The integral $\int\limits_0^\infty\frac{x^4e^x}{(e^x-1)^2} \mathrm{d}x$ How to calculate the following integral $$\int_0^\infty\frac{x^4e^x}{(e^x-1)^2}\mathrm{d}x$$
I would like to solve this integral by means of two different ways: for example, integration by parts and using Residue Theorem.  

Using
$$ \frac1{(1-x)^2}=\sum_{n=0}^\infty(n+1)x^n $$
\begin{eqnarray}
\int_0^\infty\frac{x^4e^x}{(e^x-1)^2}\mathrm{d}x&=&\int_0^\infty\frac{x^4e^{-x}}{(1-e^{-x})^2}\mathrm{d}x\\
&=&\int_0^\infty x^4e^{-x}\sum_{n=0}^\infty(n+1)e^{-nx}\mathrm{d}x\\
&=&\sum_{n=0}^\infty\int_0^\infty(n+1)x^4e^{-(n+1)x}\mathrm{d}x\\
&=&\sum_{n=0}^\infty \frac{24}{(n+1)^4}\\
&=&24\zeta(4)\\
&=&\frac{4\pi^4}{15}.
\end{eqnarray}
