TITLE: About the integral $\int_{-1}^1 \frac{1}{\pi^2+(2 \operatorname{arctanh}(x))^2} \, dx=\frac{1}{6} $
QUESTION [23 upvotes]: Here is a question that naturally arose in the study of some specific integrals. I'm curious if for such integrals are known nice real analysis tools  for calculating them (including here all possible sources
in literature that are publicaly available).  At some point I'll add my real analysis solution.
It's a question for the informative purpose rather than finding solutions, the solution is optional.
Prove that
$$\int_{-1}^1 \frac{1}{\pi^2+(2 \operatorname{arctanh}(x))^2} \, dx=\frac{1}{6}. $$
Here is a supplementary question
$$\int_{-1}^1 \frac{\log(1-x)}{\pi^2+(2 \operatorname{arctanh}(x))^2} \, dx=\frac{1}{4}+\frac{\gamma }{6}+\frac{\log (2)}{6}-2 \log (A) $$
where $A$ is Glaisher–Kinkelin constant.
for the passionates of integrals, series and limits.

REPLY [7 votes]: $\newcommand{\sech}{\operatorname{sech}}\newcommand{\arctanh}{\operatorname{arctanh}}\newcommand{\Res}{\operatorname*{Res}}$
$\Res\limits_{z=\frac\pi2i}\left(\frac{\sech^2(z)}{\pi^2+4z^2}\right)=-i\frac{3+\pi^2}{12\pi^3}$ and for $k\ge1$, $\Res\limits_{z=\frac{(2k+1)\pi}2i}\left(\frac{\sech^2(z)}{\pi^2+4z^2}\right)=\frac{8z}{\left(\pi^2+4z^2\right)^2}$. Therefore, summing the residues in the upper half-plane, we get
$$
\begin{align}
\int_{-1}^1\frac1{\pi^2+(2\arctanh(x))^2}\,\mathrm{d}x
&=\int_{-\infty}^\infty\frac1{\pi^2+4u^2}\,\mathrm{d}\tanh(u)\\
&=\int_{-\infty}^\infty\frac{\sech^2(u)}{\pi^2+4u^2}\,\mathrm{d}u\\
&=2\pi i\left(-i\frac{3+\pi^2}{12\pi^3}+\frac{i}{4\pi^3}\sum_{k=1}^\infty\frac{2k+1}{\left(k^2+k\right)^2}\right)\\
&=2\pi i\left(-i\frac{3+\pi^2}{12\pi^3}+\frac{i}{4\pi^3}\sum_{k=1}^\infty\left(\frac1{k^2}-\frac1{(k+1)^2}\right)\right)\\[4pt]
&=\frac16
\end{align}
$$
once we collapse the telescoping series.