About the integral $\int_{-1}^1 \frac{1}{\pi^2+(2 \operatorname{arctanh}(x))^2} \, dx=\frac{1}{6} $

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.

Sub $x=\tanh{u}$, $dx = \operatorname{sech^2}{u} \, du$. Then the integral is

$$\int_{-\infty}^{\infty} du \, \frac{\operatorname{sech^2}{u}}{\pi^2+4 u^2} $$

Now, use Parseval. The Fourier transforms of the pieces of the integrand are

$$\int_{-\infty}^{\infty} du \, \frac{e^{i u k}}{\pi^2+4 u^2} = \frac14 \frac{\pi}{\pi/2} e^{-\pi |k|/2} $$

$$\int_{-\infty}^{\infty} du \, \operatorname{sech^2}{u} \, e^{i u k} = \pi k \operatorname{csch}{\left (\frac{\pi k}{2} \right )}$$

so by Parseval...

$$\int_{-\infty}^{\infty} du \, \frac{\operatorname{sech^2}{u}}{\pi^2+4 u^2} = \frac12 \frac{\pi}{2 \pi} \int_{-\infty}^{\infty} dk \, k \operatorname{csch}{\left (\frac{\pi k}{2} \right )} e^{-\pi |k|/2} = \int_0^{\infty} dk \frac{k \, e^{-\pi k/2}}{e^{\pi k/2}-e^{-\pi k/2}}$$

Expand the denominator:

$$\int_{-\infty}^{\infty} du \, \frac{\operatorname{sech^2}{u}}{\pi^2+4 u^2} = \sum_{m=0}^{\infty} \int_0^{\infty} dk \, k \, e^{-(1+m) \pi k} = \frac{1}{\pi^2} \sum_{m=0}^{\infty} \frac1{\left (1+m \right )^2} = \frac1{6}$$

Approach 1:

For the first integral \begin{align} 2\int^1_{0}\frac{{\rm d}x}{\pi^2+\ln^2\left(\frac{1-x}{1+x}\right)} &=-\frac{4}{\pi}\mathrm{Im}\int^1_0\frac{{\rm d}x}{(\ln{x}+\pi i)(1+x)^2}\tag1\\ &=-\frac{4}{\pi}\mathrm{Im}\int^1_{-1}\frac{{\rm d}x}{\ln{x}(1-x)^2}\tag2\\ &=\frac{4}{\pi}\mathrm{Im}\left[\int^\pi_0\frac{ie^{i\phi}}{i\phi(1-e^{i\phi})^2}\ {\rm d}\phi-\frac{\pi i}{2}\operatorname*{Res}_{z=1}\frac{1}{\ln{z}(z-1)^2}\right]\tag3\\ &=\frac{4}{\pi}\mathrm{Im}\left[\int^\pi_0\frac{{\rm d}\phi}{-4\phi\sin^2\left(\frac{\phi}{2}\right)}-\frac{\pi i}{2}\operatorname*{Res}_{z=1}\frac{1}{\ln{z}(z-1)^2}\right]\\ &=\frac{4}{\pi}\left(-\frac{\pi}{2}\right)\operatorname*{Res}_{z=1}\left[\frac{1}{(z-1)^3}+\frac{1}{2(z-1)^2}\color{red}{-}\frac{\color{red}{1}}{\color{red}{12}(z-1)}+\mathcal{O}(1)\right]\tag4\\ &=\frac{1}{6} \end{align} Explanation:
$(1)$: Substitute $x\mapsto \dfrac{1-x}{1+x}$.
$(2)$: Substitute $x\mapsto -x$ and extend the integration interval to $[-1,1]$. The integral over $[0,1]$ contributes nothing since it is purely real.
$(3)$: Change the path of integration to the unit semicircular arc in the upper half plane. We pick up the residue at $z=1$ in the process.
$(4)$: The first integral in the previous line is purely real; Expand $\displaystyle\frac{1}{\ln{z}(z-1)^2}$ as a Laurent series.

Approach 2:

Here is another approach that avoids complex analysis. \begin{align} 2\int^1_0\frac{{\rm d}x}{\pi^2+\ln^2\left(\frac{1-x}{1+x}\right)} &=-\frac{4}{\pi}\mathrm{Im}\int^1_0\frac{{\rm d}x}{\ln(-x)(1+x)^2}\\ &=\frac{4}{\pi}\int^\infty_0\int^1_0\frac{x^s\sin(\pi s)}{(1+x)^2}\ {\rm d}x\ {\rm d}s\\ &=\frac{4}{\pi}\sum^\infty_{k=0}(-1)^k\int^1_0\int^1_0\frac{x^{s+k}\sin(\pi s)}{(1+x)^2}\ {\rm d}x\ {\rm d}s\\ &=\frac{4}{\pi}\int^1_0\sin(\pi s)\int^1_0\frac{x^{s}}{(1+x)^3}\ {\rm d}x\ {\rm d}s\\ &=\frac{4}{\pi}\int^1_0\sin(\pi s)\left[-\frac{1}{8}-\frac{s}{4}+\frac{s(s-1)}{4}\left(\psi_0\left(\frac{s}{2}\right)-\psi_0\left(\frac{s-1}{2}\right)\right)\right]\ {\rm d}s\\ &=\frac{1}{\pi}\int^1_0\sin(\pi s)s(s-1)\left[\psi_0\left(\frac{s}{2}\right)-\psi_0\left(\frac{s+1}{2}\right)\right]\ {\rm d}s\\ &=\frac{2}{\pi}\int^1_0\ln\frac{\Gamma\left(\frac{s}{2}\right)}{\Gamma\left(\frac{s+1}{2}\right)}\left(\frac{{\rm d}}{{\rm d}s}\sin(\pi s)s(1-s)\right)\ {\rm d}s \end{align} Using the fact $$\ln\frac{\Gamma\left(\frac{s}{2}\right)}{\Gamma\left(\frac{s+1}{2}\right)}=\sum_{n\in2\mathbb{N}_0+1}\left[\frac{1}{n}\cos(n\pi s)+2\frac{\gamma+\ln(2n\pi)}{n\pi}\sin(n\pi s)\right]\ \ , \ \ 0<s\le1$$ in tandem with the results \begin{align} \int^1_0\sin((2n+1)\pi s)\left(\frac{{\rm d}}{{\rm d}s}\sin(\pi s)s(1-s)\right)\ {\rm d}s &=0\\ \int^1_0\cos((2n+1)\pi s)\left(\frac{{\rm d}}{{\rm d}s}\sin(\pi s)s(1-s)\right)\ {\rm d}s &= \begin{cases} \frac{\pi}{12}+\frac{1}{4\pi},& \text{if $n=0$}\\ \frac{2n+1}{4\pi}\left(\frac{1}{(n+1)^2}-\frac{1}{n^2}\right),& \text{if $n\in\mathbb{N}$} \end{cases} \end{align} yields \begin{align} 2\int^1_0\frac{{\rm d}x}{\pi^2+\ln^2\left(\frac{1-x}{1+x}\right)} &=\frac{2}{\pi}\left[\frac{\pi}{12}+\frac{1}{4\pi}+\frac{1}{4\pi}\sum^\infty_{n=1}\left(\frac{1}{(n+1)^2}-\frac{1}{n^2}\right)\right]=\frac{1}{6} \end{align}

Supplementary Problem:

Substituting $x\mapsto\tanh{x}$ then $2x\mapsto\ln{x}$, \begin{align} \int^1_{-1}\frac{\ln(1-x)}{\pi^2+\ln^2\left(\frac{1-x}{1+x}\right)}\ {\rm d}x &=\int^1_{-1}\frac{\ln{2}}{\pi^2+\ln^2\left(\frac{1-x}{1+x}\right)}\ {\rm d}x+\int^\infty_{-\infty}\frac{\ln\left(\frac{1-\tanh{x}}{2}\right)}{(\pi^2+4x^2)\cosh^2{x}}\ {\rm d}x\\ &=\frac{\ln{2}}{6}-\int^\infty_{0}\frac{2\ln(1+x)}{(\pi^2+\ln^2{x})(1+x)^2}\ {\rm d}x\\ &=\frac{\ln{2}}{6}-\int^1_0\frac{1}{2\pi i}\int_Cf(z,x)\ {\rm d}z\ {\rm d}x \end{align} where $f(z,x)=\dfrac{2z}{(1+xz)(\ln{z}-\pi i)(1+z)^2}$ and $C$ is the keyhole contour deformed along the branch cut $[0,\infty]$. We note that $$\frac{1}{(\ln{z}-\pi i)(1+z)^2}\sim_{-1}-\frac{1}{(z+1)^3}+\frac{1}{2(z+1)^2}+\frac{1}{12(z+1)}+\mathcal{O}(1)$$ and by the residue theorem, the contour integral is \begin{align} \frac{1}{2\pi i}\int_Cf(z,x)\ {\rm d}z =&\ \operatorname*{Res}_{z=-1/x}f(z,x)+\operatorname*{Res}_{z=-1}f(z,x)\\ =&\ \frac{2\left(-\frac{1}{x}\right)}{x\left(\ln\left(-\frac{1}{x}\right)-\pi i\right)\left(1-\frac{1}{x}\right)^2}-\frac{1}{2}\left.\left(\frac{{\rm d}^2}{{\rm d}z^2}+\frac{{\rm d}}{{\rm d}z}\right)\frac{2z}{1+xz}\right|_{z=-1}-\frac{1}{6(1-x)}\\ =&\ \frac{2x}{(1-x)^3}+\frac{1}{(1-x)^2}-\frac{1}{6(1-x)}+\frac{2}{(1-x)^2\ln{x}} \end{align} Next, let $$I(s)=\int^1_0\left(\frac{2x}{(1-x)^3}+\frac{1}{(1-x)^2}-\frac{1}{6(1-x)}+\frac{2}{(1-x)^2\ln{x}}\right)x^s\ {\rm d}x$$ such that the integral we seek is $\displaystyle\frac{\ln{2}}{6}-I(0)$. Differentiating twice and integrating by parts, \begin{align} I''(s) &=1+\int^1_0\left(\frac{s^2x^{s-1}\ln^2{x}}{1-x}+\frac{2sx^{s-1}\ln{x}}{1-x}-\frac{x^s\ln^2{x}}{6(1-x)}\right)\ {\rm d}x\\ &=1-s^2\psi_2(s)-2s\psi_1(s)+\frac{1}{6}\psi_2(s+1) \end{align} and we may integrate back to obtain \begin{align} I'(s) &=\frac{1}{2}+s-s^2\psi_1(s)+\frac{1}{6}\psi_1(s+1)\\ I(s) &=L-s\ln(2\pi)-\frac{s}{2}+\frac{3s^2}{2}+\frac{1}{6}\psi_0(s+1)-s^2\psi_0(s)+2\ln{G(s+1)} \end{align} where \begin{align} L &=\lim_{s\to\infty}\left(s\ln(2\pi)+\frac{s}{2}-\frac{3s^2}{2}-\frac{1}{6}\psi_0(s+1)+s^2\psi_0(s)-2\ln{G(s+1)}\right)\\ &=\lim_{s\to\infty}\left(s\ln(2\pi)+\frac{s}{2}-\frac{3s^2}{2}-\frac{1}{6}\ln{s}+s^2\ln{s}-\frac{s}{2}-\frac{1}{12}-s^2\ln{s}+\frac{3s^2}{2}-s\ln(2\pi)+\frac{1}{6}\ln{s}-2\zeta'(-1)+\mathcal{O}\left(\frac{1}{s}\right)\right)\\ &=-2\zeta'(-1)-\frac{1}{12} \end{align} Letting $s=0$ and noting that $\psi_0(1)=-\gamma$, we arrive at $$\int^1_{-1}\frac{\ln(1-x)}{\pi^2+\ln^2\left(\frac{1-x}{1+x}\right)}\ {\rm d}x=2\zeta'(-1)+\frac{\gamma}{6}+\frac{\ln{2}}{6}+\frac{1}{12}$$

$\newcommand{\sech}{\operatorname{sech}}\newcommand{\arctanh}{\operatorname{arctanh}}\newcommand{\Res}{\operatorname*{Res}}$ $\Res\limits_{z=\frac\pi2i}\left(\frac{\sech^2(z)}{\pi^2+4z^2}\right)=-i\frac{3+\pi^2}{12\pi^3}$ 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.

After making the substitution $u = \text{arctanh}(x)$, we could use the Laplace transform $$\int_{0}^{\infty} \cos(ax) \, e^{-bx} \, dx = \frac{b}{a^{2}+b^{2}} \, , \, \text{Re} (b) >0 $$

and then switch the order of integration.

Specifically,

$$ \begin{align} \int_{-\infty}^{\infty} \frac{\text{sech}^{2}(u)}{4u^{2} + \pi^{2}} \, du &= \frac{1}{\pi} \int_{-\infty}^{\infty} \text{sech}^{2}(u) \int_{0}^{\infty} \cos(2ut)\, e^{- \pi t} \, dt \, du \\ &= \frac{1}{\pi} \int_{0}^{\infty} e^{- \pi t} \int_{-\infty}^{\infty} \text{sech}^{2} (u) \cos(2tu) \, du \, \ dt. \end{align}$$

The inside integral is basically the second Fourier transform that Ron Gordon uses in his answer.

A relatively quick way to evaluate it is to integrate the complex function $f(z) = \text{sech}^{2}(z) \, e^{2itz}$ around a rectangular contour in the upper half-plane of height $\pi$ and use the fact that $\text{sech}^{2} (z)$ is $i \pi$-periodic.

Doing so we get

$$\begin{align} \int_{-\infty}^{\infty} \text{sech}^{2}(u) \, e^{2itu} \, du - \int_{-\infty}^{\infty} \text{sech}^2(u) \, e^{2it(u+ i \pi)} \, du &= 2 \pi i \, \text{Res}[f(z), i \pi /2] \\ &= 2 \pi i \, \text{Res} [f(z+ i \pi/2), 0] \\ &= 2 \pi i \, \text{Res} \left[-\text{csch}^{2}(z) e^{2i tz} e^{-\pi t} ,0 \right] \\ &= 2 \pi i \left(-2ite^{- \pi t} \right) \end{align} $$

since $\text{csch}^{2}(z) = \frac{1}{z^{2}} + \mathcal{O}(1).$

Then combining the two integrals on the right, we get

$$ \int_{-\infty}^{\infty} \text{sech}^{2}(x) \, e^{2itu} \, du = \int_{-\infty}^{\infty} \text{sech}^{2}(x) \cos(2tu) \, du = 2 \pi t \, \frac{2 \, e^{- \pi t}}{1-e^{- 2 \pi t}} = 2 \pi t \, \text{csch} (\pi t). $$

(I couldn't' think of an approach that avoided complex analysis entirely.)

Therefore,

$$ \int_{-\infty}^{\infty} \frac{\text{sech}^{2}(u)}{\pi^{2}+4u^{2}} \, du = 2 \int_{0}^{\infty} t \, \text{csch} (\pi t) \, e^{- \pi t}\, dt. $$

After making the substitution $w = 2t$, we end up with the same integral that results from using Parseval's theorem. You can refer to Ron Gordon's answer to complete the evaluation.