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. 

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 $k$. 
