How to show that $\int_{0}^{\infty}\ln^2\left(\frac{e^x+1}{e^x-1}\right)dx=\frac{7\zeta(3)}{2}$? Inspired by the integral 
$$\int_{0}^{\infty}\ln\left(\frac{e^x+1}{e^x-1}\right)dx=\frac{\pi^2}{4}$$
I further found out that 

$$I=\int_{0}^{\infty}\ln^2\left(\frac{e^x+1}{e^x-1}\right)dx=\frac{7\zeta(3)}{2}\tag{1}$$

I am trying to prove $(1)$
Let $u=e^{-x}$,
$$u(0)=1,u(\infty)=0$$
$$du=-e^{-x}dx,du=-udx$$
$$dx=-\frac{1}{u}du$$
$$\begin{align}
I&=\int_{0}^{1}\ln^2\left(\frac{1/u+1}{1/u-1}\right)\left(\frac{1}{u}\right)du\\\\
&=\int_{0}^{1}\frac{[\ln(1+u)-\ln(1-u)]^2}{u}du\\\\
&=\int_{0}^{1}\frac{\ln^2(1+u)}{u}du-\int_{0}^{1}\frac{2\ln(1+u)\ln(1-u)}{u}du+\int_{0}^{1}\frac{\ln^2(1-u)}{u}du
\end{align}$$
I know that 
$$-\ln(1-u)=\sum_{n=0}^{\infty}\frac{u^{n+1}}{n+1}$$
$$\ln(1+u)=\sum_{n=0}^{\infty}\frac{(-1)^nu^{n+1}}{n+1}$$
What should I do from this point on? Also, I think that trig substitution is another possible way to approach this problem. 
 A: First, rewrite the integrand as$$\mathfrak{I}=\int\limits_0^{\infty}\mathrm dx\,\log^2\left(\frac {1+e^{-x}}{1-e^{-x}}\right)$$Now make the substitution $z=e^{-x}$ so that $x=-\log z$. Thus$$\begin{align*}\mathfrak{I} & =\int\limits_0^1\mathrm dz\,\frac {\log^2\left(\frac {1+z}{1-z}\right)}z\end{align*}$$From here, make the transformation $z\mapsto\tfrac {1-z}{1+z}$ and the integral transforms into$$\begin{align*}\mathfrak{I} & =\int\limits_0^1\frac {2\,\mathrm dz}{(1+z)^2}\frac {1+z}{1-z}\log^2z\\ & =2\int\limits_0^1\mathrm dz\,\frac {\log^2 z}{1-z^2}\end{align*}$$Expand the denominator as an infinite sequence and integrate by parts twice to see that$$\begin{align*}\mathfrak{I} & =2\sum\limits_{n\geq0}\int\limits_0^1\mathrm dz\, z^{2n}\log^2 z\\ & =4\sum\limits_{n\geq0}\frac 1{(2n+1)^3}\end{align*}$$It can be shown, through some manipulation, that the infinite sum is equal to$\tfrac 78\zeta(3)$. Thus$$\int\limits_0^{\infty}\mathrm dx\,\log^2\left(\frac {e^x+1}{e^x-1}\right)\color{blue}{=\frac 72\zeta(3)}$$
A: You could use the substitution $y=\log \left( \frac{e^x+1}{e^x-1}\right)$
Which then quickly leads to 
$$x=\log \left( \frac{e^y+1}{e^y-1}\right)$$
with $$\frac{dx}{dy}=-\frac{2}{e^y-e^{-y}}$$
Therefore
$$\frac{1}{2(n-1)!} \int_0^\infty \log  \left( \frac{e^x+1}{e^x-1}\right)^{n-1} \,dx=\frac{1}{(n-1)!} \int_0^\infty  \frac{x^{n-1}}{e^y-e^{-y}} \,dy=\lambda(n)$$
where $\lambda(n)=\sum_{k=1}^\infty \frac{1}{(2k-1)^n}$
$\frac{1}{(n-1)!} \int_0^\infty  \frac{x^{n-1}}{e^y-e^{-y}} \,dy$ being the standard and easily determined integral for $\lambda(n)$
