Prove: $ -\int_{-\infty}^{+\infty} \frac{e^{t}t^3}{(1-e^t)^3} dt = \pi^{2}$ $$
\mbox{Prove that:}\quad
-\int_{-\infty}^{+\infty}\frac{\mathrm{e}^{t}\, t^3}{\left(1 -\mathrm{e}^{t}\right)^{3}}\,\mathrm{d}t = \pi^{2}
$$
I tried to solve this, but it seems hard to me.

Note
\begin{align}
& -\int_{-\infty}^{+\infty} \frac{e^{t}t^3}{(1-e^t)^3} dt 
=\int_0^\infty \frac{t^3(1+e^{-t})e^{-t}}{(1-e^{-t})^3} dt\\ \overset{x=e^{-t}} = &\int_0^1 \frac{(x+1)\ \ln^3x}{(x-1)^3 }dx = 
-\int_0^1 \ln^3x d\left(\frac{x}{(x-1)^2 } \right)\\
=&3 \int_0^1 \frac{\ln^2x}{(1-x)^2} dx
= 3\int_0^1 \ln^2xd\left(\frac{1}{1-x} \right)\\
=& 6\int_0^1 \frac{\ln x}{x-1}dx=6\cdot\frac{\pi^2}6=\pi^2
\end{align}
How to integrate $\int_0^1 \frac{\ln x}{x-1}dx$
