Ways to calculate $\int_0^1 \frac{-\log x}{1+x}\ \mathrm dx$

Question

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?

Answer 1

Integrating by part, you arrive to

$$-\text{Li}_2(-x)-\log (x) \log (x+1)$$

and, using your bounds, the result is $\frac{\pi ^2}{12}$

Answer 2

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

$$\sum_{n=0}^{\infty} \frac{(-1)^n}{(n+1)^2} = \frac{\pi^2}{12}$$