How to solve the integral through a contour with branch points? Consider the integral
$$ \int^{\infty}_{0} \frac{x^{\frac{1}{3}}}{1+x^2}dx $$
on the complex plane
$$ \oint_{C} \frac{z^{\frac{1}{3}}}{1+z^2}dz $$
To find the poles $ 1+z^2=0 \Rightarrow z^2=-1 \Rightarrow z= \pm \sqrt{-1} \Rightarrow z= \pm i $ , by the residue theorems $$ \oint_{C} \frac{z^{\frac{1}{3}}}{1+z^2}dz = 2 \pi i \,\ Res f(z) $$
I know that
$$  \oint_{C} \frac{z^{\frac{1}{3}}}{1+z^2}dz = \oint_{C_R} \frac{z^{\frac{1}{3}}}{1+z^2}dz  +\oint_{C_2} \frac{z^{\frac{1}{3}}}{1+z^2}dz + \oint_{C_r} \frac{z^{\frac{1}{3}}}{1+z^2}dz + \oint_{C_1} \frac{z^{\frac{1}{3}}}{1+z^2}dz  $$
the contour used is below:

The integral in $C_R \rightarrow 0$ because $R \rightarrow \infty$ and the integral in $C_r \rightarrow 0$ because $r \rightarrow 0$.
But how can I calculate the integrals over $C_1$ and $C_2$? How can I calculate residuals?
The teacher said the result is
$$ I(1-e^{\frac{2 \pi i}{3}}) = 2 \pi i \Big[ \frac{e^{\frac{\pi i}{6}}}{2i}\Big] (1-e^{\frac{\pi i}{3}}) \quad \quad \Rightarrow  \quad \quad  I= \frac{\pi}{2 \sin(\frac{\pi}{3})} = \frac{\pi}{\sqrt{3}}$$

You may start by removing the branch point through a suitable substitution:
$$I=\int_{0}^{+\infty}\frac{x^{1/3}}{1+x^2}\,dx\stackrel{x\mapsto z^3}{=}\int_{0}^{+\infty}\frac{3x^3}{1+x^6}\,dx=\int_{0}^{1}\frac{3x^3}{1+x^6}\,dx+\int_{1}^{+\infty}\frac{3x^3}{1+x^6}\,dx$$
then realize you do not need complex analysis at all:
$$ I = 3\int_{0}^{1}\frac{x+x^3}{1+x^6}\,dx=3\int_{0}^{1}\frac{x}{1-x^2+x^4}\,dx\stackrel{x\mapsto\sqrt{u}}{=}\frac{3}{2}\int_{0}^{1}\frac{du}{1-u+u^2}=\left.\sqrt{3}\arctan\left(\frac{2u-1}{\sqrt{3}}\right)\right|_{0}^{1}$$
immediately leads to $I=\frac{\pi}{\sqrt{3}}$.
Also, all integrals of the form $\int_{0}^{+\infty}\frac{x^\alpha}{1+x^\beta}\,dx $ can be computed through the substitution $\frac{1}{1+x^\beta}\to z$, Euler's Beta function and the reflection formula for the $\Gamma$ function.
