prove $\int^\infty_0\frac x{e^x-1}dx=\frac{\pi^2}{6}$

prove $\int^\infty_0\frac x{e^x-1}dx=\frac{\pi^2}{6}$

I know that $$\int^\infty_0\frac x{e^x-1}dx=\frac{\pi^2}{6}$$ For
substitute $u=2$ into
$$\zeta(u)\Gamma(u)=\int^\infty_0\frac{x^{u-1}}{e^x-1}dx$$
However, I suspect that there is an easier proof, maybe by the use of
complex analysis. I haven't learnt zeta function yet. All I know is the
above formula and $\zeta(2)=\frac{\pi^2}{6}$. But I am wondering if we can
use the above integral to find out some of the $\zeta$'s value.