Difficult general integral definite 0 to 1 $$\int_{0}^{1} \log^2(x)\cdot x^{k+1} dx$$
I tried integration by parts but it leads to an extremely complicated computation, which didnt lead me anywhere.
Then
I tried differentiating the beta function. That was partly successful. But the problem was when I substituted $k+2$ in and then the digammas and trigamma acted out. 
Any help?
Thanks! 
Series, and ANY type EXCEPT COMPLEX ANALYSIS is welcome.

The two answers are excellent, but there is a third giving something interesting, i.e. a relation to the Gamma function:
$$\int_0^1 \log(x)^2 x^{k+1}\ dx=$$
Substituting $u=-\log(x)$:
$$\int_0^{\infty} u^2 \exp(-(k+1)u)\exp(-u)\ du=$$
Substituting $v=(k+2)u$:
$$(k+2)^{-3}\int_0^{\infty} v^2 \exp(-v)\ dv= (k+2)^{-3} \Gamma(3)=2(k+2)^{-3} $$
