TITLE: How to prove that $\int_{1}^{\sqrt{2}+1}\frac{\ln{x}}{x^{2}-1}dx=\frac{\pi^{2}}{16}-\frac{\ln^{2}\left(\sqrt{2}+1\right)}{4}$
QUESTION [9 upvotes]: How to prove that $$\int_{1}^{\sqrt{2}+1}\frac{\ln{x}}{x^{2}-1}dx=\frac{\pi^{2}}{16}-\frac{\ln^{2}\left(\sqrt{2}+1\right)}{4}?$$
I have encountered this integral recently, and I am aware that one can show how this is true with a substitution $u=\ln{x}$ and then expanding it into a series of integrals of the form $\int_{0}^{\ln{(\sqrt{2}+1)}}ue^{-nu}du$ where each integral is then calculated individually, but it feels pretty brute-forced. Is there any other way to do this? Thanks.

REPLY [9 votes]: Put
\begin{equation*}
I=\int_{1}^{\sqrt{2}+1}\dfrac{\ln x}{x^2-1}\, \mathrm{d}x =[x=1/y] = \int_{\sqrt{2}-1}^{1}\dfrac{\ln y}{y^2-1}\, \mathrm{d}y.\tag{1}
\end{equation*}
After the substitution $ y=\frac{1-z}{1+z}$ and integration by parts we have
\begin{gather*}
I = \int_{0}^{\sqrt{2}-1}\dfrac{\ln\left(\frac{1-z}{1+z}\right)}{-2z}\, \mathrm{d}z = \left[-\dfrac{1}{2}\ln(z)\ln\left(\frac{1-z}{1+z}\right)\right]_{0}^{\sqrt{2}-1}+\int_{0}^{\sqrt{2}-1}\dfrac{\ln z}{z^2-1}\, \mathrm{d}z =\\[2ex]-\dfrac{1}{2}\ln^2(\sqrt{2}+1) + \int_{0}^{\sqrt{2}-1}\dfrac{\ln z}{z^2-1}\, \mathrm{d}z.\tag{2}
\end{gather*}
If we add (1) and (2) we get
\begin{equation*}
2I = -\dfrac{1}{2}\ln^2(\sqrt{2}+1) + \int_{0}^{1}\dfrac{\ln z}{z^2-1}\, \mathrm{d}z.
\end{equation*}
Consequently
\begin{gather*}
I = -\dfrac{1}{4}\ln^2(\sqrt{2}+1) -\dfrac{1}{2}\int_{0}^{1}\left(\sum_{k=0}^{\infty}\ln(z)z^{2k}\right)\, \mathrm{d}z =\\[2ex] -\dfrac{1}{4}\ln^2(\sqrt{2}+1) -\dfrac{1}{2}\sum_{k=0}^{\infty}\int_{0}^{1}\ln(z)z^{2k}\, \mathrm{d}z =\\[2ex]
-\dfrac{1}{4}\ln^2(\sqrt{2}+1) +\dfrac{1}{2}\sum_{k=0}^{\infty}\dfrac{1}{(2k+1)^2} = \dfrac{\pi^2}{16} -\dfrac{1}{4}\ln^2(\sqrt{2}+1). 
\end{gather*}