Calculating a real integral with residues I'm attempting to compute $$\int_{-\infty}^0 \frac{x^{1/3}}{x^5-1}dx.$$
My attempt is to use the standard keyhole contour along the negative real axis (which we also take to be the branch cut for $z^{1/3}$), but I'm already having trouble simply computing the residue sum $$\sum_{k=1}^5\operatorname{Res}\left(\frac{z^{1/3}}{z^5-1}, e^{2\pi ik/5}\right) = \sum_{k=1}^5 \lim_{z\to e^{2\pi ik/5}}\frac{z^{1/3}(z-e^{2\pi ik/5})}{z^5-1}.$$
Given that this question appeared on a timed qualifying exam, I'm sure that there's a quick or clever solution to this problem that I'm not seeing. Any help would be greatly appreciated.
Edit: Another idea I had was to substitute $x \mapsto x^3$ which gives $$3\int_{-\infty}^0 \frac{x^3}{x^{15}-1}dx.$$
This relieves us of the branch cut issue at the cost of introducing many more singularities, so I'm not sure if it's actually helpful.
 A: 
...but I'm already having trouble simply computing the residue...

Hint. One may observe that, for any differentiable function $f$ near $a$, we have
$$
\lim_{z \to a} \frac{f(z)-f(a)}{z-a}=f'(a)
$$ giving here
$$
\lim_{z\to e^{2\pi ik/5}}\frac{z-e^{2\pi ik/5}}{z^5-1}=\lim_{z \to z_k} \frac{z-z_k}{z^5-z_k^5}=\frac{1}{5z_k^4}
$$ hoping this helps to obtain the given sum.
A: A classmate of mine pointed out that after making the substitution $x\mapsto -x$, the integral becomes $$ \int_0^\infty \frac{x^{1/3}}{x^5+1}\ dx, $$ which we can calculate by integrating the corresponding complex function $$ f(z) = \frac{z^{1/3}}{z^5+1}$$ (taking the principal branch of $z^{1/3}$) over the (counter-clockwise) contour $$ \Gamma_R = \gamma_R \cup C_R \cup \gamma'_R,$$ where $\gamma_R$ is the horizontal line from $0$ to $R$, $\gamma'_R$ is the line connecting $Re^{2\pi i/5}$ to $0$, and $C_R$ is the circular arc of radius $R$ connecting $R$ to $Re^{2\pi i/5}$. For $R>1$, the only singularity contained inside $\Gamma_R$ is at $e^{\pi i/5}$, so we have
$$
2\pi i\operatorname{Res}(f(z),e^{\pi i/5}) = \lim_{R\to \infty} \int_{\Gamma_R} f = \int_0^\infty \frac{x^{1/3}}{x^5+1}\ dx - \int_0^\infty \frac{(xe^{2\pi i/5})^{1/3}}{(xe^{2\pi i/5})^5 +1}e^{2\pi i/5}\ dx.
$$
Thus,
$$
\int_0^\infty \frac{x^{1/3}}{x^5+1} = \frac{2\pi i\operatorname{Res}(f(z),e^{\pi i/5})}{1-(e^{2\pi i/5})^{4/3}}.
$$
By Olivier's helpful answer, we can easily calculate this residue: $$\operatorname{Res}(f(z),e^{\pi i/5}) = \lim_{z\to e^{\pi i/5}}z^{1/3}\frac{z-e^{\pi i/5}}{z^5 - (e^{\pi i/5})^5} =\frac{(e^{\pi i/5})^{1/3}}{5(e^{\pi i/5})^4} = \frac{1}{5}e^{-11\pi i/5}.$$ The final answer can surely by simplified, but we have $$\int_0^\infty \frac{x^{1/3}}{x^5+1} = \frac{2\pi i e^{-11\pi i/5}}{5(1-e^{8\pi i/15})} \approx 0.845486.$$
