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.
 A: 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:
$$ \int_{0}^{1}\frac{\log(1-x)}{x}\,dx\stackrel{x\mapsto 1-x}{=}\int_{0}^{1}\frac{\log x}{1-x}\,dx = -\frac{4}{3}\int_{0}^{+\infty}\frac{\arctan(x)}{1+x^2}\,dx = \color{red}{-\frac{\pi^2}{6}}.$$
A: From the Maclaurin series of $\ln(1-x)$,$$f(a)=-\sum_{k\ge1}\frac1ka^k\int_0^1x^{k-1}dx=-\sum_{k\ge1}\frac{a^k}{k^2}=-\operatorname{Li}_2(a).$$In particular, $f(1)=-\zeta(2)=-\frac{\pi^2}{6}$.
A: $\int_{0}^{1}\frac{\ln(1-x)}{x} dx=-\sum_{k=1}^{\infty} \int_{0}^{1}\frac{x^{k- 
1}}{k}dx= -\sum_{k=1}^{n} \frac{1}{k^2}=-\zeta(2)=-\frac{\pi^2}{6}.$
A: $$
\begin{aligned}
\int_0^1 \frac{\ln (1-x)}{x} d x =  & \stackrel{x\mapsto 1-x}{\int_0^1 \frac{\ln x}{1-x} d x }\\
=& \int_0^1\left(\sum_{k=0}^{\infty} x^k \ln x\right) d x\\=&\sum_{k=0}^{\infty} \int_0^1 \ln x d\left(\frac{x^{k+1}}{k+1}\right) \\
=& \sum_{k=0}^{\infty}\left(\left[\frac{x^{k+1}}{k+1} \ln x\right]_0^1-\int_0^1 \frac{x^k}{k+1} d x\right) \quad \textrm{ (IBP)} \\
=&-\sum_{k=0}^{\infty} \frac{1}{(k+1)^2}\\=&-\frac{\pi^2}{6}
\end{aligned}
$$
