How to show that $\int_{0}^{\infty}{x\over e^x-1}\cdot\ln\left({e^x+1\over e^x-1}\right)\mathrm dx=\left({\pi\over 2}\right)^2\ln(2)?$ Proposed:

$$\int_{0}^{\infty}{x\over e^x-1}\cdot\ln\left({e^x+1\over e^x-1}\right)\mathrm dx=\left({\pi\over 2}\right)^2\ln(2)\tag1$$

My try:
$$u={e^x+1\over e^x-1}\implies -{2\over (u-1)(u+1)}\mathrm du=\mathrm dx$$
$$-\int_{1}^{\infty}{\ln(u)\over u+1}\ln\left({u+1\over u-1}\right){\mathrm du}\tag2$$
$$-\int_{1}^{\infty}{\ln(u)\over u+1}\ln(u+1){\mathrm du}+\int_{1}^{\infty}{\ln(u)\over u+1}\ln(u-1){\mathrm du}\tag3$$
$$\sum_{n=1}^{\infty}{(-1)^n\over n}\int_{1}^{\infty}u^n{\ln(u)\over u+1}{\mathrm du}+\int_{1}^{\infty}{\ln(u)\over u+1}\ln(u-1){\mathrm du}\tag4$$
How may one prove $(1)?$
 A: Under $t=e^{-x}$ as Jack D'Aurizio did, one has
$$ I=\int_{0}^{\infty}{x\over e^x-1}\cdot\ln\left({e^x+1\over e^x-1}\right)\mathrm dx=\int_{0}^{1}\frac{-\ln(t)\ln\left(\frac{1+t}{1-t}\right)}{1-t}\mathrm dt.$$
Define
$$ I(a)=\int_{0}^{1}\frac{-\ln(t)\ln\left(\frac{1+at}{1-t}\right)}{1-t}\mathrm dt $$
and then $I(-1)=0, I(1)=I$, and
\begin{eqnarray}
I'(a)&=&\int_{0}^{1}\frac{-t\ln(t)}{(1-t)(1+at)}\mathrm dt\\
&=&-\frac{1}{1+a}\int_0^1\bigg(\frac1{1-t}-\frac{1}{1+at}\bigg)\ln(t)\mathrm dt\\
&=&\frac{1}{1+a}\bigg(\frac{\pi^2}{6}+\frac1a\text{Li}_2(-a)\bigg).
\end{eqnarray}
So
\begin{eqnarray}
I(1)
&=&\frac16\int_{-1}^1\frac{1}{a(1+a)}\bigg(a\pi^2+6\text{Li}_2(-a)\bigg)\mathrm da\\
&=&\frac{1}{6}\int_{-1}^1\bigg(a\pi^2+6\text{Li}_2(-a)\bigg)\mathrm d\ln(\frac{|a|}{1+a})\\
&=&\frac{1}{6}\bigg(a\pi^2+6\text{Li}_2(-a)\bigg)\ln(\frac{|a|}{1+a})\bigg|_{-1}^1\\
&&-\frac16\int_{-1}^1\ln(\frac{a}{1+a})\mathrm d\bigg(a\pi^2+6\text{Li}_2(-a)\bigg)\\
&=&-\frac{\pi^2}{12}\ln2-\frac16\int_{-1}^1\ln(\frac{|a|}{1+a})(\pi^2-\frac{6\ln(1+a)}{a})\mathrm da\\
&=&-\frac{\pi^2}{12}\ln2+\frac{\pi^2}{3}\ln 2\\
&=&\frac{\pi^2}{4}\ln2.
\end{eqnarray}
A: By setting $x=\log t$ we are left with
$$ \int_{1}^{+\infty}\frac{\log t}{t(t-1)}\log\left(\frac{t+1}{t-1}\right)\,dt = \int_{0}^{1}\frac{-\log(t)\log\left(\frac{1+t}{1-t}\right)}{1-t}\,dt$$
that can be easily computed through the dilogarithm reflection formulas.
A: On the path of Xpaul and  Jack D'Aurizio,
The same old song,
$\displaystyle I=\int_{0}^{\infty}{x\over e^x-1}\ln\left({e^x+1\over e^x-1}\right)\mathrm dx$
Perform the change of variable $y=\text{e}^{-x}$,
$\begin{align}I&=\displaystyle \int_{0}^{1}\frac{\mathrm{ln}\left( y\right) \mathrm{ln}\left( y+1\right) -\mathrm{ln}\left( 1-y\right) \mathrm{ln}\left( y\right) }{y-1}dy\\
&=\int_0^1 \dfrac{\ln(1-x)\ln x}{1-x}dx-\int_0^1 \dfrac{\ln(1+x)\ln x}{1-x}dx\\
&=A-B
\end{align}$
Define, for $x\in [0;1]$,
$\begin{align}
R(x)&=\int_0^x \dfrac{\ln t}{1-t}dt\\
&=\int_0^1 \dfrac{x\ln(tx)}{1-tx}dt
\end{align}$
Observe that, 
$R(0)=0$ and $R(1)=-\dfrac{\pi^2}{6}$
$\begin{align}
B&=\int_0^1 \dfrac{\ln(1+x)\ln x}{1-x}dx\\
&=\Big[R(x)\ln(1+x)\Big]_0^1-\int_0^1 \dfrac{R(x)}{1+x}dx\\
&=-\dfrac{\pi^2}{6}\ln 2-\int_0^1 \int_0^1 \dfrac{x\ln(tx)}{(1+x)(1-tx)}dtdx\\
&=-\dfrac{\pi^2}{6}\ln 2-\int_0^1 \int_0^1 \dfrac{x\ln(t)}{(1+x)(1-tx)}dtdx-\int_0^1 \int_0^1 \dfrac{x\ln(x)}{(1+x)(1-tx)}dtdx\\
&=-\dfrac{\pi^2}{6}\ln 2+\left(\int_0^1\int_0^1 \dfrac{\ln t}{(1+x)(1+t)}dtdx-\int_0^1\int_0^1 \dfrac{\ln t}{(1+t)(1-tx)}dtdx\right)-\\
&\int_0^1 \int_0^1 \dfrac{x\ln(x)}{(1+x)(1-tx)}dtdx\\
&=-\dfrac{\pi^2}{6}\ln 2+\ln 2\int_0^1 \dfrac{\ln x}{1+x}dx+\int_0^1 \left[\dfrac{\ln(1-tx)\ln t}{t(t+1)}\right]_{x=0}^{x=1}dt+\int_0^1 \left[\dfrac{\ln(1-tx)\ln x}{1+x}\right]_{t=0}^{t=1}dx\\
&=-\dfrac{\pi^2}{6}\ln 2+\ln 2\int_0^1 \dfrac{\ln x}{1+x}dx+\int_0^1 \dfrac{\ln(1-t)\ln t}{t(t+1)}dt+\int_0^1 \dfrac{\ln(1-x)\ln x}{1+x}dx\\
&=-\dfrac{\pi^2}{6}\ln 2+\ln 2\int_0^1 \dfrac{\ln x}{1+x}dx+\int_0^1 \dfrac{\ln(1-t)\ln t}{t}dt
\end{align}$
In the latter integral perform the change of variable $x=1-t$,
$\displaystyle B=-\dfrac{\pi^2}{6}\ln 2+\ln 2\int_0^1 \dfrac{\ln x}{1+x}dx+\int_0^1 \dfrac{\ln(1-x)\ln x}{1-x}dx$
It is well known that,
$\displaystyle \int_0^1 \dfrac{\ln x}{1+x}dx=-\dfrac{\pi^2}{12}$
(Taylor's series expansion)
Therefore,
$\begin{align}I&=A-B\\
&=\dfrac{\pi^2}{6}\ln 2+\dfrac{\pi^2}{12}\ln 2\\
&=\boxed{\dfrac{\pi^2}{4}\ln 2}
\end{align}$
