$\int_0^{1}\frac{\ln(1-x)}{x}dx=I$

Question

$$\int_0^{1}\frac{\ln(1-x)}{x}dx=I$$ I tried to do it as usual but I am finding it hard to do. I have calculated the first integral as this but struck after that.

$$-\displaystyle\sum\limits_{i=1}^{\infty}\frac{1}{n^2}$$ Please help.

Answer 1

Seems you went the correct way, expanding the log into Taylor series, and got the correct expression just typed wrong index. Yours is a famous series, $$ \sum_{k=1}^\infty \frac1{k^2} = \frac{\pi^2}{6} $$