Reducing an Improper Integral The question asks to show that $\displaystyle\int_{-\infty}^\infty \log\left|\frac{1+x}{1-x}\right|\frac{dx}{x}=\pi^2.$ The question walks you through steps on how to show this, but I'm stuck on the first step which asks me to reduce the integral to $$4\int_0^1 \log\left|\frac{1+x}{1-x}\right|\frac{dx}{x}.$$ What I've done so far is show that the integrand is strictly positive and that its an even function with symmetry about $x=0.$ Further, the function blows up as $x \to \pm 1.$ Hence, I have
$$
\int_{-\infty}^\infty f(x)\,dx = 2\int_0^1 f(x)\,dx+2\int_1^\infty f(x) \, dx.
$$
I'm not sure how to handle the discontinuities. Nor am I confident that this is the correct approach at all.
 A: Enforce the substitution $x\mapsto\frac 1x$ in your latter integral to obtain
$$\small\int_1^\infty\log\left|\frac{1+x}{1-x}\right|\frac{{\rm d}x}x\stackrel{x\mapsto\frac 1x}=-\int_1^0\log\left|\frac{1+\frac1x}{1-\frac1x}\right|\frac1{\frac1x}\frac{{\rm d}x}{x^2}=\int_0^1\log\left|\frac{x+1}{x-1}\right|\frac{{\rm d}x}x=\int_0^1\log\left|\frac{1+x}{1-x}\right|\frac{{\rm d}x}x$$

Continuing by letting $x\mapsto\frac{1-x}{1+x}$ gives
$$\int_0^1\log\left|\frac{1+x}{1-x}\right|\frac{{\rm d}x}x\stackrel{x\mapsto\frac{1-x}{1+x}}=-\int_0^1\log\left|\frac1x\right|\frac{2\,{\rm d}x}{1-x^2}=-2\int_0^1\frac{\log x}{1-x^2}\,{\rm d}x$$
Using the geometric series yields
$$-2\int_0^1\frac{\log x}{1-x^2}\,{\rm d}x=-2\sum_{n\ge0}\int_0^1 x^{2n}\log x\,{\rm d}x=2\sum_{n\ge0}\frac1{(2n+1)^2}$$
Finally, since $\sum_{n\ge0}\frac1{(2n+1)^2}=\frac34\sum_{n\ge1}\frac1{n^2}=\frac34\zeta(2)$, we obtain

$$\therefore~\int_0^1\log\left|\frac{1+x}{1-x}\right|\frac{{\rm d}x}x=\frac{\pi^2}4$$

A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
With the substitution
$\ds{\pars{t = {1 + x \over 1 - x} \implies x = {t - 1 \over t + 1}}}$:
\begin{align}
&\bbox[5px,#ffd]{\int_{-\infty}^{\infty}\ln\pars{\verts{1 + x \over
1 - x}}\,{\dd x \over x}} =
2\int_{0}^{\infty}\ln\pars{\verts{1 + x \over 1 - x}}\,{\dd x \over x} =
2\int_{1}^{-1}{2\ln\pars{\verts{t}} \over -1 + t^{2}}\,\dd t
\\[5mm] = &\
-8\int_{0}^{1}{\ln\pars{t} \over 1 - t^{2}}\,\dd t =
-8\sum_{n = 0}^{\infty}\int_{0}^{1}\ln\pars{t}\,t^{2n}\,\dd t =
-8\sum_{n = 0}^{\infty}\pars{\lim_{\mu \to 0}\partiald{}{\mu}\int_{0}^{1}t^{\mu + 2n}\,\dd t}
\\[5mm] = &\
8\sum_{n = 0}^{\infty}{1 \over \pars{2n + 1}^{2}} =
8\bracks{\sum_{n = 1}^{\infty}{1 \over n^{2}} -
\sum_{n = 1}^{\infty}{1 \over \pars{2n}^{2}}} =
8\pars{{3 \over 4}\sum_{n = 1}^{\infty}{1 \over n^{2}}}
\\[5mm] = &\
8\,\pars{{3 \over 4}\,{\pi^{2} \over 6}} = \bbx{\large\pi^{2}} \\ &
\end{align}
