Evaluate $ \int_{0}^{1} \ln(x)\ln(1-x)\,dx $ Evaluate the integral, 
$$ \int_{0}^{1} \ln(x)\ln(1-x)\,dx$$ 
I solved this problem, by writing power series and then calculating the series and found the answer to be $ 2 -\zeta(2) $, but I don't think that it is best solution to this problem.   I want to know if it can be solved by any other nice/elegant method. 

\begin{align}
\int_0^1\ln(1-x)\ln x\ dx&=\int_0^1\sum_{n=1}^\infty\frac{x^n}n\ln x\ dx\\
&=\sum_{n=1}^\infty\frac{1}n\int_0^1 x^n\ln x\ dx\\
&=-\sum_{n=1}^\infty\frac{1}n\cdot\frac1{(n+1)^2}\\
&=\sum_{n=1}^\infty\left[\frac{1}n-\frac1{n+1}-\frac1{(n+1)^2}\right]\\
&=1-\left[\sum_{n=1}^\infty\frac1{n^2}-1\right]\\
&=\large\color{blue}{2-\zeta(2)=2-\frac{\pi^2}6}.
\end{align}

Note :
$\displaystyle\ \ \int_0^1 x^\alpha \ln^n x\ dx=\frac{(-1)^n n!}{(\alpha+1)^{n+1}}, \qquad\text{for }\  n=0,1,2,\ldots$
