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.
 A: Thijs solution is very elegant. But the computation can be done without appealing to any theorems (you need one if you want to justify differentiation under the integral sign). 
We have, integrating by parts twice,
\begin{align}
\int_{0}^{1} \log^2(x)\cdot x^{k+1} dx&=\left.\phantom{\int\!\!\!\!}\frac{x^{k+2}}{k+2}\,\log^2x\right|_0^1-\frac2{k+2}\left.\phantom{\int\!\!\!\!}\frac{x^{k+2}}{k+2}\log x\right|_0^1+\frac2{(k+2)^2}\int_0^1x^{k+1}dx\\
&=\frac2{(k+2)^2}\int_0^1x^{k+1}dx\\
&=\frac2{(k+2)^3}
\end{align}
A: First, consider this integral:
$$I(k)=\int_{0}^{1}x^{k+1} dx=\frac{1}{k+2}$$
The desired integral is:
$$I''(k)=\int_0^1\log^2(x)\cdot x^{k+1}dx=\frac{2}{(k+2)^3}$$
A: 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 $v=(k+2)u$:
$$(k+2)^{-3}\int_0^{\infty} v^2 \exp(-v)\ dv= (k+2)^{-3} \Gamma(3)=2(k+2)^{-3} $$
