I want to find the best way to show $\int_0^\infty\dfrac{x^{2m}}{x^{2n}+1}\,dx=\dfrac{\pi}{2n}\operatorname{csc}\left(\dfrac{2m+1}{2n}\pi\right)$, where $0\leq m<n$.
It's easy to verify some simple cases when $m$ and $n$ are small, e.g. $\int_0^\infty\dfrac{dx}{1+x^2}=\arctan|_0^{\pi/2}=\dfrac{\pi}{2}$ . In general, one may expect to use residue theorem to sum all the residues on the left hand side with the contour the upper half-disk. However, the calculation seems complicated. Is there some simple way to see it?
Apart from the Beta function approach mentioned by Julien Godawatta, this can be very nicely evaluated using the residue theorem.
The denominator has zeros in
$$\exp\left(\pi i \frac{2k+1}{2n}\right),\quad 0 \leqslant k < n,$$
and is the same on rays an angle of $\frac{\pi}{n}$ apart. The numerator differs by a constant factor of $\exp \left(\pi i\frac{2m}{n}\right)$ on such rays; we get another factor of $e^{\pi i/n}$ integrating over such a ray from the differential $dz$. So if we integrate over the boundary of a sector $\{ z : \lvert z\rvert < R,\; 0 < \arg z < \frac{\pi}{n}\}$, we find
$$2\pi i \operatorname{Res}\left(\frac{z^{2m}}{z^{2n}+1}; {\large e^{\frac{\pi i}{2n}}}\right) = \left(1 - {\large e^{\pi i\frac{2m+1}{n}}}\right)\int_0^R \frac{x^{2m}}{x^{2n}+1}\,dx + \int_{C_R} \frac{z^{2m}}{z^{2n}+1}\,dz,$$
where $C_R$ is the circular arc closing the contour. Since $m < n$, the integral over $C_R$ tends to $0$ for $R\to\infty$, leaving
$$\int_0^\infty \frac{x^{2m}}{x^{2n}+1}\,dx = \frac{2\pi i}{1 - {\large e^{\pi i\frac{2m+1}{n}}}}\operatorname{Res}\left(\frac{z^{2m}}{z^{2n}+1}; {\large e^{\frac{\pi i}{2n}}}\right).\tag{1}$$
Since the denominator has only simple zeros, the residue in a zero is
$$\operatorname{Res}\left(\frac{z^{2m}}{z^{2n}+1}; \zeta\right) = \frac{\zeta^{2m}}{2n\zeta^{2n-1}} = \frac{\zeta^{1+2m-2n}}{2n},$$
and for $\zeta = e^{\large \frac{\pi i}{2n}}$, we find the residue to be
$$\frac{e^{\pi i\large \frac{1+2m-2n}{2n}}}{2n} = -\frac{1}{2n}e^{\pi i \large\frac{2m+1}{2n}}.$$
Inserting that into $(1)$ yields
$$\int_0^\infty \frac{x^{2m}}{x^{2n}+1}\,dx = \frac{\pi}{2n\sin \left(\pi \frac{2m+1}{2n}\right)},$$
as desired.