<s>[INST] 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
$$I=\int_{0}^{\infty}\dfrac{2\ln{(1+x)}}{x}-\dfrac{2\ln{|(1-x)|}}{x}dx$$
 [/INST] 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} 
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