<s>[INST] Prove $\int_0^\infty\frac{\ln x}{x^3-1}\mathrm{d}x=\frac{4\pi^2}{27}$ Proof of the integral
$$\int_0^\infty\frac{\ln x}{x^3-1}\mathrm{d}x=\frac{4\pi^2}{27}$$
I try to substitute $u = \ln x$. Then $x = e^u,\>\mathrm{d}x = e^u\mathrm{d}u$ and the limits $(0,\infty)\to (-\infty,\infty)$.
The integral becomes $$\int_{-\infty}^\infty \frac{ue^u}{e^{3u}-1}\mathrm{d}u.$$
 [/INST] Split the integral to simplify as follows
\begin{align}\int_0^\infty\frac{\ln x}{x^3-1}{d}x
= &\int_0^1\frac{\ln x}{x^3-1}{d}x + \int_1^\infty\overset{x\to \frac1x}{\frac{\ln x}{x^3-1}}{d}x
=\int_0^1\frac{(1+x)\ln x}{x^3-1}{d}x \\
= &\int_0^1\frac{(x^2+x+1)\ln x}{x^3-1}{d}x
 - \int_0^1\overset{x^3\to x}{\frac{x^2\ln x}{x^3-1}}{d}x\\
=&\int_0^1\frac{\ln x}{x-1}{d}x-\frac19 \int_0^1\frac{\ln x}{x-1}{d}x
=\frac89 \int_0^1\frac{\ln x}{x-1}{d}x\\
=&\frac89\cdot \frac{\pi^2}6
=\frac{4\pi^2}{27}
\end{align}
Integrate $\int_0^1\frac{\ln x}{x-1}{d}x=\frac{\pi^2}6$
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