A reason for the value of $\int_{0}^{1}\log{(x)}\log{(1-x)}\,\mathrm{d}x$

Differentiation under the integral sign gives: $$ \int_{0}^{1}\log(x)\log(1-x)\,dx = \left.\frac{\partial^2}{\partial\alpha\,\partial\beta}\int_{0}^{1}x^\alpha(1-x)^{\beta}\,dx\right|_{\alpha,\beta=0} $$ hence you just have to differentiate a beta function and $\zeta(2)$ arises as $\psi'(1)$.

Differentiation is carried on through: $$ \frac{d}{dz}\,f(z) = f(z)\cdot\frac{d}{dz}\log f(z)$$ and $\psi(z)=\frac{d}{dz}\log\Gamma(z)$.

This has to do with the fact that using integration by parts you get a dilogarithm, and $\text{Li}_2(1)=\zeta(2)$, since the two defining series happen to be the same at $x=1$.

Why $\zeta(2)$ involves $\pi^2$ can be seen as a by-product of the series expansion of $\sin(x)$, as can be read about here, on the solution to the Basel problem. In short one shows that $\sin x /x$ can be factored as a product involving $n^2/\pi^2$, and comparing coefficients gives the desired sum.

Using the integration by parts, one has \begin{eqnarray} \int_0^1\ln x\ln(1-x)dx&=&x\ln x\ln(1-x)|_0^1-\int_0^1x(\frac{\ln(1-x)}{x}-\frac{\ln x}{1-x})dx\\ &=&-\int_0^1(\ln x+\ln(1-x))dx+\int_0^1\frac{\ln x}{1-x}dx\\ &=&2-\frac{\pi^2}{6}. \end{eqnarray} Here $$ \int_0^1\frac{\ln x}{1-x}dx=-\frac{\pi^2}{6} $$ is well-known.