I know that there are three simple poles
$e^{\frac{\pi\cdot i}{3}},e^{\pi\cdot i},e^{\frac{5\pi\cdot i}{3}}$
and $Res{\frac{1}{(x^3+1)}} = Res{\frac{1}{3x^2}} = Res\frac{-x}{3}$
but I don't know how to deal with that kind of contour
I gussed $e^{\frac{\pi\cdot i}{3}}$ is only pole that interior of contour
but using only residue of $e^{\frac{\pi\cdot i}{3}}$ doesn’t work.
hint will be appreciated
For the integral along the real axis, we have : $x=te^{2\pi i}=t, dx=dt$, so we have $\int_{0}^{\infty}\frac{dt}{t^3+1}$ which is our original integral ($:=I$)
For the integral along part of the circle, we have :$$x=R*e^{\frac{2\pi it}{3}}, 0 \leq t\leq 1, dx=\frac{2\pi i}{3}R*e^{\frac{2 \pi i t}{3}}$$ so we have $$\frac{2\pi i}{3}\int \frac{R*e^{\frac{2 \pi i t}{3}}}{R^3*e^{2\pi it}+1}dt$$ which tends to zero as R $\to \infty$
Now, for the last one : $x=te^{\frac{2\pi i}{3}}, dx=e^{\frac{2\pi i}{3}}dt$
Our integral is $\int_{\infty}^{0}\frac{e^{\frac{2 \pi i}{3}}dt}{t^3+1}=-e^{\frac{2 \pi i}{3}} \int^{\infty}_{0}\frac{dt}{t^3+1}= -e^{\frac{2 \pi i}{3}}*I$
So the residue theorem gives use $$(1-e^{\frac{2 \pi i}{3}})\int_{0}^{\infty}\frac{dt}{t^3+1} + 0 = 2 \pi i \sum Res(f_i)$$
Our only residue inside our domain is $x=e^{\frac{\pi i}{3}}$ If we compute our residue, we get $\frac{1}{3e^{\frac{2 \pi i}{3}}}$
Finally we have :$$\int_{0}^{\infty}\frac{dt}{t^3+1}=\frac{2 \pi i}{(1-e^{\frac{2 \pi i}{3}})}*\frac{1}{3e^{\frac{2 \pi i}{3}}}=\frac{2 \pi}{3 \sqrt{3}}$$
If I did any typo, please say it in the comments as soon as possible.
Hope this helps !