How to prove that $\int_{1}^{\sqrt{2}+1}\frac{\ln{x}}{x^{2}-1}dx=\frac{\pi^{2}}{16}-\frac{\ln^{2}\left(\sqrt{2}+1\right)}{4}$ 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. 
 A: 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*}
A: A Complete Solution Now
Consider 
$$F(s)=\int_1^s\frac{\log x}{x^2-1}\mathrm dx$$
Like before,
$$F(s)=\frac12\int_1^{s}\frac{\log x}{x-1}\mathrm dx-\frac12\int_1^{s}\frac{\log x}{x+1}\mathrm dx$$
$$F(s)=-\frac12\mathrm{Li}_2(1-s)-\frac12J(s)$$
For $J(s)$, we integrate by parts with $\mathrm dv=\frac{\mathrm dx}{1+x}$ to get 
$$J(s)=\log(s)\log(s+1)-\int_1^s\frac{\log(1+x)}{x}\mathrm dx$$
$$J(s)=\log(s)\log(s+1)+\int_2^{1+s}\frac{\log x}{1-x}\mathrm dx$$
$$J(s)=\log(s)\log(s+1)+\mathrm{Li}_2(1-x)\bigg|_2^{1+s}$$
$$J(s)=\log(s)\log(s+1)+\mathrm{Li}_2(-s)-\mathrm{Li}_2(-1)$$
Using $\mathrm{Li}_k(-1)=(2^{1-k}-1)\zeta(k)$,
$$J(s)=\log(s)\log(s+1)+\mathrm{Li}_2(-s)+\frac{\pi^2}{12}$$
Plugging in $s=1+\sqrt2$, 
$$J(1+\sqrt2)=\log(1+\sqrt2)\log(2+\sqrt2)+\mathrm{Li}_2(-1-\sqrt2)+\frac{\pi^2}{12}$$
Which, as The OP noted, becomes
$$J(1+\sqrt2)=\log^2(1+\sqrt2)+\frac12\log(2)\log(1+\sqrt2)+\mathrm{Li}_2(-1-\sqrt2)+\frac{\pi^2}{12}$$
And again as the OP noted,
$$F(1+\sqrt2)=-\frac{\pi^2}{48}-\frac14\log^2(1+\sqrt2)+\frac14\log2\log(1+\sqrt2)+\frac1{16}\log^22+\frac12\mathrm{Li}_2\bigg(\frac1{\sqrt2}\bigg)-\frac12\mathrm{Li}_2(1-\sqrt2)$$
And since 
$$\frac12\mathrm{Li}_2\bigg(\frac1{\sqrt2}\bigg)-\frac12\mathrm{Li}_2(1-\sqrt2)=\frac{\pi^2}{12}-\frac14\log2\log(1+\sqrt2)-\frac1{16}\log^22$$
We have 
$$F(1+\sqrt2)=\frac{\pi^2}{16}-\frac14\log^2(1+\sqrt2)$$
