Prove using contour integration that $\int_0^\infty \frac{\log x}{x^3-1}\operatorname d\!x=\frac{4\pi^2}{27}$ Prove using contour integration that $\displaystyle \int_0^\infty \frac{\log x}{x^3-1}\operatorname d\!x=\frac{4\pi^2}{27}$
I am at a loss at how to start this problem and which contour to pick. I have been trying to get the sector with angle $2\pi/3$ to work with a bump around the pole at $e^{i2\pi/3}$ and the origin, but I am getting 5 or 6 different integrals and it is not really getting me anywhere.

This isn't quite complex analytic, but first denote your as $$I=\int_{0}^{\infty} \frac{\ln(x)}{x^3-1} dx.$$
Consider the double integral:
$$J=\int_{0}^{\infty} \int_{0}^{\infty}  \frac{x}{(x^2+y^3)(1+x^2)}dydx.$$
We intend to evaluate $J$ and relate $J$ to $I.$
To evaluate, $J$ we integrate with respect to $y.$ You can proceed in two ways. The long way is to integrate by partial fractions, but the short way is to let $y=ux^\frac{2}{3}$ so that $dy=x^\frac{2}{3} du.$ Then we get:
$$J=\int_{0}^{\infty} \int_{0}^{\infty}  \frac{x^\frac{5}{3}}{(x^2+x^2u^3)(1+x^2)}dudx=\int_{0}^{\infty} \int_{0}^{\infty}  \frac{x^\frac{-1}{3}}{(1+u^3)(1+x^2)}dudx.$$ Using the nice formula $$\int_{0}^{\infty} \frac{t^m}{1+t^n} dt=\frac{\pi}{n}\csc\left(\frac{\pi(m+1)}{n}\right),$$
we get $$J=\frac{2\pi^2}{9}.$$
Now use Fubini's Theorem on $J$ as such:
$$J=\int_{0}^{\infty} \int_{0}^{\infty}  \frac{x}{(x^2+y^3)(1+x^2)}dxdy.$$
We will need partial fractions to integrate with respect to $x.$ Omitting the details of this computation, $$\frac{x}{(x^2+y^3)(1+x^2)}=\frac{1}{y^3-1} \left(\frac{x}{x^2+1}-\frac{x}{x^2+y^3}\right).$$ Now integrating with respect to $x$, and plugging in the endpoints, we get 
$$J=\int_{0}^{\infty} \frac{1}{y^3-1}\lim_{x\rightarrow \infty} \left(\frac{\ln(x^2+1)}{2}-\frac{\ln(x^2+y^3)}{2}\right)-\frac{1}{y^3-1}\lim_{x\rightarrow 0} \left(\frac{\ln(x^2+1)}{2}-\frac{\ln(x^2+y^3)}{2}\right)dy.$$
We get $$J=\int_{0}^{\infty} \frac{\ln(y^3)}{2(y^3-1)}dy=\int_{0}^{\infty} \frac{3\ln(y)}{2(y^3-1)}dy=\frac{3}{2}I.$$
Thus we have $$I=\frac{2}{3}J=\frac{2}{3}\left(\frac{2\pi^2}{9}\right)=\frac{4\pi^2}{27}.$$
Addendum
Consider $$I=\int_{0}^{\infty} \frac{\ln(x)}{x^3-1} dx.$$ 
Split the integral into two parts: 
$$\int_{0}^{\infty} \frac{\ln(x)}{x^3-1} dx = \int_{0}^{1} \frac{\ln(x)}{x^3-1} dx+ \int_{1}^{\infty} \frac{\ln(x)}{x^3-1} dx.$$ We can convert $\frac{\ln(x)}{x^3-1}$ into a geometric series as such (assuming $0<x<1.$) $$ \frac{\ln(x)}{x^3-1}=-\sum_{n=0}^{\infty}\ln(x)x^{3n}.$$ If we do integration by parts, $$\int_{0}^{1} \frac{\ln(x)}{x^3-1} dx= \int_{0}^{1} -\sum_{n=0}^{\infty}\ln(x)x^{3n}dx=\sum_{n=0}^{\infty} \frac{1}{(3n+1)^2}.$$ For,$$\int_{1}^{\infty} \frac{\ln(x)}{x^3-1} dx,$$ perform a $u$ substitution $x=\frac{1}{u}, dx=\frac{-1}{u^2} du,$ and convert transformed integrand into a geometric series.  You will see: 
$$\int_{1}^{\infty} \frac{\ln(x)}{x^3-1} dx=\sum_{n=0}^{\infty} \frac{1}{(3n+2)^2}=\sum_{n=-\infty}^{-1} \frac{1}{(3n+1)^2},$$ and the latter equality can be seen by simply writing out the terms of each summation.  Now, add the two computed series together and see:
$$\sum_{n=-\infty}^{\infty} \frac{1}{(3n+1)^2}= \frac{4 \pi^2}{27.}$$
