Does anybody know how to prove that?
$$ \int\limits_{0}^{\infty}\frac{x^m}{(x^n+a^n)^r}\,dx=\frac{ (-1)^{r-1}\pi a^{m+1-nr}\Gamma\left(\frac{m+1}{n}\right) }{ n\sin\left( \frac{(m+1)\pi}{n} \right)(r-1)! \Gamma\left(\frac{m+1}{n}-r+1\right) }. $$
With $ a>0,\quad m, n, r\in \mathbb N, \quad 0<m+1<nr$.
I found that long time ago into some random lecture notes.
This is on a level so that you should be able to fill in the details:
1) Let $u=(x/a)^n$.
2) Use the fact that the Beta function $B(x,y)$ is given by $$ B(x,y)=\int_0^{+\infty}\frac{t^{x-1}}{(1+t)^{x+y}}\,dt, $$ valid for positive $x$ and $y$.
3) Use the fact that $$ B(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}. $$
4) Use that $\Gamma(r)=(r-1)!$ if $r\in\mathbb N$.
5) Finally, use the Euler reflection formula, $$ \Gamma(x)\Gamma(1-x)=\frac{\pi}{\sin(\pi x)},\quad x\notin\mathbb Z. $$
If your formula is correct, you should be there now.