complex analysis - Solving the improper integral $int_0^infty frac{x^{1/3}}{1+x^2} mathrm dx$

I'm trying to solve:

$\displaystyle \int \limits_0^\infty \dfrac{x^{1/3}}{1+x^2} \mathrm dx$

I have tried contour integration with $C_R^+$ and the real line like this:

$\displaystyle \int \limits_T \dfrac{z^{1/3}}{1+z^2} \mathrm dz = \int \limits_{-\infty}^\infty \dfrac{z^{1/3}}{1+z^2}\mathrm dz + \int \limits_{C_R^+} \frac{z^{1/3}}{1+z^2} \mathrm dz$

Where the last integral tends to $0$ as $R \longrightarrow \infty$

$\text{Res}(f(z);i) = \dfrac{i^{1/3}}{2i}$

and

$\displaystyle \int \limits_{-\infty}^\infty \frac{z^{1/3}}{1+z^2}\mathrm dz = \int \limits_{0}^\infty \frac{z^{1/3}}{1+z^2} \mathrm dz + \int \limits_{-\infty}^0\frac{z^{1/3}}{1+z^2} \mathrm dz$

If i manipulate the last term by changing the limits and substitute $u=-t$ i get:

$\displaystyle \int \limits_{-\infty}^0\dfrac{z^{1/3}}{1+z^2} \mathrm dz = -\int \limits_{0}^{-\infty}\frac{z^{1/3}}{1+z^2} \mathrm dz$

If i now substitue $u=-z, u'=-1$

$\displaystyle \int \limits_{-\infty}^0\dfrac{z^{1/3}}{1+z^2} \mathrm dz = \int \limits_{0}^{\infty}\dfrac{(-u)^{1/3}}{1+u^2}\mathrm dz = (-1)^{1/3}\int \limits_{0}^{\infty}\dfrac{u^{1/3}}{1+u^2}\mathrm dz$

$\displaystyle \int \limits_{-\infty}^\infty \dfrac{z^{1/3}}{1+z^2}\mathrm dz = \left(1+e^\frac{i\pi}{3}\right) \int \limits_{0}^\infty \frac{z^{1/3}}{1+z^2} \mathrm dz $

So i end up with:

$\displaystyle \dfrac{2i\cdot\pi \cdot e^{i\cdot\pi /6}}{2i\cdot\left(1+e^\frac{i\pi}{3}\right)} = \dfrac{\pi\cdot e^{i\cdot \pi /6}}{\left(1+e^\frac{i\pi}{3}\right)} = \int \limits_{0}^\infty \dfrac{z^{1/3}}{1+z^2} \mathrm dz$ which is wrong answer