TITLE: Hints for contour integration of $\int_0^\infty \frac{x^{1/3}}{x^4+1}dx$
QUESTION [5 upvotes]: I've reading contour integration and then I came up with this one which I want to solve \begin{equation}\int_0^\infty \frac{x^{1/3}}{x^4+1}dx.\end{equation}
I saw a video where the author dealt with a similar integral as written below,
$$\int_0^\infty \frac{x^{1/2}}{x^2+1}dx,$$
but for some reasons I cannot manage to compute the integral $\int_0^\infty \frac{x^{1/3}}{x^4+1}dx$ to the correct answer, which, according to Maple is $\sqrt{3}\pi/6$. My main idea was to make a keyhole contour but I failed when calculating the answer. The residues was easy to compute. As what I see, all of them would lay inside my contour.
So can someone come with some hints so I can continue? Maybe you have the key so I can manage to compute the integral. Notice that this is not a homework problem. Please have a look at my contour!

As you can see, the $Z_1,...,Z_4$ represent my poles.
Edit: I've added my work. I took a screen dump!

REPLY [2 votes]: An alternative way to do this with a tad less computation and worrying about branch cuts is to consider a quarter circle  in the first quadrant as our contour.
Why might we  be motivated to do this? We spot that $z^4 +1 = (zi)^4 +1$; denoting the integral on the real axis by $\displaystyle{I := \int_{0}^{\infty} \frac{x^{1/3}}{x^4+1}\,\mathrm{dx}}$, the integral on the imaginary axis becomes
$\displaystyle{-\int_{0}^{\infty} \frac{(it)^{1/3}}{(it)^4 + 1} i \,\mathrm{dt} = -ie^{\frac{i \pi}{6}} I = -e^{\frac{2\pi i}{3}} I}$, a simple multiple of that on the real axis.
This also means there is only one pole enclosed in the contour (with residue $\frac{1}{4} e^{-\frac{2\pi i}{3}}$), whilst the integral over the arc vanishes in the limit of infinite radius.
Applying the residue theorem and some simple algebra, we find $I = \boxed{ \frac{\pi}{2\sqrt{3}}}$ as in the other answers.
(Incidentally, this 'wedge' method generalises to other integrals such as $\int_{0}^{\infty} \frac{\mathrm{dx}}{1+x^n}$ for $n \in \mathbb{N_{\ge 2}}$, where we might use an angle $\frac{2\pi}{n}$. )