$\int_{0}^{\infty}{\left(x^3\cdot(e^x-1)^{-1} \right)}dx$

Let $I$ be your integral.

$$I=\int_0^{\infty}\frac{x^3}{e^x-1}dx=\int_0^{\infty}x^3\sum_{k=1}^{\infty}e^{-kx}dx.$$ Here, you can use $$\begin{align}\int_0^{\infty}x^3e^{-nx}dx&=\int_{0}^{\infty}x^3\left(-\frac{1}{n}e^{-nx}\right)'dx\\&=\left[-\frac{x^3}{ne^{nx}}\right]_{0}^{\infty}-\left(-\frac 3n\right)\int_{0}^{\infty}x^2\left(-\frac 1ne^{-nx}\right)'dx\\&=0+\frac 3n\left[-\frac{x^2}{ne^{nx}}\right]_{0}^{\infty}-\frac 3n\cdot\left(-\frac 2n\right)\int_{0}^{\infty}x\left(-\frac 1ne^{-nx}\right)'dx\\&=0+\frac{3!}{n^2}\left[-\frac{x}{ne^{nx}}\right]_{0}^{\infty}-\frac{3!}{n^2}\cdot\left(-\frac 1n\right)\left[-\frac{1}{ne^{nx}}\right]_{0}^{\infty}\\&=\frac{3!}{n^4}.\end{align}$$ Hence, $$I=3!\sum_{k=1}^{\infty}\frac{1}{k^4}=3!\times \frac{π^4}{90}=\frac{π^4}{15}.$$

P.S. You can see the proof for $\sum_{k=1}^{\infty}1/k^4=\zeta (4)=\pi^4/90$ here. Also, here.