How to evalute $\int_{0}^{1}\frac{x\log x}{\log(1-x)}dx$ 
Evaluate $$I=\int_{0}^{1}\frac{x\log x}{\log(1-x)}dx$$

I try to change it to:$$I=\int_{0}^{1}\frac{(1-x)\log(1-x)}{\log x}dx$$
but I can't do next steps,help me ,thank you very much.

Partial solution
\begin{align}
I&=\int_0^1\frac{1-x}{\ln x}\ln(1-x)\ dx=\int_0^1\left(-\int_0^1x^y\ dy\right)\ln(1-x)\ dx\\
&=\int_0^1\left(-\int_0^1x^y\ln(1-x)\ dx\right)\ dy=\int_0^1\left(\sum_{n=1}^\infty\frac1n\int_0^1x^{n+y}\ dx\right)\ dy\\
&=\int_0^1\left(\sum_{n=1}^\infty\frac{1}{n(n+y+1)}\right)\ dy=\sum_{n=1}^\infty\frac1n\int_0^1\frac{dy}{n+y+1}\\
&=\sum_{n=1}^\infty\frac{\ln(n+2)-\ln(n+1)}{n}
\end{align}
