let $\displaystyle x=(\tan t)^{\frac{1}{n}}.$ then $\displaystyle dx=\frac{1}{n}(1 + \tan^2 t)(\tan t)^{\frac{1}{n} - 1}dt.$ therefore:
$\displaystyle \int_0^{\infty} \frac{dx}{1+x^{2n}}=\frac{1}{n}\int_0^{\frac{\pi}{ 2}}(\tan t)^{\frac{1}{n} - 1} dt=\frac{1}{n} \int_0^{\frac{\pi}{2}} (\sin t)^{\frac{1}{n}-1}(\cos t)^{1 - \frac{1}{n}} dt$
$\displaystyle =\frac{1}{2n}B \left(\frac{1}{2n}, 1- \frac{1}{2n} \right)=\frac{1}{2n} \Gamma \left(\frac{1}{2n} \right) \Gamma \left(1 - \frac{1}{2n} \right) = \frac{\pi}{2n \sin \left(\frac{\pi}{2n} \right)}.$
the last identity is true by
Euler's reflection formula.