I recently got stuck on evaluating the following integral. I do not know an effective substitution to use. Could you please help me evaluate:

$$\int_{0}^{\infty} \ln\left(1 - e^{-x}\right) \,\mathrm dx $$

One route to evaluating the integral is
$$-\int_0^\infty \ln(1-e^{-x})dx=\int_0^\infty\left(e^{-x}+\frac{e^{-2x}}{2}+\frac{e^{-3x}}{3}+\cdots\right)dx $$
$$=\int_0^\infty e^{-x}dx+\frac{1}{2}\int_0^\infty e^{-2x}dx+\frac{1}{3}\int_0^\infty e^{-3x}dx+\cdots$$
$$=1+\frac{1}{2}\cdot\frac{1}{2}+\frac{1}{3}\cdot\frac{1}{3}+\frac{1}{4}\cdot\frac{1}{4}\cdots $$
$$=\zeta(2)=\frac{\pi^2}{6}.$$

I don't know if a straightforward substitution could get you the answer, what with this being the Riemann zeta function and all, but you can see the integrals on the linked page and try your own hand at finding one. (Or someone else can try their hand.)
