Answer
$\pi$
Work Step by Step
Plug $p=\sqrt x \implies du=dx/2 \sqrt x$
Re-arrange the given integral as follows:
$\int_{0}^{\infty} \dfrac{dx}{(1+x)\sqrt {x}}=2\int_{0}^{\infty} \dfrac{dp}{(p^2+1)}$ ....(1)
Then, we have
$2\int_{0}^{\infty} \dfrac{dp}{(p^2+1)}=(2) \tan^{-1} (p) dp|_0^{\infty}$
Thus, equation (1) becomes
$(2) \tan^{-1} (\infty)-(2) \tan^{-1} (0)=\pi$
