This may not be that bad, but give it a go for those who wish.
Let's see what interesting methods are come up with.
$\displaystyle \int_{0}^{1}ln(x)ln(1-x)dx$
We know that $\displaystyle -\sum_{n=0}^{\infty}\frac{x^{n+1}}{n+1}=\ln(1-x)$
so then I think what you can do is this
$\displaystyle -\int{\ln(x)}\sum_{n=0}^{\infty}\frac{x^{n+1}}{n+1} dx=-\sum_{n=0}^{\infty}\frac{1}{n+1}\int\ln(x)x^{n+1}$
$\displaystyle \int_0^{1}\ln(x)\cdot{x^{n+1}}=\bigg[x^{n}\bigg[\frac{x^2\ln(x)}{n+2}-\frac{x^2}{(n+2)^2}\bigg]\bigg]\bigg|_0^{1}=\frac{-1}{(n+2)^2}$
so we have
$\displaystyle \sum_{n=0}^{\infty}\frac{1}{(n+1)(n+2)^2}=2-\frac{\pi^2}{6}$
by PFD and use of a riemann zeta table =D
NOTE: what I mean by PFD was that after I decomposed it it gave a telescoping series and a riemann zeta function, which came out to be $\displaystyle 1-\bigg(\frac{\pi^2}{6}-1\bigg)$, the latter riemann zeta was gotten using the CRC table of values
Very good. I knew you fellas would chew it up and spit it out.
Here is what I done. Which is similar to you fellas line of thought.
$\displaystyle \int_{0}^{1}ln(x)ln(1-x)dx$
$\displaystyle -\int_{0}^{1}ln(x)\sum_{k=1}^{\infty}\frac{x^{k}}{k }dx$
$\displaystyle -\sum_{k=1}^{\infty}\frac{1}{k}\int_{0}^{1}x^{k}ln( x)dx$
$\displaystyle \sum_{k=1}^{\infty}\frac{1}{k(k+1)^{2}}$
$\displaystyle \sum_{k=1}^{\infty}\frac{1}{k(k+1)}-\sum_{k=1}^{\infty}\frac{1}{(k+1)^{2}}$
$\displaystyle =1-\left(\frac{{\pi}^{2}}{6}-1\right)$
$\displaystyle =\boxed{2-\frac{{\pi}^{2}}{6}}$