How to do integral $\int_{0}^{\infty} \frac{x^4 e^{x}}{(e^x-1)^2}\,dx$ $$\int_{0}^{\infty} \frac{x^4 e^{x}}{(e^x-1)^2}\,dx$$ This integral came from the Debye theory of the molecular heat capacity of crystal. I heard that this integral form is related to Riemann Zeta function and Gamma function. Also, i heard that it can be solved by using Feynmann's Trick. How can i evaluate this integral?

The Riemann/Gamma way:$$\int_0^\infty\frac{x^4e^{-x}dx}{(1-e^{-x})^2}=\sum_{n\ge 1}n\int_0^\infty x^4e^{-nx}dx=\sum_{n\ge 1}\frac{4!}{n^4}=24\zeta(4)=\frac{4\pi^4}{15}.$$