How to prove this integral problem: $\int_0^{\infty}\frac{dx}{1+x^n}=\frac{\pi}{n}\csc\frac \pi n$? I dont know if this has already been asked.
How to prove this integral
$$\int_0^{\infty}\frac{dx}{1+x^n}=\frac{\pi}{n}\csc\frac \pi n\ {?}$$
$n\ge 2$ is a positive integer 
Frankly speaking i have no clue how to start someone please explain me. 
 A: Method 1:
Let $x=\sqrt[n]{\tan^{2}\theta }$, then
\begin{align*}
\int_{0}^{\infty }\frac{1}{1+x^{n}}\, \mathrm{d}x&=\frac{2}{n}\int_{0}^{\pi /2}\cos^{1-2/n}\theta \sin^{2/n-1}\theta \, \mathrm{d}\theta \\
&=\frac{1}{n}\mathrm{B}\left ( 1-\frac{1}{n},\frac{1}{n} \right )\\
&=\frac{1}{n}\Gamma \left ( 1-\frac{1}{n} \right )\Gamma \left ( \frac{1}{n} \right )\\
&=\frac{\pi }{n}\mathrm{csc}\frac{\pi }{n}
\end{align*}
where $\mathrm{B}\left ( \cdot  \right )$ is the Beta function and $\Gamma\left ( \cdot  \right )$ is the Gamma function.
Method 2:
\begin{align*}
&{\int_0^\infty \frac{1}{x^n+1} \:{\rm{d}}x}=\int_0^\infty\int_0^\infty e^{-(x^n+1)t} \:{\rm{d}}t\:{\rm{d}}x
\\&=\int_0^\infty e^{-t}\int_0^\infty e^{-x^n t} \:{\rm{d}}t\:{\rm{d}}x
=\int_0^\infty e^{-t}\left(\int_0^\infty e^{-x^n t}\:{\rm{d}}x\right)\:{\rm{d}}t
\\&=\frac1n\int_0^\infty t^{-\frac1n}e^{-t}\left(\int_0^\infty u^{\frac1n-1} e^{-u}{\rm{d}}u\right)\:{\rm{d}}t
=\frac1n \Gamma\left(1-\frac1n\right)\Gamma\left(\frac1n\right)
\\&=\frac{\pi }{n}\mathrm{csc}\frac{\pi }{n}
\end{align*}

More general, using the same way as Method 1 mentioned,we get
$$\int_{0}^{\infty} \frac{x^{\mu-1}}{1+x^{\nu}} \; \mathrm{d}x=\frac{\pi}{\nu} \csc \left( \frac{\pi \mu}{\nu} \right)$$
where $0< \mu < \nu $
