Prove that $\int_0^1 \frac{\ln x}{x-1} dx$ converge. 
Prove that $\int_0^1 \frac{\ln x}{x-1} dx$ converges.

We cannot apply Abel's/Dirichliet's tests here (For example, Dirichliet's test demands that for $g(x)=\ln x$, $\int_0^1 g(x)dx < \infty$ which isn't true).   
I also tried to compare the integral to another;
Since $x>\ln x$ I tried to look at  $\int_0^1 \frac{x}{x-1} dx$ but this integral diverges.
What else can I do?
EDIT:
Apparently I also need to show  that the integral equals $\sum_{n=1}^\infty \frac{1}{n^2}$.
I used WolframAlpha and figured out that the expansion of $\ln x$ at $x=1$ is $\sum_{k=1}^\infty \frac{(-1)^k(-1+x)^k}{k}$. Would that be helpful?

Since:
$$ \int_{0}^{1}\frac{dt}{1-xt}=-\frac{\log(1-x)}{x} $$
and:
$$ I=\int_{0}^{1}\frac{\log x}{x-1}\,dx = -\int_{0}^{1}\frac{\log(1-x)}{x}\,dx $$
we have:
$$ I = \iint_{(0,1)^2}\frac{1}{1-xt}\,dt\,dx = \sum_{n\geq 0}\iint_{(0,1)^2}(xt)^n\,dt\,dx = \sum_{n\geq 0}\frac{1}{(n+1)^2}=\color{red}{\zeta(2)}$$
as wanted.
