I am stuck in this integral: $\int_{\sqrt{3}}^{\infty} \frac{\ln(x-1)}{x^2-1}dx$ I reduced one of the bounty problems to sum of some integrals. One of them is the following:
$$\int_{\sqrt{3}}^{\infty} \frac{\ln(x-1)}{x^2-1}dx$$
It was a very complicated integral. And nobody gave a full computation as far as I could see.
What is my work? Well. I am really stuck after reduction. I don't know much about $Li_2$  function. I feel like that this is not so difficult if I knew that function well.
Any hint is valuable. Thanks for your good support. If you find my work not enough, please forgive me.
Note: I wanna do/understand this: https://www.wolframalpha.com/input?i=int_%28sqrt%283%29%29%5E%28infty%29+ln%28u-1%29%2F%28u%5E2-1%29
 A: We will prove
$$\int_{\sqrt{3}}^{\infty}\frac{\ln\left(x-1\right)}{x^{2}-1}dx = \frac{\pi^{2}}{12}+\frac{\ln^{2}\left(2\right)}{4}-\frac{1}{4}\left(\ln\left(\sqrt{3}-1\right)\ln\left(6\sqrt{3}-10\right)-2\operatorname{Li}_{2}\left(\frac{1-\sqrt{3}}{2}\right)\right),$$
where
$$\operatorname{Li}_{2}\left(\frac{1-\sqrt{3}}{2}\right) = -\int_{0}^{\frac{1-\sqrt{3}}{2}}\frac{\ln\left(1-t\right)}{t}dt.$$
Proof. We will first evaluate the antiderivative of $\dfrac{\ln\left(x-1\right)}{x^{2}-1}$. By partial fractions, we have
$$\frac{\ln\left(x-1\right)}{x^{2}-1} = \frac{1}{2}\left(\frac{\ln\left(x-1\right)}{x-1}-\frac{\ln\left(x-1\right)}{x+1}\right).$$
Next, we will evaluate the antiderivatives of the two expressions, excluding the $\dfrac{1}{2}$ for simplicity's sake. Also, we will ignore writing $+C$ since we are not looking for general antiderivatives despite using the $\displaystyle\int$ notation.
Trivially,
$$\int_{ }^{ }\frac{\ln\left(x-1\right)}{x-1}dx=\frac{\ln^{2}\left(x-1\right)}{2}.$$
For the next integral, we integrate by parts to get
$$\int_{ }^{ }\frac{\ln\left(x-1\right)}{x+1}dx=\ln\left(x-1\right)\ln\left(x+1\right)-\int_{ }^{ }\frac{\ln\left(1+x\right)}{x-1}dx.$$
Then using $u=x+1$ and $v=\dfrac{u}{2}$, we get
$$
\eqalign{
\int_{ }^{ }\frac{\ln\left(1+x\right)}{x-1}dx &= \int_{ }^{ }\frac{\ln\left(u+2\right)}{u}du \cr
&= \int_{ }^{ }\left(\frac{\ln\left(\frac{u}{2}+1\right)}{u}+\frac{\ln\left(2\right)}{u}\right)du \cr
&= \int_{ }^{ }\frac{\ln\left(\frac{u}{2}+1\right)}{u}du+\ln\left(2\right)\ln\left|x-1\right| \cr
&= -\int_{ }^{ }-\frac{\ln\left(1-v\right)}{v}dv+\ln\left(2\right)\ln\left|x-1\right| \cr
&=-\operatorname{Li}_{2}\left(v\right)+\ln\left(2\right)\ln\left|x-1\right| \cr
&=\ln\left(2\right)\ln\left|x-1\right|-\operatorname{Li}_{2}\left(\frac{1-x}{2}\right). \cr
}
$$
Thus, an antiderivative of the original integrand is
$$F(x):=\frac{1}{2}\left(\frac{\ln^{2}\left(x-1\right)}{2}-\left(\ln\left(x-1\right)\ln\left(x+1\right)-\left(\ln\left(2\right)\ln\left|x-1\right|-\operatorname{Li}_{2}\left(\frac{1-x}{2}\right)\right)\right)\right).$$
We can expand this antiderivative as the series
$$\frac{\pi^{2}}{12}+\frac{\ln^{2}\left(2\right)}{4}+\frac{-\ln\left(x\right)-1}{x}+\frac{1}{2x^{2}}+\frac{1-6\ln\left(x\right)}{18x^{3}}+\frac{1}{3x^{4}}+\frac{\frac{11}{100}-\frac{\ln\left(x\right)}{5}}{x^{5}}+O\left(\frac{1}{x^{6}}\right)$$
and (clearly) see that all the terms except the first two vanish as $x \to \infty$.
If $x=\sqrt{3}$, we get
$$F(\sqrt{3}) = \ \frac{1}{4}\left(\ln\left(\sqrt{3}-1\right)\ln\left(6\sqrt{3}-10\right)-2\operatorname{Li}_{2}\left(\frac{1-\sqrt{3}}{2}\right)\right).$$
We conclude this proof by using the Fundamental Theorem of Calculus to get
$$F(\infty) - F(\sqrt{3}) = \frac{\pi^{2}}{12}+\frac{\ln^{2}\left(2\right)}{4}-\frac{1}{4}\left(\ln\left(\sqrt{3}-1\right)\ln\left(6\sqrt{3}-10\right)-2\operatorname{Li}_{2}\left(\frac{1-\sqrt{3}}{2}\right)\right).$$ Q.E.D.
You can check my Desmos link if you want a little more convincing. Please let me know if there are any questions.
