It is well known that $$\int_0^1\frac{\log{x}}{1 - x}\,\mathrm{d}x = -\frac{\pi^2}{6} $$
This is generally proved by expanding the geometric series.
My question is: can this be done in reverse?
Can we evaluate the integral $\int_0^1\frac{\log{x}}{1 - x}dx$ using other method, for example, differentiation under the integral sign, and thus prove $\zeta(2) = \frac{\pi^2}6$?
Credits to Daniele Ritelli: $$\begin{eqnarray*}\zeta(2)&=&\frac{4}{3}\sum_{n=0}^{+\infty}\frac{1}{(2n+1)^2}=\color{red}{\frac{4}{3}\int_{0}^{1}\frac{\log y}{y^2-1}dy}\\&=&\frac{2}{3}\int_{0}^{1}\frac{1}{y^2-1}\left[\log\left(\frac{1+x^2 y^2}{1+x^2}\right)\right]_{x=0}^{+\infty}dy\\&=&\frac{4}{3}\int_{0}^{1}\int_{0}^{+\infty}\frac{x}{(1+x^2)(1+x^2 y^2)}dx\,dy\\&=&\frac{4}{3}\int_{0}^{1}\int_{0}^{+\infty}\frac{dx\, dz}{(1+x^2)(1+z^2)}=\frac{4}{3}\cdot\frac{\pi}{4}\cdot\frac{\pi}{2}=\color{red}{\frac{\pi^2}{6}}.\end{eqnarray*}$$
