Finding $S=\sum_2^{\infty}\frac{\ln(n+1)}{(n^2-1)}$ Is there any chance to express the series $$S=\sum_2^{\infty}\frac{\ln(n+1)}{(n^2-1)}$$ in terms of a known function?
My idea is to start from the generalized-Euler-constant function (http://arxiv.org/abs/math/0610499): $$\gamma(z)=\sum_1^{\infty}z^{n-1}\left( \frac1n-\log \frac{n+1}{n}\right)$$ Any hint?
 A: The series admits a closed form in terms of the poly-Stieltjes constants. 

Proposition. Let $$S=\sum_{n=2}^{\infty}\frac{\ln(n+1)}{n^2-1}.$$
Then
  $$
\bbox[15px,border:1px solid orange]{S=\frac{\ln 2}4 -\frac{\gamma_1}2+\frac{\gamma_1\!\left(2,0\right)}2}\tag1
$$ where 
  $$
\begin{align}
&\gamma_1 = \lim_{N\to+\infty}\left(\sum_{n=1}^N \frac{\log n}{n}-\frac12 \log^2 \!N\right) \tag2\\
&\gamma_1(a,b) = \lim_{N\to+\infty}\left(\sum_{n=1}^N \frac{\log (n+a)}{n+b}-\frac12 \log^2 \!N\right). \tag3
\end{align}
$$

Proof. We have
$$
\begin{align}
&\sum_{n=2}^{\infty}\frac{\ln(n+1)}{n^2-1}\\
&=\frac12\lim_{N \to \infty}\left(\sum_{n=2}^N\frac{\ln{(n+1)}}{n-1}-\sum_{n=2}^N\frac{\ln{(n+1)}}{n+1}\right)\\
&=\frac12\lim_{N \to \infty}\left(\sum_{n=1}^{N-1}\frac{\ln{(n+2)}}n-\sum_{n=3}^{N+1}\frac{\ln{n}}{n}\right)\\
&=\frac12\lim_{N \to \infty}\left(\sum_{n=1}^{N-1}\frac{\ln{(n+2)}}n-\frac{\ln^2 N}2\right)-\frac12\lim_{N \to \infty}\left(\sum_{n=1}^{N+1}\frac{\ln{n}}{n}-\frac{\ln^2 N}2\right)+\frac{\ln 2}4\\
&=\frac12\gamma_1\!\left(2,0\right)-\frac12\gamma_1+\frac{\ln 2}4
\end{align}
$$ as announced.
Remark. Observe that $\gamma_1(0,0)$ is just $\gamma_1$, an ordinary Stieltjes constant.
A: Not a closed form, but a numerical evaluation.
$$
\begin{align}
\sum_{k=2}^\infty\frac{\log(k+1)}{k^2-1}
&=\frac12\sum_{k=2}^\infty\log(k+1)\left(\frac1{k-1}-\frac1{k+1}\right)\\
&=\frac12\lim_{n\to\infty}\left(\sum_{k=2}^n\frac{\log(k+1)}{k-1}-\sum_{k=2}^n\frac{\log(k+1)}{k+1}\right)\\
&=\frac12\lim_{n\to\infty}\left(\sum_{k=1}^{n-1}\frac{\log(k+2)}k-\sum_{k=3}^{n+1}\frac{\log(k)}k\right)\\
&=\frac{\log(2)}4+\frac12\sum_{k=1}^\infty\frac{\log\left(1+\frac2k\right)}k\\
\end{align}
$$
Using the Euler-Maclaurin Sum Formula with an error of $O\left(\frac1{k^{23}}\right)$ and using $1000$ terms of the series, we get that
$$
\begin{align}
&\sum_{k=2}^\infty\frac{\log(k+1)}{k^2-1}\\
&=\frac{\log(2)}4+\frac12\sum_{k=1}^\infty\frac{\log\left(1+\frac2k\right)}k\\
&\approx 1.232969586009636405434158525257658981115253440057209139211078744
\end{align}
$$
