TITLE: An AMM-like integral $\int_0^1\frac{\arctan x}x\ln\frac{(1+x^2)^3}{(1+x)^2}dx$
QUESTION [12 upvotes]: How can we evaluate $$I=\int_0^1\frac{\arctan x}x\ln\frac{(1+x^2)^3}{(1+x)^2}dx=0?$$

I tried substitution $x=\frac{1-t}{1+t}$ and got
$$I=\int_0^1\frac{2 \ln \frac{2 (t^2+1)^3}{(t+1)^4} \arctan \frac{t-1}{t+1}}{t^2-1}dt\\
=\int_0^1\frac{2 \ln \frac{2 (t^2+1)^3}{(t+1)^4} (\arctan t-\frac\pi4)}{t^2-1}dt$$
I'm able to evaluate $$\int_0^1\frac{\ln \frac{2 (t^2+1)^3}{(t+1)^4}}{t^2-1}dt$$
But I have no idea where to start with the rest one.

REPLY [3 votes]: A (new) solution by Cornel Ioan Valean
It's straightforward to show by symmetry means that 

\begin{equation*}
\int_0^1 \left(\int_0^1 \frac{a x}{(1+a^2 x^2) (1+a^2 x y)} \textrm{d}x \right)\textrm{d}y=\int_0^1\frac{a x}{1+a^2 x^2}\textrm{d}x\int_0^1 \frac{1}{1+a^2 y^2}\textrm{d}y
\end{equation*}
\begin{equation*}
=\frac{\arctan(a)\log(1+a^2)}{2a^2}.
\end{equation*}

The exact flow is described in the book (Almost) Impossible Integrals, Sums, and Series, page $162$, where the only difference is that we inject a parameter $a$, that is we use $ax$ instead of $x$ and $ay$ instead of $y$.
If we multiply the opposite sides of the result above by $a$ and then integrate from $a=0$ to $a=1$, we get
\begin{equation*}
\frac{1}{2}\int_0^1 \frac{\arctan(a)\log(1+a^2)}{a}\textrm{d}a=\int_0^1\left(\int_0^1 \left(\int_0^1 \frac{a^2 x}{(1+a^2 x^2) (1+a^2 x y)} \textrm{d}x \right)\textrm{d}y\right)\textrm{d}a
\end{equation*}
\begin{equation*}
=\int_0^1\left(\int_0^1 \left(\int_0^1 \frac{a^2 x}{(1+a^2 x^2) (1+a^2 x y)} \textrm{d}y \right)\textrm{d}x\right)\textrm{d}a=\int_0^1 \left(\int_0^1 \frac{\log(1+a^2 x)}{1+a^2 x^2}\textrm{d}x\right)\textrm{d}a
\end{equation*}
\begin{equation*}
\overset{a x\mapsto x}{=}\int_0^1\left(\int_0^a \frac{\log(1+a x)}{a(1+x^2)}\textrm{d}x\right)\textrm{d}a=\int_0^1\frac{1}{1+x^2}\left(\int_x^1 \frac{\log(1+a x)}{a}\textrm{d}a\right)\textrm{d}x
\end{equation*}
\begin{equation*}
\overset{x a\mapsto a}{=}\int_0^1\frac{1}{1+x^2}\left(\int_{x^2}^x \frac{\log(1+a)}{a}\textrm{d}a\right)\textrm{d}x
\end{equation*}
\begin{equation*}
=2\int_0^1 \frac{\arctan(x)\log(1+x^2)}{x}\textrm{d}x-\int_0^1 \frac{\arctan(x)\log(1+x)}{x}\textrm{d}x,
\end{equation*} 
and the result follows.
Q.E.D.
A spectacular generalization of the main integral

\begin{equation*}
3 \int_0^x \frac{\arctan(t)\log(1+t^2)}{t} \textrm{d}t=2\int_0^x \frac{\arctan(t) \log (1+x t)}{t} \textrm{d}t
\end{equation*} 

The proof follows the strategy above where we integrate from $a=0$ to $a=r$, and $r$ is any real number.

Using the same strategy as in the main integral, we may show that

\begin{equation*}
3 \int_0^1 \frac{\arctan(x)\operatorname{Li}_2(x) }{x} \textrm{d}x+\int_0^1 \frac{\arctan(x)\operatorname{Li}_2(-x)}{x} \textrm{d}x-\int_0^1 \frac{\arctan(x)\operatorname{Li}_2\left(-x^2\right)}{x} \textrm{d}x
\end{equation*}
\begin{equation*}
=3 \zeta(2)G+\frac{45 }{16}\zeta (4)-\frac{1}{256}\psi ^{(3)}\left(\frac{1}{4}\right),
\end{equation*}

without calculating each integral separately. This problem was prepared for Romanian Mathematical Magazine.
More spectacular results 
If we use the result 
\begin{equation*}
\int_0^x\frac{\arctan(t)\log(1+t^2)}{t} \textrm{d}t-2 \int_0^1 \frac{\arctan(xt) \log (1-t)}{t}\textrm{d}t=2\sum_{n=1}^{\infty}(-1)^{n-1} \frac{x^{2n-1}}{(2n-1)^3},
\end{equation*}
which is found in the book, (Almost) Impossible Integrals, Sums, and Series,
or its extended version which exploits $\displaystyle \operatorname{Ti}_3(x)=\int_0^x\frac{\operatorname{Ti}_2(y)}{y}\textrm{d}y$, 
\begin{equation*}
\int_0^x\frac{\arctan(t)\log(1+t^2)}{t} \textrm{d}t-2 \int_0^1 \frac{\arctan(xt) \log (1-t)}{t}\textrm{d}t=2 \operatorname{Ti}_3(x),
\end{equation*}
together with the generalization above (included in the paper), we obtain the amazing results

\begin{equation*}
\int_0^x \frac{\arctan(t) \log (1+x t)}{t} \textrm{d}t-3 \int_0^1 \frac{\arctan(xt) \log (1-t)}{t}\textrm{d}t=3\sum_{n=1}^{\infty}(-1)^{n-1} \frac{x^{2n-1}}{(2n-1)^3},
\end{equation*}

and if we use the extended version to $\displaystyle \operatorname{Ti}_3(x)$, then we have 

\begin{equation*}
\int_0^x \frac{\arctan(t) \log (1+x t)}{t} \textrm{d}t-3 \int_0^1 \frac{\arctan(xt) \log (1-t)}{t}\textrm{d}t=3\operatorname{Ti}_3(x).
\end{equation*}

For example, setting $x=1$ above, we obtain the special case

\begin{equation*}
\int_0^1 \frac{\arctan(t) \log (1+t)}{t} \textrm{d}t-3 \int_0^1 \frac{\arctan(t) \log (1-t)}{t}\textrm{d}t=\frac{3}{32}\pi^3.
\end{equation*}

Let's go a bit further and notice that if we exploit the inverse relation of $\operatorname{Ti}_3(x)$, we obtain that 

\begin{equation*}
\int_0^x\frac{\arctan(t)\log(1+t^2)}{t} \textrm{d}t+\int_0^{1/x}\frac{\arctan(t)\log(1+t^2)}{t} \textrm{d}t
\end{equation*}
\begin{equation*}
-2 \int_0^1 \frac{\arctan(xt) \log (1-t)}{t}\textrm{d}t-2 \int_0^1 \frac{\arctan(t/x) \log (1-t)}{t}\textrm{d}t
\end{equation*}
\begin{equation*}
=\operatorname{sgn}(x)\left(\frac{\pi^3}{8}+\frac{\pi}{2}\log^2(|x|)\right),
\end{equation*}

and 

\begin{equation*}
\int_0^x \frac{\arctan(t) \log (1+x t)}{t} \textrm{d}t+\int_0^{1/x} \frac{\arctan(t) \log (1+t/x)}{t} \textrm{d}t
\end{equation*}
\begin{equation*}
-3 \int_0^1 \frac{\arctan(xt) \log (1-t)}{t}\textrm{d}t-3 \int_0^1 \frac{\arctan(t/x) \log (1-t)}{t}\textrm{d}t
\end{equation*}
\begin{equation*}
=\operatorname{sgn}(x)3\left(\frac{\pi^3}{16}+\frac{\pi}{4}\log^2(|x|)\right).
\end{equation*}

Let me now present a new fancy representation of $\pi$ with the results above

\begin{equation*}
\large \pi
\end{equation*}
\begin{equation*}
=2\int_0^e\frac{\arctan(t)\log(1+t^2)}{t} \textrm{d}t+2\int_0^{1/e}\frac{\arctan(t)\log(1+t^2)}{t} \textrm{d}t
\end{equation*}
\begin{equation*}
-4 \int_0^1 \frac{\arctan(et) \log (1-t)}{t}\textrm{d}t-4 \int_0^1 \frac{\arctan(t/e) \log (1-t)}{t}\textrm{d}t
\end{equation*}
\begin{equation*}
-4\int_0^1\frac{\arctan(t)\log(1+t^2)}{t} \textrm{d}t+8 \int_0^1 \frac{\arctan(t) \log (1-t)}{t}\textrm{d}t.
\end{equation*}

And another new fancy representation of $\pi$

\begin{equation*}
\large \pi
\end{equation*}
\begin{equation*}
=\frac{4}{3}\int_0^e \frac{\arctan(t) \log (1+et)}{t} \textrm{d}t+\frac{4}{3}\int_0^{1/e} \frac{\arctan(t) \log (1+t/e)}{t} \textrm{d}t
\end{equation*}
\begin{equation*}
-4 \int_0^1 \frac{\arctan(et) \log (1-t)}{t}\textrm{d}t-4 \int_0^1 \frac{\arctan(t/e) \log (1-t)}{t}\textrm{d}t
\end{equation*}
\begin{equation*}
-\frac{8}{3}\int_0^1 \frac{\arctan(t) \log (1+t)}{t} \textrm{d}t+8 \int_0^1 \frac{\arctan(t) \log (1-t)}{t}\textrm{d}t.
\end{equation*}

A note: All the results may be found in the new preprint, A symmetry-related treatment of two fascinating sums of integrals by C.I. Valean.