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?
 A: $$\begin{align} \int_0^1 \frac{\ln x}{x-1} dx &=\int_0^1 \frac{\sum_{k=1}^\infty \frac{-(-1)^k(-1+x)^k}{k}}{x-1} dx \\~\\
&=\int_0^1 \sum_{k=1}^\infty \frac{-(-1)^k(-1+x)^{k-1}}{k} dx \\~\\
&= \sum_{k=1}^\infty \frac{-(-1)^k}{k}\int_0^1(-1+x)^{k-1} dx \\~\\
&= \sum_{k=1}^\infty \frac{(-1)^k}{k}\dfrac{(-1)^k}{k} \\~\\
&= \sum_{k=1}^\infty \frac{1}{k^2} \\~\\
\end{align}$$
A: HINT:
use $$\sum_{n=0}^{\infty}x^n=\frac{1}{1-x}$$
$$\int_0^1\frac{\ln x}{x-1}dx=-\sum_{n=0}^{\infty}\int_0^1x^n\ln x dx=\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}$$
A: Note that the power series for $\ln$ may be integrated termwise within $(0,1)$.
$$\int_{\epsilon}^{1-\delta}\frac{\ln x}{x-1}dx=-\int_{\delta}^{1-\epsilon}\frac{\ln (1-x)}{x}dx \\= \int_{\delta}^{1-\epsilon}\sum_{k=1}^{\infty}\frac{x^{k-1}}{k}dx\\=\sum_{k=1}^{\infty}\int_{\delta}^{1-\epsilon}\frac{x^{k-1}}{k}dx \\=\sum_{k=1}^{\infty}\frac{1}{k^2}[(1-\epsilon)^k-\delta^k].$$
Since $\sum 1/k^2$ is convergent, then taking one-sided limits as $\delta, \epsilon \to 0$ is justified by the Abel limit theorem.
A: 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.
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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 \newcommand{\Li}[1]{\,{\rm Li}_{#1}}
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\begin{align}&\overbrace{%
\color{#66f}{\large\int_{0}^{1}\frac{\ln\pars{x}}{x - 1}\,\dd x}}
^{\ds{\dsc{x}\ \mapsto\ \dsc{1 - x}}}\ =\
=\int_{0}^{1}\ \overbrace{\frac{-\ln\pars{1 - x}}{x}}
^{\ds{=\ \dsc{\Li{2}'\pars{x}}}}\,\dd x\ =\
\int_{0}^{1}\Li{2}'\pars{x}\,\dd x=\
\overbrace{\Li{2}\pars{1}}^{\ds{=\ \dsc{\frac{\pi^{2}}{6}}}}\ -\
\overbrace{\Li{2}\pars{0}}^{\ds{=\ \dsc{0}}}
\\[5mm]&=\color{#66f}{\large\frac{\pi^{2}}{6}}
\end{align}

$\ds{\Li{\rm s}}$ is the PolyLogarithm Function where we used
  $\ds{\Li{\rm s}'\pars{x}=\frac{\Li{\rm s - 1}\pars{x}}{x}}$ and $\ds{\Li{1}\pars{x} = -\ln\pars{1 - x}}$. Note that
  $\ds{\Li{1}\pars{1}=\sum_{n=1}^{\infty}{1^{n} \over n^{2}}
     =\sum_{n=1}^{\infty}{1 \over n^{2}}=\frac{\pi^{2}}{6}}$.

