$$I(x) = \int_{-\infty}^{\infty}{\frac{x^m}{1+x^n}\, dx}$$ Usually a wedge contour with angle $\frac{\pi}{2n}$ would be used to only have one pole in the contour, but I think this can also be computed by using the entire semicircle in the upper half plane. I was trying to compute the residues in terms of $n$ and $m$, given that the poles are at $\frac{\pi}{n}(2k+1)$ :
$$\begin{align}\mathrm{Res(I)}&=\lim_{z\to e^{\frac{i\pi}{2}(2k+1)}}\frac{z^m}{1+z^n}\\&=\frac{e^{\frac{i\pi m}{2}(2k+1)}}{n(e^{\frac{i\pi}{2}(2k+1)})^{n-1}}\\&=\frac{1}{n}\left(\frac{e^{\frac{i\pi}{n}(2k+1)(m+1)}}{e^{i\pi(2k+1)}}\right)\\&=-\frac{1}{n}e^{\frac{i\pi}{2}(2k+1)(m+1)}\end{align}$$
Then summing and multiplying by $2\pi i$,
$$\begin{align}{-\frac{2\pi i}{n}\sum_{k=0}^{n-1}{e^{\frac{i\pi}{2}(2k+1)(m+1)}}}&=-\frac{2\pi i}{n}\left(\frac{e^{\frac{i\pi}{2}(m+1)}(1-e^{i\pi n(m+1)})}{1-e^{i\pi(m+1)}}\right)\\&=-\frac{2\pi i}{n}\left(\frac{1-e^{i\pi n(m+1)}}{-2i\sin(\frac{\pi}{2}(m+1))}\right)\end{align}$$
After cancelling, I’m not sure how to continue this though (mainly I’m struggling to simplify the exponential in the numerator without introducing more issues)