      | c = [[Power Rule for Derivatives]]
+}}
+{{end-eqn}}
+and let:
+{{begin-eqn}}
+{{eqn | l = \frac {\d v} {\d x}
+      | r = \frac x {\paren {\sqrt {x^2 - a^2} }^3}
+}}
+{{eqn | ll= \leadsto
+      | l = v
+      | r = \frac {-1} {\sqrt {x^2 - a^2} }
+      | c = [[Primitive of x over Root of x squared minus a squared cubed|Primitive of $\dfrac x {\paren {\sqrt {x^2 - a^2} }^3}$]]
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = \int \frac {x^2 \rd x} {\paren {\sqrt {x^2 - a^2} }^3}
+      | r = \int x \frac {x \rd x} {\paren {\sqrt {x^2 - a^2} }^3}
+      | c = 
+}}
+{{eqn | r = x \frac {-1} {\sqrt {x^2 - a^2} } - \int \frac {-1} {\sqrt {x^2 - a^2} } 1 + C
+      | c = [[Integration by Parts]]
+}}
+{{eqn | r = \frac {-x} {\sqrt {x^2 - a^2} } + \int \frac 1 {\sqrt {x^2 - a^2} } + C
+      | c = simplifying
+}}
+{{eqn | r = \frac {-x} {\sqrt {x^2 - a^2} } + \ln \size {x + \sqrt {x^2 - a^2} } + C
+      | c = [[Primitive of Reciprocal of Root of x squared minus a squared/Logarithm Form|Primitive of $\dfrac 1 {\sqrt {x^2 - a^2} }$]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of x cubed over Root of x squared minus a squared cubed}
+Tags: Primitives involving Root of x squared minus a squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {x^3 \rd x} {\paren {\sqrt {x^2 - a^2} }^3} = \sqrt {x^2 - a^2} - \frac {a^2} {\sqrt {x^2 - a^2} } + C$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int \frac {x^3 \rd x} {\paren {\sqrt {x^2 - a^2} }^3}
+      | r = \int \frac {x \paren {x^2} \rd x} {\paren {\sqrt {x^2 - a^2} }^3}
+      | c = 
+}}
+{{eqn | r = \int \frac {x \paren {x^2 - a^2 + a^2} \rd x} {\paren {\sqrt {x^2 - a^2} }^3}
+      | c = 
+}}
+{{eqn | r = \int \frac {x \paren {x^2 - a^2} \rd x} {\paren {\sqrt {x^2 - a^2} }^3} + a^2 \int \frac {x \rd x} {\paren {\sqrt {x^2 - a^2} }^3}
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \int \frac {x \rd x} {\sqrt {x^2 - a^2} } + a^2 \int \frac {x \rd x} {\paren {\sqrt {x^2 - a^2} }^3}
+      | c = simplifying
+}}
+{{eqn | r = \sqrt {x^2 - a^2} + a^2 \int \frac {x \rd x} {\paren {\sqrt {x^2 - a^2} }^3} + C
+      | c = [[Primitive of x over Root of x squared minus a squared|Primitive of $\dfrac x {\sqrt {x^2 - a^2} }$]]
+}}
+{{eqn | r = \sqrt {x^2 - a^2} + a^2 \frac {-1} {\sqrt {x^2 - a^2} } + C
+      | c = [[Primitive of x over Root of x squared minus a squared cubed|Primitive of $\dfrac x {\paren {\sqrt {x^2 - a^2} }^3}$]]
+}}
+{{eqn | r = \sqrt {x^2 - a^2} - \frac {a^2} {\sqrt {x^2 - a^2} } + C
+      | c = simplifying
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Reciprocal of x by Root of x squared minus a squared cubed}
+Tags: Primitives involving Root of x squared minus a squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {\d x} {x \paren {\sqrt {x^2 - a^2} }^3} = \frac {-1} {a^2 \sqrt {x^2 - a^2} } - \frac 1 {a^3} \arcsec \size {\frac x a} + C$
+\end{theorem}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = z
+      | r = x^2
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d z} {\d x}
+      | r = 2 x
+      | c = [[Power Rule for Derivatives]]
+}}
+{{eqn | ll= \leadsto
+      | l = \int \frac {\d x} {x \paren {\sqrt {x^2 - a^2} }^3}
+      | r = \int \frac {\d z} {2 \sqrt z \sqrt z \paren {\sqrt {z - a^2} }^3}
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \frac 1 2 \int \frac {\d z} {z \paren {\sqrt {z - a^2} }^3}
+      | c = [[Primitive of Constant Multiple of Function]]
+}}
+{{eqn | r = \frac 1 2 \paren {\frac 2 {\paren {-a^2} \sqrt {z - a^2} } + \frac 1 {-a^2} \int \frac {\d z} {z \sqrt {z - a^2} } } + C
+      | c = [[Primitive of Reciprocal of x by Power of Root of a x + b|Primitive of $\dfrac 1 {x \paren {\sqrt{a x + b} }^m}$]]
+}}
+{{eqn | r = \frac {-1} {a^2 \sqrt {z - a^2} } - \frac 1 {2 a^2} \int \frac {\d z} {z \sqrt {z - a^2} } + C
+      | c = simplifying
+}}
+{{eqn | r = \frac {-1} {a^2 \sqrt {x^2 - a^2} } - \frac 1 {2 a^2} \int \frac {2 x \rd x} {x^2 \sqrt {x^2 - a^2} } + C
+      | c = substituting for $z$
+}}
+{{eqn | r = \frac {-1} {a^2 \sqrt {x^2 - a^2} } - \frac 1 {a^2} \int \frac {\d x} {x \sqrt {x^2 - a^2} } + C
+      | c = simplifying
+}}
+{{eqn | r = \frac {-1} {a^2 \sqrt {x^2 + a^2} } - \frac 1 {a^3} \arcsec \size {\frac x a} + C
+      | c = [[Primitive of Reciprocal of x by Root of x squared minus a squared|Primitive of $\dfrac 1 {x \sqrt {x^2 - a^2} }$]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Reciprocal of x squared by Root of x squared minus a squared cubed}
+Tags: Primitives involving Root of x squared minus a squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {\d x} {x^2 \paren {\sqrt {x^2 - a^2} }^3} = \frac {-\sqrt {x^2 - a^2} } {a^4 x} - \frac x {a^4 \sqrt {x^2 - a^2} } + C$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int \frac {\d x} {x^2 \paren {\sqrt {x^2 - a^2} }^3}
+      | r = \int \frac {a^2 \rd x} {a^2 x^2 \paren {\sqrt {x^2 - a^2} }^3}
+      | c = 
+}}
+{{eqn | r = \int \frac {\paren {x^2 + a^2 - x^2} \rd x} {a^2 x^2 \paren {\sqrt {x^2 - a^2} }^3}
+      | c = 
+}}
+{{eqn | r = \frac 1 {a^2} \int \frac {-\paren {x^2 - a^2} \rd x} {a^2 x^2 \paren {\sqrt {x^2 - a^2} }^3} + \frac 1 {a^2} \int \frac {x^2 \rd x} {x^2 \paren {\sqrt {x^2 - a^2} }^3}
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \frac {-1} {a^2} \int \frac {\d x} {x^2 \sqrt {x^2 - a^2} } + \frac 1 {a^2} \int \frac {\d x} {\paren {\sqrt {x^2 - a^2} }^3}
+      | c = simplifying
+}}
+{{eqn | r = \frac {-1} {a^2} \frac {\sqrt {x^2 - a^2} } {a^2 x} + \frac 1 {a^2} \int \frac {\d x} {\paren {\sqrt {x^2 - a^2} }^3} + C
+      | c = [[Primitive of Reciprocal of x squared by Root of x squared minus a squared|Primitive of $\dfrac 1 {x^2 \sqrt {x^2 - a^2} }$]]
+}}
+{{eqn | r = \frac {-\sqrt {x^2 - a^2} } {a^4 x} + \frac 1 {a^2} \frac {-x} {a^2 \sqrt {x^2 - a^2} } + C
+      | c = [[Primitive of Reciprocal of Root of x squared minus a squared cubed|Primitive of $\dfrac 1 {\paren {\sqrt {x^2 - a^2} }^3}$]]
+}}
+{{eqn | r = \frac {-\sqrt {x^2 - a^2} } {a^4 x} - \frac x {a^4 \sqrt {x^2 - a^2} } + C
+      | c = simplification
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Reciprocal of x cubed by Root of x squared minus a squared cubed}
+Tags: Primitives involving Root of x squared minus a squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {\d x} {x^3 \paren {\sqrt {x^2 - a^2} }^3} = \frac 1 {2 a^2 x^2 \sqrt {x^2 - a^2} } - \frac 3 {2 a^4 \sqrt {x^2 - a^2} } + \frac 3 {2 a^5} \arcsec \size {\frac x a} + C$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int \frac {\d x} {x^3 \paren {\sqrt {x^2 - a^2} }^3}
+      | r = \int \frac {a^2 \rd x} {a^2 x^3 \paren {\sqrt {x^2 - a^2} }^3}
+      | c = 
+}}
+{{eqn | r = \int \frac {\paren {x^2 + a^2 - x^2} \rd x} {a^2 x^3 \paren {\sqrt {x^2 - a^2} }^3}
+      | c = 
+}}
+{{eqn | r = \frac 1 {a^2} \int \frac {-\paren {x^2 - a^2} \rd x} {x^3 \paren {\sqrt {x^2 - a^2} }^3} + \frac 1 {a^2} \int \frac {x^2 \rd x} {x^3 \paren {\sqrt {x^2 - a^2} }^3}
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \frac {-1} {a^2} \int \frac {\d x} {x^3 \sqrt {x^2 - a^2} } + \frac 1 {a^2} \int \frac {\d x} {x \paren {\sqrt {x^2 - a^2} }^3}
+      | c = simplifying
+}}
+{{eqn | r = \frac {-1} {a^2} \paren {\frac {\sqrt {x^2 - a^2} } {2 a^2 x^2} + \frac 1 {2 a^3} \arcsec \size {\frac x a} } + \frac 1 {a^2} \int \frac {\d x} {x \paren {\sqrt {x^2 - a^2} }^3} + C
+      | c = [[Primitive of Reciprocal of x cubed by Root of x squared minus a squared|Primitive of $\dfrac 1 {x^3 \sqrt {x^2 - a^2} }$]]
+}}
+{{eqn | r = \frac {-1} {a^2} \paren {\frac {\sqrt {x^2 - a^2} } {2 a^2 x^2} + \frac 1 {2 a^3} \arcsec \size {\frac x a} }
+      | c = 
+}}
+{{eqn | o = 
+      | ro= +
+      | r = \frac 1 {a^2} \paren {\frac {-1} {a^2 \sqrt {x^2 - a^2} } - \frac 1 {a^3} \arcsec \size {\frac x a} } + C
+      | c = [[Primitive of Reciprocal of x by Root of x squared minus a squared cubed|Primitive of $\dfrac 1 {x \paren {\sqrt {x^2 - a^2} }^3}$]]
+}}
+{{eqn | r = \frac 1 {2 a^2 x^2 \sqrt {x^2 - a^2} } - \frac 3 {2 a^4 \sqrt {x^2 - a^2} } + \frac 3 {2 a^5} \arcsec \size {\frac x a} + C
+      | c = simplification
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Root of x squared minus a squared cubed}
+Tags: Primitives involving Root of x squared minus a squared
+
+\begin{theorem}
+:$\displaystyle \int \paren {\sqrt {x^2 - a^2} }^3 \rd x = \frac {x \paren {\sqrt {x^2 - a^2} }^3} 4 - \frac {3 a^2 x \sqrt {x^2 - a^2} } 8 + \frac {3 a^4} 8 \ln \size {x + \sqrt {x^2 - a^2} } + C$
+\end{theorem}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = z
+      | r = x^2
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d z} {\d x}
+      | r = 2 x
+      | c = [[Power Rule for Derivatives]]
+}}
+{{eqn | ll= \leadsto
+      | l = \int \paren {\sqrt {x^2 - a^2} }^3 \rd x
+      | r = \int \frac {\paren {\sqrt {z - a^2} }^3} {2 \sqrt z} \rd x
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \frac {\sqrt z \paren {\sqrt {z - a^2} }^3} 4 - \frac {3 a^2} 8 \int \frac {\sqrt {z - a^2} } {\sqrt z} \rd x + C
+      | c = [[Primitive of Power of p x + q over Root of a x + b|Primitive of $\dfrac {\paren {p x + q}^n} {\sqrt{a x + b} }$]]
+}}
+{{eqn | r = \frac {\sqrt z \paren {\sqrt {z - a^2} }^3} 4 - \frac {3 a^2} 8 \paren {\sqrt z \sqrt {z - a^2} - \frac {a^2} 2 \int \frac {\d x} {\sqrt z \sqrt {z - a^2} } } + C
+      | c = [[Primitive of Root of p x + q over Root of a x + b|Primitive of $\dfrac {\sqrt {p x + q} } {\sqrt{a x + b} }$]]
+}}
+{{eqn | r = \frac {x \paren {\sqrt {x^2 - a^2} }^3} 4 - \frac {3 a^2} 8 x \sqrt {x^2 + a^2} + \frac {3 a^4} 8 \int \frac {\d x} {\sqrt {x^2 - a^2} } + C
+      | c = substituting for $z$ and simplifying
+}}
+{{eqn | r = \frac {x \paren {\sqrt {x^2 - a^2} }^3} 4 - \frac {3 a^2 x \sqrt {x^2 - a^2} } 8 + \frac {3 a^4} 8 \ln \size {x + \sqrt {x^2 - a^2} } + C
+      | c = [[Primitive of Reciprocal of Root of x squared minus a squared/Logarithm Form|Primitive of $\dfrac 1 {\sqrt {x^2 - a^2} }$]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of x by Root of x squared minus a squared cubed}
+Tags: Primitives involving Root of x squared minus a squared
+
+\begin{theorem}
+:$\displaystyle \int x \paren {\sqrt {x^2 - a^2} }^3 \rd x = \frac {\paren {\sqrt {x^2 - a^2} }^5} 5 + C$
+\end{theorem}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = z
+      | r = x^2 - a^2
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d z} {\d x}
+      | r = 2 x
+      | c = [[Power Rule for Derivatives]]
+}}
+{{eqn | ll= \leadsto
+      | l = \int x \paren {\sqrt {x^2 - a^2} }^3 \rd x
+      | r = \int \frac {z^{3/2} } 2 \rd z
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \frac 1 2 \frac {z^{5/2} } {\frac 5 2} + C
+      | c = [[Primitive of Power]]
+}}
+{{eqn | r = \frac {z^{5/2} } 5 + C
+      | c = simplifying
+}}
+{{eqn | r = \frac {\paren {\sqrt {x^2 - a^2} }^5} 5 + C
+      | c = substituting for $z$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of x squared by Root of x squared minus a squared cubed}
+Tags: Primitives involving Root of x squared minus a squared
+
+\begin{theorem}
+:$\displaystyle \int x^2 \paren {\sqrt {x^2 - a^2} }^3 \rd x = \frac {x \paren {\sqrt {x^2 - a^2} }^5} 6 + \frac {a^2 x \paren {\sqrt {x^2 - a^2} }^3} {24} - \frac {a^4 x \sqrt {x^2 - a^2} } {16} + \frac {a^6} {16} \ln \size {x + \sqrt {x^2 - a^2} } + C$
+\end{theorem}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = z
+      | r = x^2
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d z} {\d x}
+      | r = 2 x
+      | c = [[Power Rule for Derivatives]]
+}}
+{{eqn | ll= \leadsto
+      | l = \int x^2 \paren {\sqrt {x^2 - a^2} }^3 \rd x
+      | r = \int \frac {\sqrt z \paren {z - a^2}^{3/2} } 2 \rd z
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \frac {\sqrt z \paren {z - a^2}^{5/2} } 6 + \frac {a^2} {12} \int \frac {\paren {z - a^2}^{3/2} } {\sqrt z} \rd z + C
+      | c = [[Primitive of Power of p x + q by Root of a x + b|Primitive of $\paren {p x + q}^n \sqrt{a x + b}$]]
+}}
+{{eqn | r = \frac {x \paren {x^2 - a^2}^{5/2} } 6 + \frac {a^2} 6 \int \paren {x^2 - a^2}^{3/2} \rd x + C
+      | c = substituting for $z$ and simplifying
+}}
+{{eqn | r = \frac {x \paren {x^2 - a^2}^{5/2} } 6
+}}
+{{eqn | o =
+      | ro= +
+      | r = \frac {a^2} 6 \paren {\frac {x \paren {\sqrt {x^2 - a^2} }^3} 4 - \frac {3 a^2 x \sqrt {x^2 - a^2} } 8 + \frac {3 a^4} 8 \ln \size {x + \sqrt {x^2 - a^2} } } + C
+      | c = [[Primitive of Root of x squared minus a squared cubed|Primitive of $\paren {\sqrt {x^2 - a^2} }^3$]]
+}}
+{{eqn | r = \frac {x \paren {x^2 - a^2}^{5/2} } 6 + \frac {a^2 x \paren {\sqrt {x^2 - a^2} }^3} {24}
+}}
+{{eqn | o =
+      | ro= -
+      | r = \frac {a^4 x \sqrt {x^2 - a^2} } {16} + \frac {a^6} {16} \ln \size {x + \sqrt {x^2 - a^2} } + C
+      | c = simplifying
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of x cubed by Root of x squared minus a squared cubed}
+Tags: Primitives involving Root of x squared minus a squared
+
+\begin{theorem}
+:$\displaystyle \int x^3 \paren {\sqrt {x^2 - a^2} }^3 \rd x = \frac {\paren {\sqrt {x^2 - a^2} }^7} 7 + \frac {a^2 \paren {\sqrt {x^2 - a^2} }^5} 5 + C$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int x^3 \paren {\sqrt {x^2 - a^2} }^3 \rd x
+      | r = \int x \paren {x^2} \paren {\sqrt {x^2 - a^2} }^3 \rd x
+      | c = 
+}}
+{{eqn | r = \int x \paren {x^2 - a^2 + a^2} \paren {\sqrt {x^2 - a^2} }^3 \rd x
+      | c = [[Primitive of Power]]
+}}
+{{eqn | r = \int x \paren {x^2 - a^2} \paren {\sqrt {x^2 - a^2} }^3 \rd x + a^2 \int x \paren {\sqrt {x^2 - a^2} }^3 \rd x
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \int x \paren {\sqrt {x^2 - a^2} }^5 \rd x + a^2 \int x \paren {\sqrt {x^2 - a^2} }^3 \rd x
+      | c = simplifying
+}}
+{{end-eqn}}
+Let:
+{{begin-eqn}}
+{{eqn | l = z
+      | r = x^2 - a^2
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d z} {\d x}
+      | r = 2 x
+      | c = [[Power Rule for Derivatives]]
+}}
+{{eqn | ll= \leadsto
+      | o = 
+      | r = \int x \paren {\sqrt {x^2 - a^2} }^5 \rd x + a^2 \int x \paren {\sqrt {x^2 - a^2} }^3 \rd x
+      | c = 
+}}
+{{eqn | r = \int \frac {\sqrt z z^{5/2} } {2 \sqrt z} \rd z + a^2 \int \frac {\sqrt z z^{3/2} } {2 \sqrt z} \rd z
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r =  \frac 1 2 \int z^{5/2} \rd z + \frac {a^2} 2 \int z^{3/2} \rd z
+      | c = simplifying
+}}
+{{eqn | r = \frac 1 2 \frac {z^{7/2} } {7/2} + \frac {a^2} 2 \frac {z^{5/2} } {5/2} + C
+      | c = [[Primitive of Power]]
+}}
+{{eqn | r = \frac {z^{7/2} } 7 + a^2 \frac {z^{5/2} } 5 + C
+      | c = simplifying
+}}
+{{eqn | r = \frac {\paren {\sqrt {x^2 - a^2} }^7} 7 + \frac {a^2 \paren {\sqrt {x^2 - a^2} }^5} 5 + C
+      | c = substituting for $z$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Root of x squared minus a squared cubed over x}
+Tags: Primitives involving Root of x squared minus a squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {\paren {\sqrt {x^2 - a^2} }^3} x \rd x = \frac {\paren {\sqrt {x^2 - a^2} }^3} 3 - a^2 \sqrt {x^2 - a^2} + a^3 \arcsec \size {\frac x a} + C$
+\end{theorem}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = z
+      | r = x^2
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d z} {\d x}
+      | r = 2 x
+      | c = [[Power Rule for Derivatives]]
+}}
+{{eqn | ll= \leadsto
+      | l = \int \frac {\paren {\sqrt {x^2 - a^2} }^3} x \rd x
+      | r = \int \frac {\paren {\sqrt {z - a^2} }^3} {2 z} \rd z
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \frac {\paren {\sqrt {z - a^2} }^3} 3 - \frac {a^2} 2 \int \frac {\sqrt {z - a^2} } z \rd z
+      | c = [[Primitive of Power of Root of a x + b over x|Primitive of $\dfrac {\paren {\sqrt {a x + b} }^m} x$]]
+}}
+{{eqn | r = \frac {\paren {\sqrt {z - a^2} }^3} 3 - \frac {a^2} 2 \paren {2 \sqrt {z - a^2} - a^2 \int \frac {\d z} {z \sqrt {z - a^2} } }
+      | c = [[Primitive of Root of a x + b over x|Primitive of $\dfrac {\sqrt {a x + b} } x$]]
+}}
+{{eqn | r = \frac {\paren {\sqrt {x^2 - a^2} }^3} 3 - a^2 \sqrt {x^2 - a^2} + a^4 \int \frac {\d x} {x \sqrt {x^2 - a^2} }
+      | c = substituting for $z$ and simplifying
+}}
+{{eqn | r = \frac {\paren {\sqrt {x^2 - a^2} }^3} 3 - a^2 \sqrt {x^2 - a^2} + a^3 \arcsec \size {\frac x a} + C
+      | c = [[Primitive of Reciprocal of x by Root of x squared minus a squared|Primitive of $\dfrac 1 {x \sqrt {x^2 - a^2} }$]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Root of x squared minus a squared cubed over x squared}
+Tags: Primitives involving Root of x squared minus a squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {\paren {\sqrt {x^2 - a^2} }^3} {x^2} \rd x = \frac {-\paren {\sqrt {x^2 - a^2} }^3} x + \frac{3 x \sqrt {x^2 - a^2} } 2 - \frac {3 a^2} 2 \ln \size {x + \sqrt {x^2 - a^2} } + C$
+\end{theorem}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = z
+      | r = x^2
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d z} {\d x}
+      | r = 2 x
+      | c = [[Power Rule for Derivatives]]
+}}
+{{eqn | ll= \leadsto
+      | l = \int \frac {\paren {\sqrt {x^2 - a^2} }^3} {x^2} \rd x
+      | r = \int \frac {\paren {\sqrt {z - a^2} }^3} {2 z \sqrt z} \rd z
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \frac {-\paren {\sqrt {z - a^2} }^3} {\sqrt z} + \frac 3 2 \int \frac {\sqrt {z - a^2} } {\sqrt z} \rd z
+      | c = [[Primitive of Power of a x + b over Power of p x + q/Formulation 3|Primitive of $\dfrac {\paren {a x + b}^m} {\paren {p x + q}^n}$: Formulation 3]]
+}}
+{{eqn | r = \frac {-\paren {\sqrt {x^2 - a^2} }^3} x + 3 \int \sqrt {x^2 - a^2} \rd x
+      | c = substituting for $z$ and simplifying
+}}
+{{eqn | r = \frac {-\paren {\sqrt {x^2 - a^2} }^3} x - \frac {3 x \sqrt {x^2 - a^2} } 2 - \frac {3 a^2} 2 \ln \size {x + \sqrt {x^2 - a^2} } + C
+      | c = [[Primitive of Root of x squared minus a squared|Primitive of $\sqrt {x^2 - a^2}$]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Root of x squared minus a squared cubed over x cubed}
+Tags: Primitives involving Root of x squared minus a squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {\paren {\sqrt {x^2 - a^2} }^3} {x^3} \rd x = \frac {-\paren {\sqrt {x^2 - a^2} }^3} {2 x^2} + \frac {3 \sqrt {x^2 - a^2} } 2 - \frac {3 a} 2 \arcsec \size {\frac x a} + C$
+\end{theorem}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = z
+      | r = x^2
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d z} {\d x}
+      | r = 2 x
+      | c = [[Power Rule for Derivatives]]
+}}
+{{eqn | ll= \leadsto
+      | l = \int \frac {\paren {\sqrt {x^2 - a^2} }^3} {x^3} \rd x
+      | r = \int \frac {\paren {\sqrt {z - a^2} }^3} {2 z^2} \rd z
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \frac 1 2 \paren {\frac {-\paren {\sqrt {z - a^2} }^3} z + \frac 3 2 \int \frac {\sqrt {z - a^2} } z \rd z}
+      | c = [[Primitive of Power of a x + b over Power of p x + q/Formulation 3|Primitive of $\dfrac {\paren {a x + b}^m} {\paren {p x + q}^n}$: Formulation 3]]
+}}
+{{eqn | r = \frac {-\paren {\sqrt {x^2 - a^2} }^3} {2 x^2} + \frac 3 2 \int \frac {\sqrt {x^2 - a^2} } x \rd x
+      | c = substituting for $z$ and simplifying
+}}
+{{eqn | r = \frac {-\paren {\sqrt {x^2 - a^2} }^3} {2 x^2} + \frac 3 2 \paren {\sqrt {x^2 - a^2} - a \arcsec \size {\frac x a} } + C
+      | c = [[Primitive of Root of x squared minus a squared over x|Primitive of $\dfrac {\sqrt {x^2 - a^2} } x$]]
+}}
+{{eqn | r = \frac {-\paren {\sqrt {x^2 - a^2} }^3} {2 x^2} + \frac {3 \sqrt {x^2 - a^2} } 2 - \frac {3 a} 2 \arcsec \size {\frac x a} + C
+      | c = simplifying
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of x over Root of a squared minus x squared}
+Tags: Primitives involving Root of a squared minus x squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {x \rd x} {\sqrt {a^2 - x^2} } = -\sqrt {a^2 - x^2} + C$
+\end{theorem}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = z^2
+      | r = a^2 - x^2
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = 2 z \frac {\d z} {\d x}
+      | r = -2 x
+      | c = [[Chain Rule for Derivatives]], [[Power Rule for Derivatives]]
+}}
+{{eqn | ll= \leadsto
+      | l = \int \frac {x \rd x} {\sqrt {a^2 - x^2} }
+      | r = \int -\frac {z \rd z} z
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = -\int \rd z
+      | c = 
+}}
+{{eqn | r = -z + C
+      | c = [[Primitive of Constant]]
+}}
+{{eqn | r = -\sqrt {a^2 - x^2} + C
+      | c = substituting for $z$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of x cubed over Root of a squared minus x squared}
+Tags: Primitives involving Root of a squared minus x squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {x^3 \rd x} {\sqrt {a^2 - x^2} } = \frac {\left({\sqrt {a^2 - x^2} }\right)^3} 3 - a^2 \sqrt {a^2 - x^2} + C$
+\end{theorem}
+
+\begin{proof}
+With a view to expressing the problem in the form:
+:$\displaystyle \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$
+let:
+{{begin-eqn}}
+{{eqn | l = u
+      | r = x^2
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \frac {\d u} {\d x}
+      | r = 2 x
+      | c = [[Power Rule for Derivatives]]
+}}
+{{end-eqn}}
+and let:
+{{begin-eqn}}
+{{eqn | l = \frac {\d v} {\d x}
+      | r = \frac x {\sqrt {a^2 - x^2} }
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = v
+      | r = -\sqrt {a^2 - x^2}
+      | c = [[Primitive of x over Root of a squared minus x squared|Primitive of $\dfrac x {\sqrt {a^2 - x^2} }$]]
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = \int \frac {x^3 \rd x} {\sqrt {a^2 - x^2} }
+      | r = \int x^2 \frac {x \rd x} {\sqrt {a^2 - x^2} }
+      | c = 
+}}
+{{eqn | r = -x^2 \sqrt {a^2 - x^2} - \int 2 x \left({-\sqrt {a^2 - x^2} }\right) \rd x
+      | c = [[Integration by Parts]]
+}}
+{{eqn | r = -x^2 \sqrt {a^2 - x^2} + 2 \int x \left({\sqrt {a^2 - x^2} }\right) \rd x
+      | c = simplification
+}}
+{{eqn | r = -x^2 \sqrt {a^2 - x^2} + 2 \left({-\frac {\left({\sqrt {a^2 - x^2} }\right)^3} 3 }\right) + C
+      | c = [[Primitive of x by Root of a squared minus x squared|Primitive of $x \sqrt {a^2 - x^2}$]]
+}}
+{{eqn | r = \left({a^2 - x^2 - a^2}\right) \sqrt {a^2 - x^2} - \frac {2 \left({\sqrt {a^2 - x^2} }\right)^3} 3 + C
+      | c = 
+}}
+{{eqn | r = \left({\sqrt {a^2 - x^2} }\right)^3 - a^2 \sqrt {a^2 - x^2} - \frac {2 \left({\sqrt {a^2 - x^2} }\right)^3} 3 + C
+      | c = 
+}}
+{{eqn | r = \frac {\left({\sqrt {a^2 - x^2} }\right)^3} 3 - a^2 \sqrt {a^2 - x^2} + C
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Reciprocal of x squared by Root of a squared minus x squared}
+Tags: Primitives involving Root of a squared minus x squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {\mathrm d x} {x^2 \sqrt {a^2 - x^2} } = \frac {-\sqrt {a^2 - x^2} } {a^2 x} + C$
+\end{theorem}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = z
+      | r = x^2
+}}
+{{eqn | ll= \implies
+      | l = \frac {\mathrm d z} {\mathrm d x}
+      | r = 2 x
+      | c = [[Power Rule for Derivatives]]
+}}
+{{eqn | ll= \implies
+      | l = \int \frac {\mathrm d x} {x^2 \sqrt {a^2 - x^2} }
+      | r = \int \frac {\mathrm d z} {2 z \sqrt z \sqrt {a^2 - z} }
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \frac 1 2 \int \frac {\mathrm d z} {z^{3/2} \sqrt {a^2 - z} }
+      | c = 
+}}
+{{end-eqn}}
+Using [[Primitive of Reciprocal of Power of x by Root of a x + b|Primitive of $ \dfrac 1{x^m \sqrt{a x + b} }$]]:
+:$\displaystyle \int \frac {\mathrm d x} {x^m \sqrt{a x + b} } = -\frac {\sqrt{a x + b} } {\left({m - 1}\right) b x^{m-1} } - \frac {\left({2 m - 3}\right) a} {\left({2 m - 2}\right) b} \int \frac {\mathrm d x} {x^{m - 1} \sqrt{a x + b} }$
+Setting:
+{{begin-eqn}}
+{{eqn | l = x
+      | o = :=
+      | r = -z
+}}
+{{eqn | l = m
+      | o = :=
+      | r = \frac 3 2
+}}
+{{eqn | l = a
+      | o = :=
+      | r = 1
+}}
+{{eqn | l = b
+      | o = :=
+      | r = a^2
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = \frac 1 2 \int \frac {\mathrm d z} {\left({-z}\right)^{3/2} \sqrt {-z + a^2} }
+      | r = \frac {-\sqrt{-z + a^2} } {2 \left({\left({\frac 3 2}\right) - 1}\right) a^2 \left({-z}\right)^{\left({3/2}\right) - 1} } - \frac {2 \left({\frac 3 2}\right) - 3} {2 \left({2 \left({\frac 3 2}\right) - 2}\right) a^2} \int \frac {\mathrm d z} {\left({-z}\right)^{\left({\frac 3 2}\right) - 1} \sqrt{-z + a^2} } + C
+      | c = [[Primitive of Reciprocal of Power of x by Root of a x + b|Primitive of $ \dfrac 1{x^m \sqrt{a x + b} }$]]
+}}
+{{eqn | r = \frac {-\sqrt{a^2 - z} } {a^2 z^{1/2} } - 0 + C
+      | c = simplifying
+}}
+{{eqn | r = \frac {-\sqrt {a^2 - x^2} } {a^2 x} + C
+      | c = substituting back for $z$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Reciprocal of x cubed by Root of a squared minus x squared}
+Tags: Primitives involving Root of a squared minus x squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {\d x} {x^3 \sqrt {a^2 - x^2} } = \frac {-\sqrt {a^2 - x^2} } {2 a^2 x^2} - \frac 1 {2 a^3} \map \ln {\frac {a + \sqrt {a^2 - x^2} } x} + C$
+\end{theorem}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = z
+      | r = x^2
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d z} {\d x}
+      | r = 2 x
+      | c = [[Power Rule for Derivatives]]
+}}
+{{eqn | ll= \leadsto
+      | l = \int \frac {\d x} {x^3 \sqrt {a^2 - x^2} }
+      | r = \int \frac {\d z} {2 z^{3/2} \sqrt z \sqrt {a^2 - z} }
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \frac 1 2 \int \frac {\d z} {z^2 \sqrt {-z + a^2} }
+      | c = 
+}}
+{{end-eqn}}
+Using [[Primitive of Reciprocal of Power of x by Root of a x + b|Primitive of $ \dfrac 1{x^m \sqrt{a x + b} }$]]:
+:$\displaystyle \int \frac {\d x} {x^m \sqrt {a x + b} } = -\frac {\sqrt{a x + b} } {\paren {m - 1} b x^{m - 1} } - \frac {\paren {2 m - 3} a} {\paren {2 m - 2} b} \int \frac {\d x} {x^{m - 1} \sqrt{a x + b} }$
+Setting:
+{{begin-eqn}}
+{{eqn | l = x
+      | o = :=
+      | r = z
+}}
+{{eqn | l = m
+      | o = :=
+      | r = 2
+}}
+{{eqn | l = a
+      | o = :=
+      | r = -1
+}}
+{{eqn | l = b
+      | o = :=
+      | r = a^2
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = \frac 1 2 \int \frac {\d z} {z^2 \sqrt {-z + a^2} }
+      | r = \frac {-\sqrt {-z + a^2} } {2 \paren {2 - 1} a^2 z^{2 - 1} } - \frac {\paren {2 \paren 2 - 3} \paren {-1} } {2 \paren {2 \paren 2 - 2} a^2} \int \frac {\d z} {z^{2 - 1} \sqrt {-z + a^2} } + C
+      | c = [[Primitive of Reciprocal of Power of x by Root of a x + b|Primitive of $ \dfrac 1{x^m \sqrt{a x + b} }$]]
+}}
+{{eqn | r = \frac {-\sqrt {a^2 - z} } {2 a^2 z} - \frac 1 {4 a^2} \int \frac {\d z} {z \sqrt{a^2 - z} } + C
+      | c = simplifying
+}}
+{{eqn | r = \frac {-\sqrt {a^2 - x^2} } {2 a^2 x^2} + \frac 1 {4 a^2} \paren {\int \frac {2 x \rd x} {x^2 \sqrt {a^2 - x^2} } } + C
+      | c = substituting for $z$
+}}
+{{eqn | r = \frac {-\sqrt {a^2 - x^2} } {2 a^2 x^2} + \frac 1 {2 a^2} \paren {\int \frac {\d x} {x \sqrt {a^2 - x^2} } } + C
+      | c = simplification
+}}
+{{eqn | r =\frac {-\sqrt {a^2 - x^2} } {2 a^2 x^2} + \frac 1 {2 a^3} \paren {-\map \ln {\frac {a + \sqrt {a^2 - x^2} } x} } + C
+      | c = [[Primitive of Reciprocal of x by Root of a squared minus x squared/Logarithm Form|Primitive of $\dfrac 1 {x \sqrt {a^2 - x^2} }$]]
+}}
+{{eqn | r =\frac {-\sqrt {a^2 - x^2} } {2 a^2 x^2} - \frac 1 {2 a^3} \map \ln {\frac {a + \sqrt {a^2 - x^2} } x} + C
+      | c = simplifying
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Root of a squared minus x squared}
+Tags: Primitives involving Root of a squared minus x squared
+
+\begin{theorem}
+:$\displaystyle \int \sqrt {a^2 - x^2} \rd x = \frac {x \sqrt {a^2 - x^2} } 2 + \frac {a^2} 2 \arcsin \frac x a + C$
+\end{theorem}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = x
+      | r = a \sin \theta
+}}
+{{eqn | n = 1
+      | ll= \leadsto
+      | l = \frac {\d x} {\d \theta}
+      | r = a \cos \theta
+      | c = [[Derivative of Sine Function]]
+}}
+{{end-eqn}}
+Also:
+{{begin-eqn}}
+{{eqn | l = x
+      | r = a \sin \theta
+}}
+{{eqn | ll= \leadsto
+      | l = a^2 - x^2
+      | r = a^2 - a^2 \sin^2 \theta
+      | c = 
+}}
+{{eqn | r = a^2 \paren {1 - \sin^2 \theta}
+      | c = 
+}}
+{{eqn | r = a^2 \cos^2 \theta
+      | c = [[Sum of Squares of Sine and Cosine]]
+}}
+{{eqn | n = 2
+      | ll= \leadsto
+      | l = \sqrt {a^2 - x^2}
+      | r = a \cos \theta
+      | c = 
+}}
+{{end-eqn}}
+and:
+{{begin-eqn}}
+{{eqn | l = x
+      | r = a \sin \theta
+}}
+{{eqn | n = 3
+      | ll= \leadsto
+      | l = \theta
+      | r = \map \arcsin {\frac x a}
+      | c = {{Defof|Real Arcsine}}
+}}
+{{end-eqn}}
+Thus:
+{{begin-eqn}}
+{{eqn | l = \int \sqrt {a^2 - x^2} \rd x
+      | r = \int \sqrt {a^2 - x^2} \, a \cos \theta \rd \theta
+      | c = [[Integration by Substitution]] from $(1)$
+}}
+{{eqn | r = \int a^2 \cos^2 \theta \rd \theta
+      | c = substituting for $\sqrt {a^2 - x^2}$ from $(2)$
+}}
+{{eqn | r = a^2 \int \cos^2 \theta \rd \theta
+      | c = [[Primitive of Constant Multiple of Function]]
+}}
+{{eqn | r = a^2 \frac {\sin \theta \cos \theta + \theta} 2 + C
+      | c = [[Primitive of Square of Cosine Function/Corollary|Primitive of Square of Cosine Function: Corollary]]
+}}
+{{eqn | r = \frac 1 2 a \sin \theta \, a \cos \theta + \frac {a^2 \theta} 2 + C
+      | c = rearranging
+}}
+{{eqn | r = \frac 1 2 x a \cos \theta + \frac {a^2 \theta} 2 + C
+      | c = substituting $x = a \sin \theta$
+}}
+{{eqn | r = \frac 1 2 x \sqrt {a^2 - x^2} + \frac {a^2 \theta} 2 + C
+      | c = substituting $\sqrt {a^2 - x^2} = a \cos \theta$ from $(2)$
+}}
+{{eqn | r = \frac {x \sqrt {a^2 - x^2} } 2 + \frac {a^2} 2 \map \arcsin {\frac x a} + C
+      | c = substituting for $\theta = \map \arcsin {\dfrac x a}$ from $(3)$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of x by Root of a squared minus x squared}
+Tags: Primitives involving Root of a squared minus x squared
+
+\begin{theorem}
+:$\displaystyle \int x \sqrt {a^2 - x^2} \ \mathrm d x = \frac {-\left({\sqrt {a^2 - x^2} }\right)^3} 3 + C$
+\end{theorem}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = z
+      | r = x^2
+}}
+{{eqn | ll= \implies
+      | l = \frac {\mathrm d z} {\mathrm d x}
+      | r = 2 x
+      | c = [[Power Rule for Derivatives]]
+}}
+{{eqn | ll= \implies
+      | l = \int x \sqrt {a^2 - x^2} \ \mathrm d x
+      | r = \int \frac {\sqrt z \sqrt {a^2 - z} \ \mathrm d z} {2 \sqrt z}
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \frac 1 2 \int \sqrt {a^2 - z} \ \mathrm d z
+      | c = [[Primitive of Constant Multiple of Function]]
+}}
+{{eqn | r = \frac 1 2 \frac {2 \left({\sqrt {a^2 - z} }\right)^3} {-3} + C
+      | c = [[Primitive of Root of a x + b|Primitive of $\sqrt {a x + b}$]]
+}}
+{{eqn | r = \frac {-\left({\sqrt {a^2 - x^2} }\right)^3} 3 + C
+      | c = substituting for $z$ and simplifying
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of x squared by Root of a squared minus x squared}
+Tags: Primitives involving Root of a squared minus x squared
+
+\begin{theorem}
+:$\displaystyle \int x^2 \sqrt {a^2 - x^2} \rd x = \frac {-x \paren {\sqrt {a^2 - x^2} }^3} 4 + \frac {a^2 x \sqrt {a^2 - x^2} } 8 + \frac {a^4} 8 \sinh^{-1} \frac x a + C$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = z
+      | r = x^2
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d z} {\d x}
+      | r = 2 x
+      | c = [[Power Rule for Derivatives]]
+}}
+{{eqn | ll= \leadsto
+      | l = \int x^2 \sqrt {a^2 - x^2} \rd x
+      | r = \int \frac {z \sqrt {a^2 - z} \rd z} {2 \sqrt z}
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \frac 1 2 \int \sqrt z \sqrt {a^2 - z} \rd z
+      | c = [[Primitive of Constant Multiple of Function]]
+}}
+{{eqn | r = \frac 1 2 \paren {\frac {-2 z + a^2} 4 \sqrt z \sqrt {a^2 - z} + \frac {a^4} 8 \int \frac {\d z} {\sqrt z \sqrt {a^2 - z} } } + C
+      | c = [[Primitive of Root of a x + b by Root of p x + q|Primitive of $\sqrt {\paren {a x + b} \paren {p x + q} }$]]
+}}
+{{eqn | r = \frac {-2 z + a^2} 8 \sqrt z \sqrt {a^2 - z} + \frac {a^4} {16} \paren {2 \sinh^{-1} \sqrt {\frac z {a^2} } } + C
+      | c = [[Primitive of Reciprocal of Root of a x + b by Root of p x + q|Primitive of $\dfrac 1 {\sqrt {\paren {a x + b} \paren {p x + q} } }$]]
+}}
+{{eqn | r = \frac {-2 z + 2 a^2} 8 \sqrt z \sqrt {a^2 - z} + \frac {a^2} 8 \sqrt z \sqrt {a^2 - z} + \frac {a^4} 8 \sinh^{-1} \sqrt {\frac z {a^2} } + C
+      | c = 
+}}
+{{eqn | r = \frac {a^2 - z} 4 \sqrt z \sqrt {a^2 - z} + \frac {a^2} 8 \sqrt z \sqrt {a^2 - z} + \frac {a^4} 8 \sinh^{-1} \frac {\sqrt z} a + C
+      | c = 
+}}
+{{eqn | r = \frac {-x \paren {\sqrt {a^2 - x^2} }^3} 4 + \frac {a^2 x \sqrt {x^2 - a^2} } 8 + \frac {a^4} 8 \sinh^{-1} \frac x a + C
+      | c = substituting for $z$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of x cubed by Root of a squared minus x squared}
+Tags: Primitives involving Root of a squared minus x squared
+
+\begin{theorem}
+:$\displaystyle \int x^3 \sqrt {a^2 - x^2} \ \mathrm d x = \frac {\left({\sqrt {a^2 - x^2} }\right)^5} 5 - \frac {a^2 \left({\sqrt {a^2 - x^2} }\right)^3} 3 + C$
+\end{theorem}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = z
+      | r = x^2
+}}
+{{eqn | ll= \implies
+      | l = \frac {\mathrm d z} {\mathrm d x}
+      | r = 2 x
+      | c = [[Power Rule for Derivatives]]
+}}
+{{eqn | ll= \implies
+      | l = \int x^3 \sqrt {a^2 - x^2} \ \mathrm d x
+      | r = \int \frac {z^{3/2} \sqrt {a^2 - z} \ \mathrm d z} {2 \sqrt z}
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \frac 1 2 \int z \sqrt {a^2 - z} \ \mathrm d z
+      | c = [[Primitive of Constant Multiple of Function]]
+}}
+{{eqn | r = \frac 1 2 \left({\frac {2 \left({-3 z - 2 a^2}\right)} {15} \left({\sqrt {a^2 - z} }\right)^3 }\right) + C
+      | c = [[Primitive of x by Root of a x + b|Primitive of $x \sqrt {\left({a x + b}\right)}$]]
+}}
+{{eqn | r = \frac {-3 z - 2 a^2} {15} \left({\sqrt {a^2 - z} }\right)^3 + C
+      | c = simplification
+}}
+{{eqn | r = \frac {-3 x^2 - 2 a^2} {15} \left({\sqrt {a^2 - x^2} }\right)^3 + C
+      | c = substituting for $z$
+}}
+{{eqn | r = \frac {3 a^2 - 3 x^2 - 5 a^2} {15} \left({\sqrt {a^2 - x^2} }\right)^3 + C
+      | c = 
+}}
+{{eqn | r = \frac {3 \left({a^2 - x^2}\right) \left({\sqrt {a^2 - x^2} }\right)^3 - 5 a^2 \left({\sqrt {a^2 - x^2} }\right)^3} {15} + C
+      | c = 
+}}
+{{eqn | r = \frac {\left({\sqrt {a^2 - x^2} }\right)^5} 5 - \frac {a^2 \left({\sqrt {a^2 - x^2} }\right)^3} 3 + C
+      | c = simplification
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Root of a squared minus x squared over x}
+Tags: Primitives involving Root of a squared minus x squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {\sqrt {a^2 - x^2} } x \rd x = \sqrt {a^2 - x^2} - a \, \map \ln {\frac {a + \sqrt {a^2 - x^2} } x} + C$
+\end{theorem}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = z
+      | r = x^2
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d z} {\d x}
+      | r = 2 x
+      | c = [[Power Rule for Derivatives]]
+}}
+{{eqn | ll= \leadsto
+      | l = \int \frac {\sqrt {a^2 - x^2} } x \rd x
+      | r = \int \frac {\sqrt {a^2 - z} \rd z} {2 \sqrt z \sqrt z}
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \frac 1 2 \int \frac {\sqrt {a^2 - z} \rd z} z
+      | c = [[Primitive of Constant Multiple of Function]]
+}}
+{{eqn | r = \frac 1 2 \paren {2 \sqrt {a^2 - z} + a^2 \int \frac {\d z} {z \sqrt {a^2 - z} } } + C
+      | c = [[Primitive of Root of a x + b over x|Primitive of $\dfrac {\sqrt {a x + b} } x$]]
+}}
+{{eqn | r = \sqrt {a^2 - x^2} + \frac {a^2} 2 \int \frac {2 x \rd x} {x^2 \sqrt {a^2 - x^2} } + C
+      | c = substituting for $z$
+}}
+{{eqn | r = \sqrt {a^2 - x^2} + a^2 \int \frac {\d x} {x \sqrt {a^2 - x^2} } + C
+      | c = simplifying
+}}
+{{eqn | r = \sqrt {a^2 - x^2} - a^2 \paren {-\frac 1 a \, \map \ln {\frac {a + \sqrt {a^2 - x^2} } x} } + C
+      | c = [[Primitive of Reciprocal of x by Root of a squared minus x squared/Logarithm Form|Primitive of $\dfrac 1 {x \sqrt {a^2 - x^2} }$]]
+}}
+{{eqn | r = \sqrt {a^2 - x^2} - a \, \map \ln {\frac {a + \sqrt {a^2 - x^2} } x} + C
+      | c = simplification
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Root of a squared minus x squared over x squared}
+Tags: Primitives involving Root of a squared minus x squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {\sqrt {a^2 - x^2} } {x^2} \rd x = \frac {-\sqrt {a^2 - x^2} } x - \arcsin \frac x a + C$
+\end{theorem}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = z
+      | r = x^2
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d z} {\d x}
+      | r = 2 x
+      | c = [[Power Rule for Derivatives]]
+}}
+{{eqn | ll= \leadsto
+      | l = \int \frac {\sqrt {a^2 - x^2} } {x^2} \rd x
+      | r = \int \frac {\sqrt {a^2 - z} \rd z} {2 z \sqrt z}
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \frac 1 2 \int \frac {\sqrt {a^2 - z} \rd z} {z^{3/2} }
+      | c = [[Primitive of Constant Multiple of Function]]
+}}
+{{eqn | r = \frac 1 2 \paren {\frac {-\sqrt {a^2 - z} } {\frac 1 2 \sqrt z} - \frac 1 2 \int \frac {\d z} {\sqrt z \sqrt {a^2 - z} } } + C
+      | c = [[Primitive of Root of a x + b over Power of x/Formulation 1|Primitive of $\dfrac {\sqrt {a x + b} } {x^m}$]]
+}}
+{{eqn | r = \frac {-\sqrt {a^2 - x^2} } x - \frac 1 2 \int \frac {2 x \rd x} {x \sqrt {a^2 - x^2} } + C
+      | c = substituting for $z$
+}}
+{{eqn | r = \frac {-\sqrt {a^2 - x^2} } x - \int \frac {\d x} {\sqrt {a^2 - x^2} } + C
+      | c = simplifying
+}}
+{{eqn | r = \frac {-\sqrt {a^2 - x^2} } x - \arcsin \frac x a + C
+      | c = [[Primitive of Reciprocal of Root of a squared minus x squared/Arcsine Form|Primitive of $\dfrac 1 {\sqrt {a^2 - x^2} }$]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Reciprocal of Root of a squared minus x squared cubed}
+Tags: Primitives involving Root of a squared minus x squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {\mathrm d x} {\left({\sqrt {a^2 - x^2} }\right)^3} = \frac x {a^2 \sqrt {a^2 - x^2} } + C$
+\end{theorem}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = z
+      | r = x^2
+}}
+{{eqn | ll= \implies
+      | l = \frac {\mathrm d z} {\mathrm d x}
+      | r = 2 x
+      | c = [[Power Rule for Derivatives]]
+}}
+{{eqn | ll= \implies
+      | l = \int \frac {\mathrm d x} {\left({\sqrt {a^2 - x^2} }\right)^3}
+      | r = \int \frac {\mathrm d z} {2 \sqrt z \left({\sqrt {a^2 - z} }\right)^3}
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \frac 1 2 \int \frac {\mathrm d z} {\left({a^2 - z}\right) \sqrt z \sqrt {a^2 - z} }
+      | c = [[Primitive of Constant Multiple of Function]]
+}}
+{{eqn | r = \frac 1 2 \left({\frac {2 \sqrt z} {a^2 \sqrt {a^2 - z} } }\right) + C
+      | c = [[Primitive of Reciprocal of p x + q by Root of a x + b by Root of p x + q|Primitive of $\dfrac 1 {\left({p x + q}\right) \sqrt{\left({a x + b}\right) \left({p x + q}\right)} }$]]
+}}
+{{eqn | r = \frac 1 2 \left({\frac {2 x} {a^2 \sqrt {a^2 - x^2} } }\right) + C
+      | c = substituting for $z$
+}}
+{{eqn | r = \frac x {a^2 \sqrt {a^2 - x^2} } + C
+      | c = simplifying
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of x over Root of a squared minus x squared cubed}
+Tags: Primitives involving Root of a squared minus x squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {x \ \mathrm d x} {\left({\sqrt {a^2 - x^2} }\right)^3} = \frac 1 {\sqrt {a^2 - x^2} } + C$
+\end{theorem}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = z
+      | r = a^2 - x^2
+}}
+{{eqn | ll= \implies
+      | l = \frac {\mathrm d z} {\mathrm d x}
+      | r = -2 x
+      | c = [[Power Rule for Derivatives]]
+}}
+{{eqn | ll= \implies
+      | l = \int \frac {x \ \mathrm d x} {\left({\sqrt {a^2 - x^2} }\right)^3}
+      | r = \int \frac {x \mathrm d z} {-2 x z^{3/2} }
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \frac 1 2 \int -z^{-3/2} \ \mathrm d z
+      | c = [[Primitive of Constant Multiple of Function]]
+}}
+{{eqn | r = \frac 1 2 \left({\frac {z^{1/2} } {\frac 1 2}  }\right) + C
+      | c = [[Primitive of Power]]
+}}
+{{eqn | r = \frac 1 {\sqrt z} + C
+      | c = simplifying
+}}
+{{eqn | r = \frac 1 {\sqrt {a^2 - x^2} } + C
+      | c = substituting for $z$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of x squared over Root of a squared minus x squared cubed}
+Tags: Primitives involving Root of a squared minus x squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {x^2 \rd x} {\paren {\sqrt {a^2 - x^2} }^3} = \frac x {\sqrt {a^2 - x^2} } - \arcsin \frac x a + C$
+\end{theorem}
+
+\begin{proof}
+With a view to expressing the problem in the form:
+:$\displaystyle \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$
+let:
+{{begin-eqn}}
+{{eqn | l = u
+      | r = x
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d u} {\d x}
+      | r = 1
+      | c = [[Power Rule for Derivatives]]
+}}
+{{end-eqn}}
+and let:
+{{begin-eqn}}
+{{eqn | l = \frac {\d v} {\d x}
+      | r = \frac x {\paren {\sqrt {a^2 - x^2} }^3}
+}}
+{{eqn | ll= \leadsto
+      | l = v
+      | r = \frac 1 {\sqrt {a^2 - x^2} }
+      | c = [[Primitive of x over Root of a squared minus x squared cubed|Primitive of $\dfrac x {\paren {\sqrt {a^2 - x^2} }^3}$]]
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = \int \frac {x^2 \rd x} {\paren {\sqrt {a^2 - x^2} }^3}
+      | r = \int x \frac {x \rd x} {\paren {\sqrt {a^2 - x^2} }^3}
+      | c = 
+}}
+{{eqn | r = x \frac 1 {\sqrt {a^2 - x^2} } - \int \frac 1 {\sqrt {a^2 - x^2} } 1 + C
+      | c = [[Integration by Parts]]
+}}
+{{eqn | r = \frac x {\sqrt {a^2 - x^2} } - \int \frac 1 {\sqrt {a^2 - x^2} } + C
+      | c = simplifying
+}}
+{{eqn | r = \frac x {\sqrt {a^2 - x^2} } - \arcsin \frac x a + C
+      | c = [[Primitive of Reciprocal of Root of a squared minus x squared/Arcsine Form|Primitive of $\dfrac 1 {\sqrt {a^2 - x^2} }$]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Reciprocal of x by Root of a squared minus x squared cubed}
+Tags: Primitives involving Root of a squared minus x squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {\mathrm d x} {x \left({\sqrt {a^2 - x^2} }\right)^3} = \frac 1 {a^2 \sqrt {a^2 - x^2} } - \frac 1 {a^3} \ln \left({\frac {a + \sqrt {a^2 - x^2} } x}\right) + C$
+\end{theorem}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = z
+      | r = x^2
+}}
+{{eqn | ll= \implies
+      | l = \frac {\mathrm d z} {\mathrm d x}
+      | r = 2 x
+      | c = [[Power Rule for Derivatives]]
+}}
+{{eqn | ll= \implies
+      | l = \int \frac {\mathrm d x} {x \left({\sqrt {a^2 - x^2} }\right)^3}
+      | r = \int \frac {\mathrm d z} {2 \sqrt z \sqrt z \left({\sqrt {a^2 - z} }\right)^3}
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \frac 1 2 \int \frac {\mathrm d z} {z \left({\sqrt {a^2 - z} }\right)^3}
+      | c = [[Primitive of Constant Multiple of Function]]
+}}
+{{eqn | r = \frac 1 2 \left({\frac 2 {a^2 \sqrt {a^2 - z} } + \frac 1 {a^2} \int \frac {\mathrm d z} {z \sqrt {a^2 - z} } }\right) + C
+      | c = [[Primitive of Reciprocal of x by Power of Root of a x + b|Primitive of $\dfrac 1 {x \left({\sqrt{a x + b} }\right)^m }$]]
+}}
+{{eqn | r = \frac 1 {a^2 \sqrt {a^2 - z} } + \frac 1 {2 a^2} \int \frac {\mathrm d z} {z \sqrt {a^2 - z} } + C
+      | c = simplifying
+}}
+{{eqn | r = \frac 1 {a^2 \sqrt {a^2 - x^2} } + \frac 1 {2 a^2} \int \frac {2 x \mathrm d x} {x^2 \sqrt {a^2 - x^2} } + C
+      | c = substituting for $z$
+}}
+{{eqn | r = \frac 1 {a^2 \sqrt {a^2 - x^2} } + \frac 1 {a^2} \int \frac {\mathrm d x} {x \sqrt {a^2 - x^2} } + C
+      | c = simplifying
+}}
+{{eqn | r = \frac1 {a^2 \sqrt {a^2 - x^2} } - \frac 1 {a^3} \ln \left({\frac {a + \sqrt {a^2 - x^2} } x}\right) + C
+      | c = [[Primitive of Reciprocal of x by Root of a squared minus x squared|Primitive of $\dfrac 1 {x \sqrt {a^2 - x^2} }$]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Reciprocal of x squared by Root of a squared minus x squared cubed}
+Tags: Primitives involving Root of a squared minus x squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {\mathrm d x} {x^2 \left({\sqrt {a^2 - x^2} }\right)^3} = \frac {-\sqrt {a^2 - x^2} } {a^4 x} + \frac x {a^4 \sqrt {a^2 - x^2} } + C$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int \frac {\mathrm d x} {x^2 \left({\sqrt {a^2 - x^2} }\right)^3}
+      | r = \int \frac {a^2 \ \mathrm d x} {a^2 x^2 \left({\sqrt {a^2 - x^2} }\right)^3}
+      | c = 
+}}
+{{eqn | r = \int \frac {\left({a^2 - x^2 + x^2}\right) \ \mathrm d x} {a^2 x^2 \left({\sqrt {a^2 - x^2} }\right)^3}
+      | c = 
+}}
+{{eqn | r = \frac 1 {a^2} \int \frac {\left({a^2 - x^2}\right) \ \mathrm d x} {a^2 x^2 \left({\sqrt {a^2 - x^2} }\right)^3} + \frac 1 {a^2} \int \frac {x^2 \ \mathrm d x} {x^2 \left({\sqrt {a^2 - x^2} }\right)^3}
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \frac 1 {a^2} \int \frac {\mathrm d x} {x^2 \sqrt {a^2 - x^2} } + \frac 1 {a^2} \int \frac {\mathrm d x} {\left({\sqrt {a^2 - x^2} }\right)^3}
+      | c = simplifying
+}}
+{{eqn | r = \frac 1 {a^2} \frac {-\sqrt {a^2 - x^2} } {a^2 x} + \frac 1 {a^2} \int \frac {\mathrm d x} {\left({\sqrt {x^2 - a^2} }\right)^3} + C
+      | c = [[Primitive of Reciprocal of x squared by Root of a squared minus x squared|Primitive of $\dfrac 1 {x^2 \sqrt {a^2 - x^2} }$]]
+}}
+{{eqn | r = \frac {-\sqrt {a^2 - x^2} } {a^4 x} + \frac 1 {a^2} \frac x {a^2 \sqrt {a^2 - x^2} } + C
+      | c = [[Primitive of Reciprocal of Root of a squared minus x squared cubed|Primitive of $\dfrac 1 {\left({\sqrt {a^2 - x^2} }\right)^3}$]]
+}}
+{{eqn | r = \frac {-\sqrt {a^2 - x^2} } {a^4 x} + \frac x {a^4 \sqrt {x^2 - a^2} } + C
+      | c = simplification
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Reciprocal of x cubed by Root of a squared minus x squared cubed}
+Tags: Primitives involving Root of a squared minus x squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {\mathrm d x} {x^3 \left({\sqrt {a^2 - x^2} }\right)^3} = \frac {-1} {2 a^2 x^2 \sqrt {a^2 - x^2} } + \frac 3 {2 a^4 \sqrt {a^2 - x^2} } - \frac 3 {2 a^5} \ln \left({\frac {a + \sqrt {a^2 - x^2} } x}\right) + C$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int \frac {\mathrm d x} {x^3 \left({\sqrt {a^2 - x^2} }\right)^3}
+      | r = \int \frac {a^2 \ \mathrm d x} {a^2 x^3 \left({\sqrt {a^2 - x^2} }\right)^3}
+      | c = 
+}}
+{{eqn | r = \int \frac {\left({a^2 - x^2 + x^2}\right) \ \mathrm d x} {a^2 x^3 \left({\sqrt {a^2 - x^2} }\right)^3}
+      | c = 
+}}
+{{eqn | r = \frac 1 {a^2} \int \frac {\left({a^2 - x^2}\right) \ \mathrm d x} {x^3 \left({\sqrt {a^2 - x^2} }\right)^3} + \frac 1 {a^2} \int \frac {x^2 \ \mathrm d x} {x^3 \left({\sqrt {a^2 - x^2} }\right)^3}
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \frac 1 {a^2} \int \frac {\mathrm d x} {x^3 \sqrt {a^2 - x^2} } + \frac 1 {a^2} \int \frac {\mathrm d x} {x \left({\sqrt {a^2 - x^2} }\right)^3}
+      | c = simplifying
+}}
+{{eqn | r = \frac 1 {a^2} \left({\frac {-\sqrt {a^2 - x^2} } {2 a^2 x^2} - \frac 1 {2 a^3} \ln \left({\frac {a + \sqrt {a^2 - x^2} } x}\right)}\right) + \frac 1 {a^2} \int \frac {\mathrm d x} {x \left({\sqrt {x^2 - a^2} }\right)^3} + C
+      | c = [[Primitive of Reciprocal of x cubed by Root of a squared minus x squared|Primitive of $\dfrac 1 {x^3 \sqrt {a^2 - x^2} }$]]
+}}
+{{eqn | r = \frac 1 {a^2} \left({\frac {-\sqrt {a^2 - x^2} } {2 a^2 x^2} - \frac 1 {2 a^3} \ln \left({\frac {a + \sqrt {a^2 - x^2} } x}\right)}\right)
+      | c = 
+}}
+{{eqn | o = 
+      | ro= +
+      | r = \frac 1 {a^2} \left({\frac 1 {a^2 \sqrt {a^2 - x^2} } - \frac 1 {a^3} \ln \left({\frac {a + \sqrt {a^2 - x^2} } x}\right)}\right) + C
+      | c = [[Primitive of Reciprocal of x by Root of a squared minus x squared cubed|Primitive of $\dfrac 1 {x \left({\sqrt {a^2 - x^2} }\right)^3}$]]
+}}
+{{eqn | r = \frac {-1} {2 a^2 x^2 \sqrt {a^2 - x^2} } + \frac 3 {2 a^4 \sqrt {a^2 - x^2} } - \frac 3 {2 a^5} \ln \left({\frac {a + \sqrt {a^2 - x^2} } x}\right) + C
+      | c = simplification
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Root of a squared minus x squared cubed}
+Tags: Primitives involving Root of a squared minus x squared
+
+\begin{theorem}
+:$\displaystyle \int \paren {\sqrt {a^2 - x^2} }^3 \rd x = \frac {x \paren {\sqrt {a^2 - x^2} }^3} 4 + \frac {3 a^2 x \sqrt {a^2 - x^2} } 8 + \frac {3 a^4} 8 \arcsin \frac x a + C$
+\end{theorem}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = z
+      | r = x^2
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d z} {\d x}
+      | r = 2 x
+      | c = [[Power Rule for Derivatives]]
+}}
+{{eqn | ll= \leadsto
+      | l = \int \paren {\sqrt {a^2 - x^2} }^3 \rd x
+      | r = \int \frac {\paren {\sqrt {a^2 - z} }^3} {2 \sqrt z} \rd x
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \frac {\sqrt z \paren {\sqrt {a^2 - z} }^3} 4 + \frac {3 a^2} 8 \int \frac {\sqrt {a^2 - z} } {\sqrt z} \rd x + C
+      | c = [[Primitive of Power of p x + q over Root of a x + b|Primitive of $\dfrac {\paren {p x + q}^n} {\sqrt{a x + b} }$]]
+}}
+{{eqn | r = \frac {\sqrt z \paren {\sqrt {a^2 - z} }^3} 4 + \frac {3 a^2} 8 \paren {\sqrt z \sqrt {a^2 - z} + \frac {a^2} 2 \int \frac {\d x} {\sqrt z \sqrt {a^2 - z} } } + C
+      | c = [[Primitive of Root of p x + q over Root of a x + b|Primitive of $\dfrac {\sqrt{p x + q} } {\sqrt{a x + b} }$]]
+}}
+{{eqn | r = \frac {x \paren {\sqrt {a^2 - x^2} }^3} 4 + \frac {3 a^2} 8 x \sqrt {a^2 - x^2} + \frac {3 a^4} 8 \int \frac {\d x} {\sqrt {a^2 - x^2} } + C
+      | c = substituting for $z$ and simplifying
+}}
+{{eqn | r = \frac {x \paren {\sqrt {a^2 - x^2} }^3} 4 + \frac {3 a^2 x \sqrt {a^2 - x^2} } 8 + \frac {3 a^4} 8 \arcsin \frac x a + C
+      | c = [[Primitive of Reciprocal of Root of a squared minus x squared/Arcsine Form|Primitive of $\dfrac 1 {\sqrt {a^2 - x^2} }$]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of x by Root of a squared minus x squared cubed}
+Tags: Primitives involving Root of a squared minus x squared
+
+\begin{theorem}
+:$\displaystyle \int x \left({\sqrt {a^2 - x^2} }\right)^3 \ \mathrm d x = \frac {-\left({\sqrt {a^2 - x^2} }\right)^5} 5 + C$
+\end{theorem}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = z
+      | r = a^2 - x^2
+}}
+{{eqn | ll= \implies
+      | l = \frac {\mathrm d z} {\mathrm d x}
+      | r = -2 x
+      | c = [[Power Rule for Derivatives]]
+}}
+{{eqn | ll= \implies
+      | l = \int x \left({\sqrt {a^2 - x^2} }\right)^3 \ \mathrm d x
+      | r = \int \frac {z^{3/2} } {-2} \ \mathrm d z
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \frac 1 2 \frac {-z^{5/2} } {\frac 5 2} + C
+      | c = [[Primitive of Power]]
+}}
+{{eqn | r = \frac {-z^{5/2} } 5 + C
+      | c = simplifying
+}}
+{{eqn | r = \frac {\left({-\sqrt {a^2 - x^2} }\right)^5} 5 + C
+      | c = substituting for $z$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of x squared by Root of a squared minus x squared cubed}
+Tags: Primitives involving Root of a squared minus x squared
+
+\begin{theorem}
+:$\displaystyle \int x^2 \paren {\sqrt {a^2 - x^2} }^3 \rd x = \frac {-x \paren {\sqrt {a^2 - x^2} }^5} 6 + \frac {a^2 x \paren {\sqrt {a^2 - x^2} }^3} {24} + \frac {a^4 x \sqrt {a^2 - x^2} } {16} + \frac {a^6} {16} \arcsin \frac x a + C$
+\end{theorem}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = z
+      | r = x^2
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d z} {\d x}
+      | r = 2 x
+      | c = [[Power Rule for Derivatives]]
+}}
+{{eqn | ll= \leadsto
+      | l = \int x^2 \paren {\sqrt {a^2 - x^2} }^3 \rd x
+      | r = \int \frac {\sqrt z \paren {a^2 - z}^{3/2} } 2 \rd z
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \frac {\sqrt z \paren {a^2 - z}^{5/2} } {-6} + \frac {-a^2} {-12} \int \frac {\paren {a^2 - z}^{3/2} } {\sqrt z} \rd z + C
+      | c = [[Primitive of Power of p x + q by Root of a x + b|Primitive of $\paren {p x + q}^n \sqrt{a x + b}$]]
+}}
+{{eqn | r = \frac {-x \paren {a^2 - x^2}^{5/2} } 6 + \frac {a^2} 6 \int \paren {a^2 - x^2}^{3/2} \rd x + C
+      | c = substituting for $z$ and simplifying
+}}
+{{eqn | r = \frac {-x \paren {a^2 - x^2}^{5/2} } 6
+}}
+{{eqn | o =
+      | ro= +
+      | r = \frac {a^2} 6 \paren {\frac {x \paren {\sqrt {a^2 - x^2} }^3} 4 + \frac {3 a^2 x \sqrt {a^2 - x^2} } 8 + \frac {3 a^4} 8 \arcsin \frac x a} + C
+      | c = [[Primitive of Root of a squared minus x squared cubed|Primitive of $\paren {\sqrt {a^2 - x^2} }^3$]]
+}}
+{{eqn | r = \frac {-x \paren {a^2 - x^2}^{5/2} } 6 + \frac {a^2 x \paren {\sqrt {a^2 - x^2} }^3} {24}
+}}
+{{eqn | o =
+      | ro= +
+      | r = \frac {a^4 x \sqrt {a^2 - x^2} } {16} + \frac {a^6} {16} \arcsin \frac x a + C
+      | c = simplifying
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of x cubed by Root of a squared minus x squared cubed}
+Tags: Primitives involving Root of a squared minus x squared
+
+\begin{theorem}
+:$\displaystyle \int x^3 \paren {\sqrt {a^2 - x^2} }^3 \rd x = \frac {\paren {\sqrt {a^2 - x^2} }^7} 7 - \frac {a^2 \paren {\sqrt {a^2 - x^2} }^5} 5 + C$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int x^3 \paren {\sqrt {a^2 - x^2} }^3 \rd x
+      | r = \int x \paren {x^2} \paren {\sqrt {a^2 - x^2} }^3 \rd x
+      | c = 
+}}
+{{eqn | r = \int x \paren {x^2 - a^2 + a^2} \paren {\sqrt {a^2 - x^2} }^3 \rd x
+      | c = [[Primitive of Power]]
+}}
+{{eqn | r = -\int x \paren {a^2 - x^2} \paren {\sqrt {a^2 - x^2} }^3 \rd x + a^2 \int x \paren {\sqrt {a^2 - x^2} }^3 \rd x
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = -\int x \paren {\sqrt {a^2 - x^2} }^5 \rd x + a^2 \int x \paren {\sqrt {a^2 - x^2} }^3 \rd x
+      | c = simplifying
+}}
+{{end-eqn}}
+Let:
+{{begin-eqn}}
+{{eqn | l = z
+      | r = a^2 - x^2
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d z} {\d x}
+      | r = -2 x
+      | c = [[Power Rule for Derivatives]]
+}}
+{{eqn | ll= \leadsto
+      | o = 
+      | r = -\int x \paren {\sqrt {a^2 - x^2} }^5 \rd x + a^2 \int x \paren {\sqrt {a^2 - x^2} }^3 \rd x
+      | c = 
+}}
+{{eqn | r = -\int \frac {\sqrt z z^{5/2} } {-2 \sqrt z} \rd z + a^2 \int \frac {\sqrt z z^{3/2} } {-2 \sqrt z} \rd z
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \frac 1 2 \int z^{5/2} \rd z - \frac {a^2} 2 \int z^{3/2} \rd z
+      | c = simplifying
+}}
+{{eqn | r = \frac 1 2 \frac {z^{7/2} } {7/2} - \frac {a^2} 2 \frac {z^{5/2} } {5/2} + C
+      | c = [[Primitive of Power]]
+}}
+{{eqn | r = \frac {z^{7/2} } 7 - a^2 \frac {z^{5/2} } 5 + C
+      | c = simplifying
+}}
+{{eqn | r = \frac {\paren {\sqrt {a^2 - x^2} }^7} 7 - \frac {a^2 \paren {\sqrt {a^2 - x^2} }^5} 5 + C
+      | c = substituting for $z$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Reciprocal of a x squared plus b x plus c}
+Tags: Primitives involving a x squared plus b x plus c, Primitive of Reciprocal of a x squared plus b x plus c, Primitives involving Reciprocals
+
+\begin{theorem}
+:$\displaystyle \int \frac {\d x} {a x^2 + b x + c} = \begin{cases}
+\dfrac 2 {\sqrt {4 a c - b^2} } \, \map \arctan {\dfrac {2 a x + b} {\sqrt {4 a c - b^2} } } + C & : b^2 - 4 a c < 0 \\
+\dfrac 1 {\sqrt {b^2 - 4 a c} } \ln \size {\dfrac {2 a x + b - \sqrt {b^2 - 4 a c} } {2 a x + b + \sqrt {b^2 - 4 a c} } } + C & : b^2 - 4 a c > 0 \\
+\dfrac {-2} {2 a x + b} + C & : b^2 = 4 a c
+\end{cases}$
+\end{theorem}
+
+\begin{proof}
+First:
+{{begin-eqn}}
+{{eqn | l = a x^2 + b x + c
+      | r = \frac {\paren {2 a x + b}^2 + 4 a c - b^2} {4 a}
+      | c = [[Completing the Square]]
+}}
+{{eqn | n = 1
+      | ll= \leadsto
+      | l = \int \frac {\d x} {a x^2 + b x + c}
+      | r = \int \frac {4 a \rd x} {\paren {2 a x + b}^2 - \paren {b^2 - 4 a c} }
+      | c = 
+}}
+{{end-eqn}}
+Put: $z = 2 a x + b$
+{{begin-eqn}}
+{{eqn | l = z
+      | r = 2 a x + b
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d z} {\d x}
+      | r = 2 a
+      | c = [[Derivative of Power]]
+}}
+{{end-eqn}}
+Let $D = b^2 - 4 a c$.
+Thus:
+{{begin-eqn}}
+{{eqn | l = \int \frac {\d x} {a x^2 + b x + c}
+      | r = \int \frac {4 a \rd x} {\paren {2 a x + b}^2 - \paren {b^2 - 4 a c} }
+      | c = from $(1)$
+}}
+{{eqn | r = \int \frac {2 \rd z} {z^2 - D}
+      | c = [[Integration by Substitution]]
+}}
+{{end-eqn}}
+=== [[Primitive of Reciprocal of a x squared plus b x plus c/Negative Discriminant|Negative Discriminant]] ===
+{{:Primitive of Reciprocal of a x squared plus b x plus c/Negative Discriminant}}
+{{qed|lemma}}
+=== [[Primitive of Reciprocal of a x squared plus b x plus c/Positive Discriminant|Positive Discriminant]] ===
+{{:Primitive of Reciprocal of a x squared plus b x plus c/Positive Discriminant}}
+{{qed|lemma}}
+=== [[Primitive of Reciprocal of a x squared plus b x plus c/Zero Discriminant|Zero Discriminant]] ===
+{{:Primitive of Reciprocal of a x squared plus b x plus c/Zero Discriminant}}
+{{qed}}
+\end{proof}
\ No newline at end of file