Hints for contour integration of $\int_0^\infty \frac{x^{1/3}}{x^4+1}dx$ 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!

 A: You're right that once we've explained why the arcs vanish we're left with$$\left(1-\exp\frac{2\pi i}{3}\right)\int_0^\infty\frac{x^{1/3}}{x^4+1}dx=2\pi i\sum_{j=0}^3\lim_{z\to e^{\pi i/4+j\pi i/2}}\frac{z^{4/3}-e^{\pi i/4+j\pi i/2}z^{1/3}}{z^4+1}.$$By L'Hôpital's rule, the sum is$$\sum_j\lim_{z\to e^{\pi i/4+j\pi i/2}}\frac{\tfrac43z^{1/3}-\tfrac13e^{\pi i/4+j\pi i/2}z^{-2/3}}{4z^3}=\tfrac14\sum_j\lim_{z\to e^{\pi i/4+j\pi i/2}}e^{-2\pi i/3-j\pi i/4},$$because$$\frac{\tfrac43z^{1/3}-\tfrac13z\cdot z^{-2/3}}{4z^3}=\tfrac14z^{-8/3}.$$Now you just need to sum this geometric series, then multiply by $\frac{2\pi i}{1-\exp\frac{2\pi i}{3}}$.
A: I'm sorry - a bit long for a comment.
You should correctly choose the phase of the roots. If you made the cut from $0$ to $\infty$ you cannot cross it - the phases of your roots should remain positive (the root $e^{\frac{-\pi i}{4}}$ is not allowed).
You go from positive part of $X$ axis counter clockwise (in the positive direction), so your roots are $e^{\frac{\pi i}{4}}e^{\frac{2\pi i k}{4}}$, where $k=0, 1, 2, 3$.
Evaluation in this way (with $4$ poles) is good as a learning exercise, but to make the task easier I would recommend to make change of the variable in the initial integral:
$$I=\int_0^\infty \frac{x^{1/3}}{x^4+1}dx=\frac{1}{4}\int_0^\infty \frac{x^{1/3}x^{-3}}{x^4+1}d(x^4)=\frac{1}{4}\int_0^\infty \frac{t^{-2/3}}{t+1}dt$$
Now you have only one pole, but the approach with the keyhole contour is still applicable.
Or you can also use the Beta-function to evaluate integral.
Make the change $x=\frac{1}{1+t}$, and you get
$$I=\frac{1}{4}\int_0^1 (1-x)^{-2/3}x^{-1/3}dx=\frac{1}{4}B(1/3;2/3)=\frac{1}{4}\Gamma(1/3)\Gamma(2/3)=\frac{1}{4}\frac{\pi}{\sin\frac{\pi}{3}}=\frac{\pi}{2\sqrt3}$$
A: 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 $\frac{2\pi}{n}$. )
