$\displaystyle \int_{0}^{\infty}\frac{x^2}{e^x-1}dx$
We normally like to see some work done. Anyway, you are going to need to rewrite the integral like so
$\displaystyle \lim_{t\to \infty} \int_0^t\! x^2 (e^x -1)^{-1}\, dx$
and then integrate by parts. Let $\displaystyle u = x^2$ and then $\displaystyle dv = (e^x-1)^{-1}\, dx$.
See what you can do with that.
With the successive substitutions $\displaystyle x= \ln \xi$ and $\displaystyle \xi = \frac{1}{t}$ the integral becomes...
$\displaystyle \displaystyle \int_{0}^{\infty} \frac{x^{2}}{e^{x}-1}\ dx = \int_{0}^{1} \frac{\ln^{2} t}{1-t}\ dt$ (1)
Now with a little of patience You can verify that is...
$\displaystyle \displaystyle \int_{0}^{1} t^{m}\ \ln^{n} t\ dt = (-1)^{n}\ \frac{n!}{(m+1)^{n+1}}$ (2)
... and because for $\displaystyle |t|<1$ is...
$\displaystyle \displaystyle \frac{1}{1-t} = \sum_{m=0}^{\infty} t^{m}$ (3)
... combining (1), (2) and (3) we arrive to write...
$\displaystyle \displaystyle \int_{0}^{1} \frac{\ln^{2} t}{1-t}\ dt = \sum_{m=0}^{\infty} \int_{0}^{1} t^{m}\ \ln^{2} t\ dt = 2\ \sum_{m=0}^{\infty} \frac{1}{(m+1)^{3}} = 2\ \zeta (3)$ (4)
... where $\displaystyle \zeta(*)$ is the 'Riemann Zeta Function'...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$