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.  

$$I=\int_0^\infty\frac{x^4e^x}{(e^x-1)^2}\ dx\overset{e^{-x}=y}{=}\int_0^1\frac{\ln^4x}{(1-x)^2}\ dx=\sum_{n=1}^\infty n\int_0^1x^{n-1}\ln^4x\ dx=\sum_{n=1}^\infty\frac{4!}{n^4}=4!\zeta(4)$$
