Is there an elementary proof for $\int_{0}^{\infty} \frac{x^{r}}{x^{m}+1} d x=\frac{\pi}{m} \csc \frac{(r+1) \pi}{m} $, where $m>r+1>0$? I am recently investigating integrals with rational integrand such as $$
\int_{0}^{\infty} \frac{P(x)}{\left(x^{m}+1\right)^{n}} d x,
$$
where $P(x)$ is a polynomial, $m$ and $n$ are natural numbers.
After trying several methods, I realize that I have to prove a fundamental integral
$$\int_{0}^{\infty} \frac{x^{r}}{x^{m}+1} d x=\frac{\pi}{m} \csc \frac{(r+1) \pi}{m},$$
where $m>r+1>0$.
When I tried to evaluate it by Gamma and Beta functions, I eventually need the Euler Reflection Theorem to complete the proof.
My question: Is there an elementary proof for the integral?
Your proofs and suggestions are highly appreciated.
 A: Letting $\displaystyle \frac{1}{t}=x^{m}+1$, then $
\displaystyle d x=\frac{1}{m} \left(\frac{1}{t}-1\right)^{\frac{1}{m}-1}\left(-\frac{1}{t^{2}}\right) d t.
$
Consequently
$$
\begin{aligned}
\int_{0}^{\infty} \frac{x^{r}}{x^{m}+1} d x &=-\int_{1}^{0} \frac{\left(\frac{1}{t}-1\right)^{\frac{r}{m}}}{\frac{1}{t}} \frac{1}{m t^{2}}\left(\frac{1}{t}-1\right)^{\frac{1}{m}-1} d t \\
&=\frac{1}{m} \int_{0}^{1} \frac{(1-t)^{\frac{r}{m}}(1-t)^{\frac{1}{m}-1}}{t^{\frac{r+1}{m}}} d t \\
&=\frac{1}{m} \int_{0}^{1} t^{-\frac{r+1}{m}}(1-t)^{\frac{r+1}{m}-1} d t \\
&=\frac{1}{m} B\left(1-\frac{r+1}{m}, \frac{r+1}{m}\right)
\end{aligned}
$$
By the property of Beta function,
$$
B(x, y)=\frac{\Gamma(x) \Gamma(y)}{\Gamma(x+y)},
$$
where $\operatorname{Re}(x)>0$ and $\operatorname{Re}(y)>0,$
we have
$$
\int_{0}^{\infty} \frac{x^{r}}{x^{m}+1} d x=\frac{\Gamma\left(1-\frac{r+1}{m}\right) \Gamma\left(\frac{r+1}{m}\right)}{m\Gamma(1)}
$$
Using the Euler’s Reflection Theorem,
$$
\Gamma(1-z) \Gamma(z)=\pi \csc (\pi z),
$$
where $z\notin Z$,
we can now conclude that
$$\boxed{\int_{0}^{\infty} \frac{x^{r}}{x^{m}+1} d x=\frac{\pi}{m} \csc \frac{(r+1) \pi}{m}}$$
