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 $$\int_{-\infty}^\infty \frac{ue^u}{e^{3u}-1}\mathrm{d}u.$$

Writing
$$\frac 1{x^3-1}=\frac{1}{(a-1) (a-b) (x-a)}+\frac{1}{(b-1) (b-a) (x-b)}+\frac{1}{(a-1) (b-1) (x-1)}$$ where
$$a=-\frac{1}{2}-\frac{i \sqrt{3}}{2} \qquad \text{and} \qquad b=-\frac{1}{2}+\frac{i \sqrt{3}}{2}$$ we face three integrals
$$I(c)=\int \frac{\log(x)}{x-c}\,dx=\text{Li}_2\left(\frac{x}{c}\right)+\log (x) \log \left(1-\frac{x}{c}\right)$$
$$J(c)=\int_0^t \frac{\log(x)}{x-c}\,dx=\text{Li}_2\left(\frac{t}{c}\right)+\log (t) \log \left(1-\frac{t}{c}\right)$$
Recombining the three terms, computing at the bounds and using the values of the polylogarithms leads to
$$\int_0^\infty\frac{\ln(x)}{x^3-1}dx=\frac{5 \pi ^2}{54}-\left(-\frac{\pi ^2}{18} \right)=\frac{4 \pi ^2}{27}$$
