TITLE: Calculate $\int_0^1 \frac{-\ln(1-x)}{x} d x$ without $\zeta(2)$.
QUESTION [5 upvotes]: Can we calculate the following integral without the need of $\zeta(2)$, I actually believe that this can be a method to find the accurate value of $\sum_{n\geq 1}n^{-2}$. 
$$\int_0^1 -\frac{\log (1-x)}{x} \ \mathrm{d}x$$

Any help is appreciated , just act as you do not know that the sum of reciprocal of squares equal $ \frac{\pi^2}{6}$.
I hope this topic is not duplicated.

REPLY [3 votes]: $$\int_0^1-\frac{\ln (1-x)}{x}\,dx=\int_0^1\int_0^1\frac{1}{1-xy}\,dy\,dx$$
Now got to page 10 of this article.