I'm having troubles evaluating this improper integral

Show: $\displaystyle \int_0^{\infty} \frac{1}{x^3+1} dx = \frac{2\pi}{3\sqrt{3}}$ using residues and a closed contour.

So, i used a closed contour C that only contains the singularity $\displaystyle z= e^{\frac{i\pi}{3}}$. I let R be the radius of the contour and I let R tend to infinity

So, $\displaystyle C = C_{+} + C_{R} + C_{\frac{2\pi}{3}} $

where $\displaystyle C_{+} = \{z=r: 0<r<R, 1<R<\infty \}$

$\displaystyle C_{R} = \{ z = Re^{i\theta}: 0\leq\theta\leq\frac{2\pi}{3}, 1<R<\infty \}$

$\displaystyle C_{\frac{2\pi}{3}} = \{z = re^{\frac{i\pi}{3}}: 0<r<R, 1<R<\infty \} $

I let $\displaystyle I = \lim_{R\to\infty}{\int_0^{R} \frac{1}{z^3 +1}dz}, I_{R} = \int_{C_R}\frac{1}{z^3 +1}dz, I_{\frac{2\pi}{3}} = \int_{C_{\frac{2\pi}{3}}}\frac{1}{z^3 +1}dz$

By Integral theorems, I get $\displaystyle I_{R} = 0$ as $\displaystyle R\to\infty$ and $\displaystyle I_{\frac{2\pi}{3}} = - e^{\frac{i\pi}{3}}* I $

By the Cauchy Residue Theorem, I get

$\displaystyle (1 - e^{\frac{i\pi}{3}})* I = 2\pi i{Res}_{z=e^{\frac{i\pi}{3}}} \frac{1}{z^3+1}$$\displaystyle = 2\pi i\lim_{z\to e^{\frac{i\pi}{3}}} ((1 - e^{\frac{i\pi}{3}})\frac{1}{z^3+1}) $

but I can't get the answer to evaluate. I just wanna know if there is something I'm missing in complex arithmetic, or is it my set-up that is wrong