<s>[INST] How to calculate $\int_0^1 \frac{\ln(1-x)}{x} dx$? I have tried to let$f(a)=\int_0^1\frac{\ln(1-ax)}{x} dx$, which means   
$f'(x)=-a\int_0^1 \frac{1}{x(1-ax)} dx=a\int_0^1 \left[\frac{a}{ax-1}-\frac{1}{x}\right] dx$. 
However, I am stuck at here.
 [/INST] Let us consider 
$$ I = \int_{0}^{+\infty}\frac{\arctan(x)}{1+x^2}\,dx =\frac{1}{2}\left[\arctan^2(x)\right]_{0}^{+\infty}=\frac{\pi^2}{8}$$
and the parametric integral
$$ I(a) = \int_{0}^{+\infty}\frac{\arctan(ax)}{1+x^2}\,dx. $$
We have $I(0)=0$ and $I(1)=I=\frac{\pi^2}{8}$. By differentiation under the integral sign
$$\frac{\pi^2}{8} = \int_{0}^{1}\int_{0}^{+\infty}\frac{x}{(1+x^2)(1+a^2 x^2)}\,dx\,da$$
and by partial fraction decomposition $\frac{x}{(1+x^2)(1+a^2 x^2)}=\frac{1}{1-a^2}\cdot\left(\frac{x}{1+a^2 x^2}-\frac{x}{1+x^2}\right)$, so
$$ \frac{\pi^2}{8}=\int_{0}^{1}\frac{\log(a)}{a^2-1}\,da=-\sum_{n\geq 0}\int_{0}^{1}a^{2n}\log(a)\,da=\sum_{n\geq 0}\frac{1}{(2n+1)^2}. $$
By termwise integration of a Maclaurin series we have
$$ \int_{0}^{1}\frac{\log(x)}{1-x}\,dx = -\sum_{n\geq 1}\frac{1}{n^2}, $$
where the absolute convergence of the RHS allows to state
$$ S=\sum_{n\geq 1}\frac{1}{n^2}=\sum_{m\geq 1}\frac{1}{(2m)^2}+\sum_{m\geq 0}\frac{1}{(2m+1)^2}=\frac{1}{4}S+\frac{\pi^2}{8}. $$
Connecting the dots:
$$ \int_{0}^{1}\frac{\log(1-x)}{x}\,dx\stackrel{x\mapsto 1-x}{=}\int_{0}^{1}\frac{\log x}{1-x}\,dx = -\frac{4}{3}\int_{0}^{+\infty}\frac{\arctan(x)}{1+x^2}\,dx = \color{red}{-\frac{\pi^2}{6}}.$$
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