Integral involving a logarithm and a linear rational function $$\int_{0}^{1} \frac{\log x}{x-1}dx$$
I was wondering: is it possible to evaluate this integral with real methods? Playing around with a series expansion I was able to recognize that the integral is equal to $\zeta(2)$, but since I don't know how to evaluate that without the parseval identity (I haven't really studied it yet, so it would be cheating :P) that road isn't feasible. I was thinking maybe of some tricks using differentiation under the integral sign or the method used to evaluate $\displaystyle\int_{0}^{\infty}\frac{\sin x}{x}$ (recognize that $\displaystyle\int_{0}^{a}\frac{\sin x}{x}dx=\int_{0}^{a}dx\int_{0}^{\infty}e^{-xy}\sin x\ dy$ and use the Fubini-Tonelli Theorem) but I couldn't get anywhere with those. 
 A: Note that
$$
\int_z^1\frac{\ln x}{1-x}\ dx=-\text{Li}_2(1-z)=-\sum_{n=1}^\infty\frac{(1-z)^n}{n^2}.
$$
In this case, we have $z=0$ and
$$
\int_0^1\frac{\ln x}{1-x}\ dx=-\sum_{n=1}^\infty\frac{1}{n^2}=-\zeta(2)=\large\color{blue}{-\frac{\pi^2}{6}}.
$$

Another way to evaluate the integral is using substitution $u=1-x$ and the integral turns out to be
$$
\int_0^1\frac{\ln (1-u)}{u}\ du.\tag1
$$
Use Maclaurin series  for Natural logarithm:
$$
\ln(1-u)=-\sum_{n=1}^\infty \frac{u^{n}}{n}.\tag2\\
$$
Substituting $\,(2)$ to $\,(1)$, yields
$$
\begin{align}
\int_0^1\frac{\ln(1-u)}{u}\,du&=-\int_0^1\sum_{n=1}^\infty \frac{u^{n}}{un}\,du\\
&=-\sum_{n=1}^\infty\int_0^1 \frac{u^{n-1}}{n}\,du\\
&=-\sum_{n=1}^\infty \left.\frac{u^{n}}{n^2}\right|_{u=0}^1\\
&=-\sum_{n=1}^\infty \frac{1}{n^2}.\tag3
\end{align}
$$
The infinite series in $(3)$ is defined as Riemann zeta function $\,\zeta (2)=\large\dfrac{\pi^2}{6}$.
A: Another way: it's actually easier to expand the denominator: $-\frac{1}{1-x} = \sum_{k=0}^{\infty} x^k$ and since the bounds on the integral are $0$ and $1$, $x^k$ converges uniformly, and you can interchange summation and integration. You get an integral of the form 
$$
I_k = - \int_{0}^{1}x^k \log x dx
$$
which are easily solved by parts, and then sum over $k$. 
