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

Question

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.

Answer 1

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.