<s>[INST] Integral Involving Harmonic Numbers: $\int_{\sqrt{3}}^{\infty}\frac{\ln\left(x^{4}-1\right)}{x^{2}-1}dx$ (Motivation) In an attempt to answer this question, I got stuck on evaluating a certain integral. I have made up the following conjecture:
$$\int_{\sqrt{3}}^{\infty}\frac{\ln\left(x^{4}-1\right)}{x^{2}-1}dx = \frac{\pi^{2}}{4}+\frac{3}{2}\ln\left(2\right)\ln\left(\frac{\sqrt{3}+1}{\sqrt{3}-1}\right).$$
(Attempt) Let $H_n$ denote the n-th harmonic number $\displaystyle H_{n}=\sum_{k=1}^{n}\frac{1}{k}$. I will warn this process gets ugly, but it is the best I have so far. Expanding the integrand as a series, we get:
$$
\eqalign{
\int_{\sqrt{3}}^{\infty}\frac{\ln\left(x^{4}-1\right)}{x^{2}-1}dx &= \sum_{n=0}^{\infty}\int_{\sqrt{3}}^{\infty}\frac{4\ln\left(x\right)-H_{n}}{x^{4n+2}}dx+\sum_{n=0}^{\infty}\int_{\sqrt{3}}^{\infty}\frac{4\ln\left(x\right)-H_{n}}{x^{4n+4}}dx, \cr
}
$$
which simplifies down to
$$\sum_{n=0}^{\infty}\left(\frac{H_{n}\sqrt{3}^{-4n-1}}{-4n-1}-\frac{4\ln\left(\sqrt{3}\right)\sqrt{3}^{-4n-1}}{-4n-1}+\frac{4\sqrt{3}^{-4n-1}}{\left(-4n-1\right)^{2}}\right)+\sum_{n=0}^{\infty}\left(\frac{H_{n}\sqrt{3}^{-4n-3}}{-4n-3}-\frac{4\ln\left(\sqrt{3}\right)\sqrt{3}^{-4n-3}}{-4n-3}+\frac{4\sqrt{3}^{-4n-3}}{\left(-4n-3\right)^{2}}\right).$$
Since both series converge, we can split up the first series like
$$-\sum_{n=0}^{\infty}\frac{H_{n}\sqrt{3}^{-4n-1}}{4n+1}+4\ln\left(\sqrt{3}\right)\sum_{n=0}^{\infty}\frac{\sqrt{3}^{-4n-1}}{4n+1}+4\sum_{n=0}^{\infty}\frac{\sqrt{3}^{-4n-1}}{\left(4n+1\right)^{2}}$$
and the second series like
$$-\sum_{n=0}^{\infty}\frac{H_{n}\sqrt{3}^{-4n-3}}{4n+3}+4\ln\left(\sqrt{3}\right)\sum_{n=0}^{\infty}\frac{\sqrt{3}^{-4n-3}}{4n+3}+4\sum_{n=0}^{\infty}\frac{\sqrt{3}^{-4n-3}}{\left(4n+3\right)^{2}}.$$
Next, I found that
$$\sum_{n=0}^{\infty}\frac{\sqrt{3}^{-4n-1}}{4n+1}=\frac{\pi}{12}+\frac{1}{2}\operatorname{arctanh}\left(\frac{1}{\sqrt{3}}\right)$$
and
$$\sum_{n=0}^{\infty}\frac{\sqrt{3}^{-4n-3}}{4n+3}=\frac{1}{2}\operatorname{arctanh}\left(\frac{1}{\sqrt{3}}\right)-\frac{\pi}{12}.$$
However, after trying for a while, I am out of ideas for evaluating the other sums.
I realize I skipped a lot of steps, but that is because I don't want this question to be too long. So for your convenience, I put all of these into Desmos, so I believe my process is correct so far based on numerical approximations.
(Question) Does anyone have an idea of how to evaluate the integral in question, or how to evaluate the sums I am stuck on? Any hints and ideas are appreciated.
(Miscellaneous) Here are some other ideas I have:
$$
\eqalign{
\int_{\sqrt{3}}^{\infty}\frac{\ln\left(x^{4}-1\right)}{x^{2}-1}dx &= \int_{\operatorname{arcsec}\left(\sqrt{3}\right)}^{\frac{\pi}{2}}\frac{\ln\left(\left(\sec x\right)^{4}-1\right)}{\sec^{2}x-1}\sec\left(x\right)\tan\left(x\right)dx \cr
\int_{\sqrt{3}}^{\infty}\frac{\ln\left(x^{4}-1\right)}{x^{2}-1}dx &= \int_{0}^{\infty}\frac{\ln\left(\left(x+\sqrt{3}\right)^{4}-1\right)}{\left(x+\sqrt{3}\right)^{2}-1}dx.
}
$$
Maybe I could construct a keyhole contour for the last integral?
 [/INST] This is not a complete solution but an extended comment.
I was particularly interested to find out what makes the lower integration border $\sqrt{3}$ so special as a prerequisite for an appreciable simplification of the final expression of the integral.
Hence I started with the same decomposition of our integral into the sum of three integrals as @dfnu, but now with a general lower integration border $a$, where $a>1$ to ensure convergence:
$$i_1(a) =\int_a^{\infty } \frac{\log (x-1)}{x^2-1} \, dx$$
$$i_2(a) =\int_a^{\infty } \frac{\log (x+1)}{x^2-1} \, dx$$
and looked particularly at the integral
$$i_3(a) = \int_{a}^{\infty}\frac{\log(x^2+1)}{x^2-1}$$
After a lengthy journey I found the following expression
$$\begin {align} i_3(a) = \frac{\pi ^2}{4}-2 \sum _{k=1}^{\infty } \frac{\left(\frac{\sqrt{a^2+1}}{\sqrt{2}}\right)^k \sin \left(\frac{\pi  k}{4}\right) \sin \left(k \tan ^{-1}(a)\right)}{k^2}\\+\frac{1}{2} \log \left(\frac{a+1}{a-1}\right) \log \left(a^2+1\right)+\frac{1}{2} \pi  \tan ^{-1}\left(\frac{1}{a}\right)\end{align}$$
Now to the special role of $\sqrt{3}$. The only value of $a=\sqrt{m}$ with positive integer $m$ making $\tan ^{-1}\left(\frac{1}{a}\right)$ a rational multiple of $\pi$ is $\sqrt{3}$, giving $\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)=\frac{\pi }{6}.$
The complete integral evaluates to
$$\begin{align} i(a) & =  \int_{a}^{\infty}\frac {\log(x^4-1)}{x^2-1} \\ 
& = \frac{1}{2} \pi  \tan ^{-1}\left(\frac{1}{a}\right)+\frac{\pi ^2}{3} \text{(*1*)}
\\
& -\frac{1}{4} \log ^2(a-1)+\frac{1}{2} \log ^2(a+1)
\\
& -\frac{1}{2} \log (2) \log (a-1)+\frac{\log ^2(2)}{4}
\\ 
& + \frac{1}{2} \log \left(\frac{a+1}{a-1}\right) \log \left(a^2+1\right) \text{(*2*)}
\\ 
& +\frac{1}{2} \text{Li}_2\left(\frac{1-a}{2}\right)+\frac{1}{2} \text{Li}_2\left(\frac{2}{a+1}\right) 
\\ 
& +\frac{1}{2} \left(-\text{Li}_2\left(\left(\frac{1}{2}+\frac{i}{2}\right) (a-i)\right)
\\ 
+  \text{Li}_2\left(\left(-\frac{1}{2}+\frac{i}{2}\right) (a-i)\right)
\\
-\text{Li}_2\left(\left(\frac{1}{2}-\frac{i}{2}\right) (a+i)\right)
\\
+\text{Li}_2\left(\left(-\frac{1}{2}-\frac{i}{2}\right) (a+i)\right)\right) 
\end{align}$$
As the two lines marked $(*1*)$ and $(*2*)$ are already similar to the final result for $a=\sqrt {3}$ there should be a lot of cancellation from the polylog functions.
(to be completed)
</s>