(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?
Best Answer
I will answer my own question by proving
$$\int_{\sqrt{3}}^{\infty}\frac{\ln\left(x^{4}-1\right)}{x^{2}-1}dx = \frac{\pi^{2}}{4}+\frac{3}{2}\ln\left(2\right)\ln\left(2+\sqrt{3}\right).$$
Let the integral in question equal $I$. Letting $x \to \dfrac{1-x}{1+x}$, we rewrite the integral as
$$I = \int_{\sqrt{3}-2}^{-1}\frac{\ln\left(\left(\frac{1-x}{1+x}\right)^{4}-1\right)}{\left(\frac{1-x}{1+x}\right)^{2}-1}\left(\frac{-2}{\left(1+x\right)^{2}}\right)dx.$$
Doing some simplifications, we get
$$I = \frac{3\ln\left(2\right)}{2}\int_{\sqrt{3}-2}^{-1}\frac{1}{x}dx+\frac{1}{2}\int_{\sqrt{3}-2}^{-1}\frac{\ln\left(-x\right)}{x}dx+\frac{1}{2}\int_{\sqrt{3}-2}^{-1}\frac{\ln\left(1+x^{2}\right)}{x}dx-\frac{1}{2}\int_{\sqrt{3}-2}^{-1}\frac{\ln\left(\left(x+1\right)^{4}\right)}{x}dx.$$
Trivially, we can solve the first two integrals. For the last two, we can use the dilogarithm definition. Thus,
$$I = \frac{3\ln\left(2\right)}{2}\left(-\ln\left(2-\sqrt{3}\right)\right)+\frac{1}{2}\left(-\frac{1}{2}\ln^{2}\left(2-\sqrt{3}\right)\right)+\frac{1}{2}\left(\frac{1}{24}\left(12\operatorname{Li}_2 \left(4\sqrt{3}-7\right)+\pi^{2}\right)\right)-\frac{1}{2}\left(4\left(\operatorname{Li}_2 \left(2-\sqrt{3}\right)-\frac{\pi^{2}}{6}\right)\right)$$
which simplifies down to
$$I = \ln\left(\frac{1}{2-\sqrt{3}}\right)\left(\frac{1}{4}\ln\left(2-\sqrt{3}\right)+\frac{3}{2}\ln\left(2\right)\right)+\frac{17\pi^{2}}{48}+\frac{1}{4}\operatorname{Li}_2 \left(-\left(2-\sqrt{3}\right)^{2}\right)-2\operatorname{Li}_2 \left(2-\sqrt{3}\right)$$
From this answer, we can prove that
$$\frac{1}{4}\operatorname{Li}_2 \left(-\left(2-\sqrt{3}\right)^{2}\right)-2\operatorname{Li}_2 \left(2-\sqrt{3}\right) = \frac{\ln^{2}\left(2-\sqrt{3}\right)}{4}-\frac{5\pi^{2}}{48}.$$
Combining the results, we conclude that the integral $I$ is
$$\int_{\sqrt{3}}^{\infty}\frac{\ln\left(x^{4}-1\right)}{x^{2}-1}dx = \frac{\pi^{2}}{4}+\frac{3}{2}\ln\left(2\right)\ln\left(2+\sqrt{3}\right).$$