HINT
The integral
$$ \mathcal{I}= \int_0^\infty \frac{1}{1+x^n} dx,$$
can be equivalently expressed as
$$ \mathcal{I} = \frac{1}{n} \int^1_0 t^{\left(1-\frac{1}{n}\right) - 1} \left(1-t\right)^{\left(\frac{1}{n}\right)-1} dt = \frac{1}{n} B \left(1-\frac{1}{n},\frac{1}{n} \right),$$
where $B(x,y)$ is the Beta function. You can then make use of the identity
$$ B(x,y) = \frac{\Gamma(x) \Gamma(y)}{\Gamma(x+y)} = \frac{\Gamma\left( 1 - \frac{1}{n}\right) \Gamma \left(\frac{1}{n} \right)}{\Gamma(1)}, $$
where $\Gamma$ denotes the Gamma function and $\Gamma(1) = 1$. It can be shown that
$$ \Gamma\left( 1 - \frac{1}{n}\right) \Gamma \left(\frac{1}{n} \right) = \frac{\pi}{ \sin(\pi/n)}. $$
Hence, the integral takes the form
$$ \mathcal{I}= \int_0^\infty \frac{1}{1+x^n} dx = \left( \frac{\pi}{n} \right) \frac{1}{\sin(\pi/n)},$$
where the limit follows immediately.
The integral is obtained following the substitution
$$ t = \frac{1}{1+x^n}, $$
and making use of the fact
$$ dx = -\frac{1}{n} \left(\frac{1}{t(1-t)} \right) \left( \frac{1-t}{t}\right)^{1/n} dt.$$