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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.

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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$$\int\limits_0^{\infty}\mathrm dx\,\log^2\left(\frac {e^x+1}{e^x-1}\right)\color{blue}{=\frac 72\zeta(3)}$$

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    $\begingroup$ \begin{align}\frac{2}{1-x^2}=\frac{2}{1-x}-\frac{2x}{1-x^2}\end{align} therefore the integral is expressible in term of $\int_0^1 \frac{\ln^2 x}{1-x}\,dx$ (the change of variable $y=x^2$ is required) $\endgroup$ – FDP Feb 20 at 17:05
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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 $\lambda(n)$

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