Let

$\displaystyle \displaystyle I(a)=\int_0^{\infty}\frac{dx}{1+x^a}$

Let $\displaystyle 1+x^a=u$ and so $\displaystyle du=a x^{a-1}$ or $\displaystyle dx=\frac{1}{a}\left(u-1\right)^{\frac{1}{a}-1}$. Thus,

$\displaystyle \displaystyle I(a)=\frac{1}{a}\int_1^{\infty}\frac{(u-1)^{\frac{1}{a}-1}}{u}\text{ }du$

Let $\displaystyle z=\frac{1}{u}$ to get that

$\displaystyle \displaystyle \begin{aligned}I(a) &=\frac{1}{a}\int_0^1 z^{\frac{-1}{a}}\left(1-z\right)^{\frac{1}{a}-1}\\ &= \frac{1}{a}B\left(1-\frac{1}{a},\frac{1}{a}\right)\\ &= \frac{1}{a}\frac{\Gamma\left(\frac{1}{a}\right)\Ga mma\left(1-\frac{1}{a}\right)}{\Gamma(1)}\\ &=\frac{1}{a}\Gamma\left(\frac{1}{a}\right)\Gamma\ left(1-\frac{1}{a}\right)\\ &= \frac{\pi}{a\sin\left(\frac{\pi}{a}\right)}\end{al igned}$