Evaluating $\int_0^1 (\ln(x)\ln(1-x))^n dx$

Question

Interested in evaluating:

$\displaystyle\int_0^1 (\ln(x)\ln(1-x))^n \mathrm{d}x,$ where $n \in \Bbb Z^+.$

I don't really know how to tackle this problem for $n > 1.$

I was able to get to this step: $\displaystyle\int_0^1 (\ln(x)\ln(1-x))^n \mathrm{d}x= \int_0^1\left(\sum_{n=1}^\infty\frac{x^n}n\ln x\right)^n \mathrm{d}x$.

This question is related to: Evaluate $ \int_{0}^{1} \ln(x)\ln(1-x)\,dx $

Answer 1

$$B(x,y)=\int_0^1 t^{x-1}(1-t)^{y-1}dt$$ Where $B(x,y)$ is the Beta Function. Thus $$\int_0^1 \left(\ln(t)\ln(1-t)\right)^ndt=\left(\frac{d}{dy}\frac{d}{dx}\right)^nB(x,y)|_{(x,y)=(1,1)}$$