 Solution

Let $\omega =e^{2\pi i/3}$ and integrate $1/(1+z^3)$ around the positively oriented sector from argument zero to $2\pi /3$, with outer radius $R>1$. The single pole inside is $z_0=e^{i\pi /3}$, with residue $1/(3z_0^2)=1/(3\omega )$. The outward real radial segment contributes $I_R={\int }_0^{R}dx/(1+x^3)$; the return radial segment contributes $-\omega I_R$ because $(\omega x{)}^3=x^3$.

On the circular arc, $|1+z^3|\ge R^3-1$, so its integral has modulus at most $(2\pi R/3)/(R^3-1)\to 0$. Thus the residue theorem yieldsSince $\omega (1-\omega )=\omega -{\omega }^2=i\sqrt{3}$,The ordinary improper integral converges at both endpoints, so no principal-value interpretation is needed here.