 Solution

Integrate $(1+z^p{)}^{-1}$ around the wedge $0\le \arg z\le 2\pi /p$. The radial integrals differ by the factor $e^{2\pi i/p}$, the arc vanishes, and the wedge contains the simple pole $e^{i\pi /p}$ with residue $-[p{e}^{-i\pi /p}{]}^{-1}$. Solving the resulting identity gives $I=(\pi /p)\csc (\pi /p)$.