TITLE: Evaluate $\int\frac{1+x+\sqrt{1+x^2}}{\sqrt{x+1}+\sqrt{x}}\,dx$
QUESTION [15 upvotes]: Question:

Solve the integral
  $$
\int\frac{1+x+\sqrt{1+x^2}}{\sqrt{x+1}+\sqrt{x}}\,dx
$$

My solution:
Multiply both the numerator and denominator by $\sqrt{x+1}-\sqrt{x}$. This changes the integral to
$$
\begin{align}
\int&\left(1+x+\sqrt{1+x^2}\right)\left(\sqrt{x+1}-\sqrt{x}\right)\,dx\\
&= \int{(1+x)^{3/2}}\,dx-\int{\sqrt{(1+x)(1+x^2)}}\,dx-\int{\sqrt{x}(1+x)}\,dx-\int{\sqrt{x}\sqrt{1+x^2}}\,dx\\
& = \frac{2}{5}(1+x)^{5/2}-I-\frac{2}{3}x^{3/2}-\frac{2}{5}x^{5/2}-J
\end{align}
$$
where $I = \int{(1+x)^{1/2}(1+x^2)^{1/2}}\,dx$ and $J = \int{x^{1/2}(1+x^2)^{1/2}}\,dx$.
How can I solve the integrals $I$ and $J$?

REPLY [4 votes]: For any real number of $x$ ,
When $|x|\leq1$ ,
$J=\int x^\frac{1}{2}(1+x^2)^\frac{1}{2}$
$=\int x^\frac{1}{2}\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!x^{2n}}{4^n(n!)^2(1-2n)}dx$
$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!x^{2n+\frac{1}{2}}}{4^n(n!)^2(1-2n)}dx$
$=\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!x^{2n+\frac{3}{2}}}{4^n(n!)^2(1-2n)\left(2n+\dfrac{3}{2}\right)}+C$
$=\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!x^{2n+\frac{3}{2}}}{2^{2n-1}(n!)^2(1-2n)(4n+3)}+C$
$I=\int(1+x)^\frac{1}{2}(1+x^2)^\frac{1}{2}$
$=\int(x+1)^\frac{1}{2}\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!x^{2n}}{4^n(n!)^2(1-2n)}dx$
$=\int(x+1)^\frac{1}{2}\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!(x+1-1)^{2n}}{4^n(n!)^2(1-2n)}dx$
$=\int(x+1)^\frac{1}{2}\sum\limits_{n=0}^\infty\sum\limits_{k=0}^{2n}\dfrac{(-1)^{-n}(2n)!C_k^{2n}(-1)^{2n-k}(x+1)^k}{4^n(n!)^2(1-2n)}dx$
$=\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^{2n}\dfrac{(-1)^{n-k}((2n)!)^2(x+1)^{k+\frac{1}{2}}}{4^n(n!)^2k!(2n-k)!(1-2n)}dx$
$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^{2n}\dfrac{(-1)^{n-k}((2n)!)^2(x+1)^{k+\frac{3}{2}}}{4^n(n!)^2k!(2n-k)!(1-2n)\left(k+\frac{3}{2}\right)}+C$
$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^{2n}\dfrac{(-1)^{n-k}((2n)!)^2(x+1)^{k+\frac{3}{2}}}{2^{2n-1}(n!)^2k!(2n-k)!(1-2n)(2k+3)}+C$
When $|x|\geq1$ ,
$J=\int x^\frac{1}{2}(1+x^2)^\frac{1}{2}$
$=\int x^\frac{1}{2}\left(x^2\left(\dfrac{1}{x^2}+1\right)\right)^\frac{1}{2}$
$=\int x^\frac{3}{2}\left(1+\dfrac{1}{x^2}\right)^\frac{1}{2}$
$=\int x^\frac{3}{2}\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!x^{-2n}}{4^n(n!)^2(1-2n)}dx$
$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!x^{\frac{3}{2}-2n}}{4^n(n!)^2(1-2n)}dx$
$=\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!x^{\frac{5}{2}-2n}}{4^n(n!)^2(1-2n)\left(\dfrac{5}{2}-2n\right)}+C$
$=\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!}{2^{2n-1}(n!)^2(2n-1)(4n-5)x^{2n-\frac{5}{2}}}+C$