The integral is $$ \int_{-\infty}^0 \frac{x^{1/3}}{x^5-1}\mathrm dx $$
I have tried taking the typical half circle contour and finding the enclosed residues: $$ \text{Res}(f,1)=\frac{1}{5}\\ \text{Res}(f,e^{\frac{2\pi i}{5}})=\frac{e^{\frac{8\pi i}{15}}}{5}\\ \text{Res}(f,e^{\frac{4\pi i}{5}})=\frac{e^{\frac{-2\pi i}{15}}}{5} $$ By Jordan's lemma, the arc contributes nothing. However, I have a problem with the non integrable singularity at 1 on $\mathbb{R}^+$.
Can I deviate to avoid it while also keeping track of the contribution from the positive real axis? There doesn't seem to be great symmetry at $1$ thanks to the numerator. Is my contour not a good choice?
edit: Also tried changing variables to evaluate on the positive real axis, but this again introduces the bad singularity.
Assuming that $x^{1/3}$ is defined as $-(-x)^{1/3}$ on $\mathbb{R}^-$, the change of variable $x=-z^3$ bring the given integral in the form $$ \int_{0}^{+\infty}\frac{3z^3}{z^{15}+1}\,dz $$ and by setting $\frac{1}{1+z^{15}}=u$ the previous integral can be computed through Euler's Beta function. By exploiting the reflection formula for the $\Gamma$ function, $\Gamma(z)\,\Gamma(1-z)=\frac{\pi}{\sin(\pi z)}$, such integral simplifies to $$ \frac{\pi}{5\sin\frac{4\pi}{15}}=\frac{\pi}{5\cos\frac{7\pi}{30}}=\frac{4\pi}{5\sqrt{7-\sqrt{5}+\sqrt{6 \left(5-\sqrt{5}\right)}}}.$$