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. 

I've found a solution that is interesting, but probably not elegant, and definitely not short.
$I = \displaystyle\int_0^1 \ln(x)\ln(1 - x) dx$
Basic results: 


*

*$\lim\limits_{n \to 0} \dfrac{x^n - 1}{n} = \log x$, or $\lim\limits_{n \to 1}\dfrac{x^{n-1} - 1}{n - 1} = \log x$.

*$\dfrac{d}{dn}\beta(n, n) = 2\beta(n, n)(\psi_0(n) - \psi_0(2n))$ where $\psi_0(n)$ is the digamma function.

*$\dfrac{d^2}{dn^2}\beta(n, n) = 4\beta(n, n)(\psi_0(n) - \psi_0(2n))^2 + 2\beta(n, n)(\psi_1(n) - 2\psi_1(2n))$, where $\psi(1)(n)$ is the polygamma function.

*$\psi_0(1) - \psi_0(2) = -1$ according to the recurrence relation.

*$\psi_1(2) = \psi_1(1) - 1$ according to the recurrence relation.

*$\psi_1(1) = \zeta(2)$.


Solution:
$\begin{align}
I & = \lim\limits_{n \to 1} \displaystyle\int_0^1 \dfrac{(x^{n - 1} - 1)((1 - x)^{n - 1} - 1)}{(n - 1)^2} dx\\
& = \lim\limits_{n \to 1}\displaystyle\int_0^1 \dfrac{x^{n-1}(1-x)^{n-1} - x^{n - 1} - (1-x)^{n-1} + 1}{(n-1)^2} dx\\
& = \lim\limits_{n \to 1} \dfrac{\beta(n,n) - \frac{1}{n} - \frac{1}{n} + 1}{(n-1)^2}\\
& = \lim\limits_{n \to 1} \dfrac{\beta(n,n)(\psi_0(n)-\psi_0(2n)) + \frac{2}{n^2}}{2(n-1)} \quad [\text{l'Hospital's rule}]\\
& = \lim\limits_{n \to 1} \dfrac{4\beta(n,n)(\psi_0(n)-\psi_0(2n))^2 + 2\beta(n,n)(\psi_1(n)-2\psi_1(2n))- \frac{4}{n^3}}{2}\quad [\text{l'Hospital's rule}]\\
& = 2\beta(1, 1)(\psi_0(1) - \psi_0(2))^2 + \beta(1, 1)(\psi_1(1) - 2\psi_1(2)) - 2\\
& = 2(-1)^2 + 1(\psi_1(1) - 2\psi_1(1) + 2) - 2\\
& = 2 - \psi_1(1)\\
& = 2-\zeta(2)
\end{align}$
