How to compute the integral $\int_0^1 \frac{x\ln x}{\ln (1-x)}dx$? 
How to compute the integral $\int_0^1 \frac{x\ln x}{\ln (1-x)}dx$?

We can write 
\begin{align*}
\int_0^1 \frac{x\ln x}{\ln (1-x)}dx=\int_0^1\frac{(1-u)\ln (1-u)}{\ln u}du
=-\int_0^1\frac{1-u}{\ln u}\sum_{n=1}^\infty \frac{u^n}{n}du
\end{align*}
Seeing
\begin{align*}
\int_0^1\frac{u^{a-1}-u^{b-1}}{\ln u}du=\ln\frac{a}{b},\forall a,b>0,
\end{align*}
we get 
\begin{align*}
\int_0^1 \frac{x\ln x}{\ln (1-x)}dx=\sum_{n=1}^\infty \frac{1}{n}\ln \frac{n+2}{n+1}.
\end{align*}
I have tried hard to compute the series above ,but without any progress.

One thig we could do is to expand as series
$$\frac{x}{\log (1-x)}=-1+\frac{x}{2}+\frac{x^2}{12}+\frac{x^3}{24}+\frac{19 x^4}{720}+\frac{3
   x^5}{160}+\frac{863 x^6}{60480}+\frac{275 x^7}{24192}+\frac{33953
   x^8}{3628800}+O\left(x^9\right)$$ where the coefficients correspond to the absolute value of Gregory  coefficients, that is to say
$$\frac{x \log (x)}{\log (1-x)}=-\log(x)+\sum_{n=1}^\infty |G_n| x^n \log(x)$$ Integrating termwise
$$\int_0^1\frac{x \log (x)}{\log (1-x)}\,dx=1-\sum_{n=1}^\infty |G_n|\frac{H_{n+1}}{n+1}$$ which generates the sequence
$$\left\{1,\frac{7}{8},\frac{187}{216},\frac{2983}{3456},\frac{372419}{432000},\frac{
   186097}{216000},\frac{1994053}{2315250},\frac{2721985571}{3161088000},\frac{1984
   061289739}{2304433152000},\cdots\right\}$$ which seems to converge quite fast.
