Integrate \(\int_{0}^{\infty} \frac{x^{4}}{(\mathrm{e}^{x}-1)^{2}} \mathrm{d} x \)

How to evaluate
\begin{equation}
\int\limits_{0}^{\infty} \frac{x^{4}}{(\mathrm{e}^{x}-1)^{2}} \mathrm{d} x
\end{equation}
was a question on Mathematics Stack Exchange which for some inexplicable reason has been closed. Here is another solution.

Let
\begin{equation}
f(x) = \frac{1}{(\mathrm{e}^{x}-1)^{2}}
\end{equation}
Then the Mellin transform of the function is
\begin{equation}
\mathcal{M}[f(x)](s) = \int\limits_{0}^{\infty} \frac{x^{s-1}}{(\mathrm{e}^{x}-1)^{2}} \mathrm{d} x
= \Gamma(s)[\zeta(s-1) – \zeta(s)]
\end{equation}
For \(s=5\), we have
\begin{equation}
\int\limits_{0}^{\infty} \frac{x^{4}}{(\mathrm{e}^{x}-1)^{2}} \mathrm{d} x = \Gamma(5)[\zeta(4) – \zeta(5)]
= \frac{4\pi^{4}}{15} – 24\zeta(5)
\end{equation}

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