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.
 A: 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}
A: I am not sure that I could find a closed formula for the result.
Beside numerical integration, I should use the classical series expansion of $\log(1-x)$ and use the long division to get
$$\frac 1 {\log(1-x)}=-\frac{1}{x}+\frac{1}{2}+\frac{x}{12}+\frac{x^2}{24}+\frac{19 x^3}{720}+\frac{3
   x^4}{160}+\frac{863 x^5}{60480}+\frac{275 x^6}{24192}+O\left(x^7\right)$$ making
$$\frac {x \log(x)} {\log(1-x)}=\log(x) \left(-1+\frac{x}{2}+\frac{x^2}{12}+\frac{x^3}{24}+\frac{19 x^4}{720}+\frac{3
   x^5}{160}+\frac{863 x^6}{60480}+\frac{275 x^7}{24192}+O\left(x^8\right)\right)$$ and now we face the problem of $$I_n=\int x^n \log(x) \,dx=\frac{x^{n+1} ((n+1) \log (x)-1)}{(n+1)^2}$$ that is to say
$$J_n=\int_0^1 x^n \log(x) \,dx=-\frac{1}{(n+1)^2}$$ Using this truncated series, we should end with $\frac{2721985571}{3161088000}\approx 0.861091$ while the numerical integration would give $0.860620$.
A: Different approach 
let $I$ denotes our integral $\int_0^1\frac{(1-x)\ln(1-x)}{\ln x}\ dx$ and let $I_n=\int_0^1\frac{(1-x^n)\ln(1-x)}{\ln x}\ dx,\quad I_0=0$ and $I_1=I$
$$I^{\large'}_n=-\int_0^1x^{n}\ln(1-x)\ dx=\sum_{k=1}^\infty \frac{1}{k}\int_0^1 x^{n+k} \ dx=\sum_{k=1}^\infty\frac{1}{k(n+k+1)}$$
Then $$I=I_1=\int_0^1I^{\large'}_n\ dn=\sum_{k=1}^\infty\frac1k\int_0^1\frac{dn}{n+k+1}=\sum_{k=1}^\infty\frac{{\ln(k+2)-\ln(k+1)}}{k}$$
A: One way to do this is going by Taylor series expansion which gives the indefinite integral$$I\mbox{ (indefinite)}=x(1-\log{x})+O(x^2)$$To find the definite integral, I recommend using the computational intelligence (like Wolfram, Mathlab, etc.) because using the expansion you can only get an approx value which is $$I\approx 0.8606...$$and there is no standard result for the definite integral.
