TITLE: Ways to calculate $\int_0^1 \frac{-\log x}{1+x}\ \mathrm dx$
QUESTION [6 upvotes]: I came across the integral
$$
\int_0^1 \frac{-\log x}{1+x}\ \mathrm dx = \frac{\pi^2}{12},
$$
which can be calculated as $\frac 1 2 \zeta(2)$ using analytic number theory.
I'm interested if this integral can be calculated in any other interesting, possibly more elementary ways?

REPLY [4 votes]: You can simply expand the $1/(1+x)$ piece into its equivalent geometric series and get
$$-\sum_{n=0}^{\infty} (-1)^n \int_0^1 dx \, x^n \log{x}$$
One may show using integration by parts that the integral in the sum is simply $-1/(n+1)^2$.  The result is simply
$$\sum_{n=0}^{\infty} \frac{(-1)^n}{(n+1)^2} = \frac{\pi^2}{12}$$