How prove this $I=\int_{0}^{\infty}\frac{1}{x}\ln{\left(\frac{1+x}{1-x}\right)^2}dx=\pi^2$

Prove this
$$I=\int_{0}^{\infty}\dfrac{1}{x}\ln{\left(\dfrac{1+x}{1-x}\right)^2}dx=\pi^2$$

My try: let
$$I=\int_{0}^{\infty}\dfrac{2\ln{(1+x)}}{x}-\dfrac{2\ln{|(1-x)|}}{x}dx$$

Solutions Collecting From Web of "How prove this $I=\int_{0}^{\infty}\frac{1}{x}\ln{\left(\frac{1+x}{1-x}\right)^2}dx=\pi^2$"

Here is an approach. Using the change of variables $\frac{1+x}{1-x}=y$ gives

$$ I=\int_{0}^{\infty}\dfrac{1}{x}\ln{\left(\left(\dfrac{1+x}{1-x}\right)^2\right)}dx=2\int_{-1}^{1}\frac{\ln(y^2)}{1-y^2}dy=8\int_{0}^{1}\frac{\ln y}{1-y^2}dy $$

$$ = 8\sum_{k=0}^{\infty} \int_{0}^{1} y^{2k} \ln y\, dy = 8\sum_{k=0}^{\infty} \frac{1}{(2k+1)^2} = 8.\frac{\pi^2}{8} = \pi^2.$$

First perform the bilinear transformation:
$$x={\frac {1-t}{t+1}}$$
to get:
$$\int _{0}^{\infty }\!\ln \left( {\frac { \left( 1+x \right) ^{2}}{
\left( 1-x \right) ^{2}}} \right)\dfrac{1}{x} {dx}=4\,\int _{0}^{1}\!{
\frac {\ln \left( {t}^{2} \right) }{{t}^{2}-1}}{dt}=8\,\int _{0}^{1}\!{
\frac {\ln \left(t \right) }{{t}^{2}-1}}{dt} $$
then, from this answer, consider the integral:
$$I(m)=\int _{0}^{1}\!{\frac { \ln\left( t \right) ^{m-1}}{
{t}^{2}-1}}{dt} \quad:\quad \mathfrak{R}(m)>1 $$
and the substitution $t=e^{-u}$:
$$\begin{aligned}
\int _{0}^{1}\!{\frac { \ln\left( t \right) ^{m-1}}{
{t}^{2}-1}}{dt}=& \left( -1 \right) ^{m-1}\int _{0}^{\infty }\!{\frac {
{u}^{m-1}{{\rm e}^{-u}}}{-1+{{\rm e}^{-2\,u}}}}{du}\\
=&\left( -1 \right) ^{m-1}
\int _{0}^{\infty }\!-{\frac {{u}^{m-1}}{-1+{{\rm e}^{u}}}}+{\frac {{u
}^{m-1}}{-1+{{\rm e}^{2\,u}}}}{du}\\
=&\left( -1 \right) ^{m-1}\left(
1- \dfrac{1}{2^m} \right)
\int _{0}^{\infty }\!{\frac {{u}^{m-1}}{-1+{{\rm e}^{u}}}}{du}\\
=&\left( -1 \right) ^{m}\left(
1- \dfrac{1}{2^m} \right)
\Gamma \left( m \right) \zeta \left( m \right)
\end{aligned}$$
where we have used Riemann’s integral representation of the zeta function and we also made the substitution $u\rightarrow\frac{u}{2}$ in the second term of the second line to pass to line three (having noted that convergence of both terms individually is assured by comparison with Riemanns integral).

One way to evaluate the Riemann zeta function at even integers is to use the Fourier series for Bernoulli polynomials. For example, calculating the Fourier series for the second Bernoulli polynomial tells you that:
$$\displaystyle{x}^{2}-x+1/6=\frac{1}{\pi^2}\sum _{n=1}^{\infty }{\frac {\cos \left( nx \right) }{{n
}^{2}}}\quad : \quad-\pi<x<\pi$$

and evaluating this series at $x=0$ then tells you that $\zeta(2)=\frac{\pi^2}{6}$ which together with $\Gamma(2)=1!=1$ leads to: $$I(2)=\frac{\pi^2}{8}$$ and the result follows.

Allow me to present an approach that uses dilogarithms. Split the integral up into 2 and substitute $x \mapsto \dfrac{1}{x}$ for the second integral. This yields
\begin{align}
\int^\infty_0\frac{1}{x}\ln\left(\frac{1+x}{1-x}\right)^2dx
&=\int^1_0\frac{1}{x}\ln\left(\frac{1+x}{1-x}\right)^2dx+\int^\infty_1\frac{1}{x}\ln\left(\frac{1+x}{1-x}\right)^2dx\\
&=4\int^1_0\frac{\ln(1+x)-\ln(1-x)}{x}dx\\
&=4\left(\operatorname{Li}_2(1)-\operatorname{Li}_2(-1)\right)\\
&=4\left(\frac{\pi^2}{6}+\frac{\pi^2}{12}\right)\\
&=\pi^2
\end{align}

We can use an argument from Complex Analysis and the calculation of the integral is pretty simple and neat. Note that
$$ \arctan z=\frac{1}{2i}\ln\frac{i-z}{i+z}. $$
So we have
$$ \ln\frac{1+x}{1-x}=\ln\frac{i+ix}{i-ix}=-\ln\frac{i-ix}{i+ix}=-2i\arctan(ix) $$
and hence
\begin{eqnarray}
I&=&\int_{0}^{\infty}\dfrac{1}{x}\ln{\left(\dfrac{1+x}{1-x}\right)^2}dx\\
&=&2\int_{0}^{\infty}\dfrac{1}{x}\ln{\left(\dfrac{1+x}{1-x}\right)}dx\\
&=&-4i\int_{0}^{\infty}\dfrac{1}{x}\arctan(ix)dx\\
&=&-4i\left(\arctan(ix)\ln x\bigg|_0^\infty-\int_0^\infty\frac{i\ln x}{-x^2+1}dx\right)\\
&=&4\int_0^\infty\frac{\ln x}{x^2-1}dx=-8\int_0^1\frac{\ln x}{1-x^2}dx\\
&=&-8\int_0^1\sum_{n=0}^\infty x^{2n}\ln xdx=-8\sum_{n=0}^\infty \frac{1}{(2n+1)^n}\\
&=&\pi^2.
\end{eqnarray}