Answer
$$\pi$$
Work Step by Step
Simplify the improper integral $\int_{0}^{\infty} \dfrac{1}{\sqrt x (x+1)} dx$
Let us consider that $I=\int_{0}^{\infty} \dfrac{1}{\sqrt x (x+1)} dx$
Plug $a=\sqrt x \implies dx=2 a \ da$
Integrate the integral by using limits.
$I=2 \int_0^{\infty} \dfrac{2}{a^{2}+1} da \\=2 \arctan(a)|_0^{\infty} \\=2 \arctan(\sqrt x)|_0^{\infty} \\= 2( \dfrac{\pi}{2}-0)\\=\pi$
