Improper integral of $\frac{x}{e^x-1}$

Make a change or variables, $e^{-x}=u$, and write the integral as

$$\int_0^\infty \frac{x e^{-x}}{1 - e^{-x}}dx$$

Now substituting gives

$$-\int_0^1 \frac{\log u}{1 - u}du $$

This is a known integral that evaluates to $\dfrac{\pi^2}{6}$.

If you want to prove it, you can use the dilogarithm. Let $1-u=x$, so that

$$-\int_0^1 \frac{\log(1 - x)}{x}dx$$

Now, since we're working on $(0,1)$ it is legitimate to use

$$\frac{{ - \log \left( {1 - x} \right)}}{x} = \sum_{n = 1}^\infty {\frac{{{x^{n - 1}}}}{n}} $$

Integrating termwise gives

$$\int_0^t \frac{-\log(1 - x)}{x} dx = \sum_{n = 1}^\infty \frac{t^n}{n^2} = \mathrm{Li}_2 (t)$$

Evaluating at $t=1$ gives

$$ -\int_0^1 \frac{\log(1 - x)}{x} dx = \sum_{n = 1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$$

Any text book which dicusses the Riemann Zeta function will likely have this.

We have for $\Re(z) \gt 1$, that

$$ \zeta(z) \Gamma(z) = \int_{0}^{\infty} \frac{t^{z-1}}{e^t - 1} \text{d}t$$

Your integral is therefore

$$\zeta(2)\Gamma(2) = \frac{\pi^2}{6}$$

For an online discussion, see this: http://www.math.utah.edu/~milicic/zeta.pdf