TITLE: Conjecture $\sum_{n=1}^\infty\frac{\ln(n+2)}{n\,(n+1)}\,\stackrel{\color{gray}?}=\,{\large\int}_0^1\frac{x\,(\ln x-1)}{\ln(1-x)}\,dx$
QUESTION [13 upvotes]: Numerical calculations suggest that
$$\sum_{n=1}^\infty\frac{\ln(n+2)}{n\,(n+1)}\,\stackrel{\color{gray}?}=\,\int_0^1\frac{x\,(\ln x-1)}{\ln(1-x)}\,dx=1.553767373413083673460727...$$
How can we prove it? Is there a closed form expression for this value?

REPLY [4 votes]: Additionnally, by applying Theorem $2$ (here), one obtains a closed form in terms of the poly-Stieltjes constants.
Proposition. We have

$$
\begin{align}
\int_0^1\frac{x\,(\ln x-1)}{\ln(1-x)}\,dx= \ln 2+\gamma_1(2,0)-\gamma_1(1,0)
\tag1
\\\\\sum_{n=1}^\infty\frac{\ln(n+2)}{n\,(n+1)}=\ln 2+\gamma_1(2,0)-\gamma_1(1,0)
\tag2
\end{align}$$ 

where 

$$
\gamma_1(a,b) = \lim_{N\to+\infty}\left(\sum_{n=1}^N \frac{\log (n+a)}{n+b}-\frac{\log^2 \!N}2\right).
$$