Answer
$\dfrac{\pi}{6}$
Work Step by Step
Let us consider that $I=\int_{12}^{\infty} \dfrac{1}{\sqrt x(x+4)} dx$
Plug $a=\sqrt x \implies dx=2a \ da$
Integrate the integral by using limits.
$I=\lim\limits_{z \to \infty}\int_{12}^{z} \dfrac{1}{\sqrt x(x+4)} dx\\=\lim\limits_{z \to \infty}[\arctan (\dfrac{\sqrt x}{2})]_{12}^z\\=\lim\limits_{z \to \infty}[\arctan (\dfrac{\sqrt z}{2})-\lim\limits_{z \to \infty}[\arctan (\dfrac{\sqrt {12}}{2})]\\=\dfrac{\pi}{2}-\dfrac{\pi}{3}\\=\dfrac{\pi}{6}$
