Answer
$$\pi $$
Work Step by Step
$$\eqalign{
& \int_0^\infty {\frac{2}{{\sqrt x \left( {x + 4} \right)}}dx} \cr
& \frac{2}{{\sqrt x \left( {x + 4} \right)}}{\text{ has an infinite discontinuity at }}x = 0,{\text{so we can write}} \cr
& \int_0^\infty {\frac{2}{{\sqrt x \left( {x + 4} \right)}}dx} = \mathop {\lim }\limits_{a \to {0^ + }} \int_a^1 {\frac{2}{{\sqrt x \left( {x + 4} \right)}}} dx + \mathop {\lim }\limits_{b \to \infty } \int_1^b {\frac{2}{{\sqrt x \left( {x + 4} \right)}}} dx \cr
& \cr
& *{\text{Integrating }}\int {\frac{2}{{\sqrt x \left( {x + 4} \right)}}dx} \cr
& {\text{Let }}{u^2} = x,{\text{ }}2udu = dx \cr
& \int {\frac{2}{{\sqrt x \left( {x + 4} \right)}}dx} = \int {\frac{2}{{\sqrt {{u^2}} \left( {{u^2} + 4} \right)}}\left( {2u} \right)du} = \int {\frac{4}{{{u^2} + 4}}du} \cr
& = \frac{4}{2}{\tan ^{ - 1}}\left( {\frac{u}{2}} \right) + C \cr
& = 2{\tan ^{ - 1}}\left( {\frac{{\sqrt x }}{2}} \right) + C,{\text{ then}} \cr
& \cr
& \mathop {\lim }\limits_{a \to {0^ + }} \int_a^1 {\frac{2}{{\sqrt x \left( {x + 4} \right)}}} dx + \mathop {\lim }\limits_{b \to \infty } \int_1^b {\frac{2}{{\sqrt x \left( {x + 4} \right)}}} dx \cr
& = \mathop {\lim }\limits_{a \to {0^ + }} \left[ {2{{\tan }^{ - 1}}\left( {\frac{{\sqrt x }}{2}} \right)} \right]_a^1 + \mathop {\lim }\limits_{b \to \infty } \left[ {2{{\tan }^{ - 1}}\left( {\frac{{\sqrt x }}{2}} \right)} \right]_1^b \cr
& = 2\mathop {\lim }\limits_{a \to {0^ + }} \left[ {{{\tan }^{ - 1}}\left( {\frac{{\sqrt 1 }}{2}} \right) - {{\tan }^{ - 1}}\left( {\frac{{\sqrt a }}{2}} \right)} \right] \cr
& + 2\mathop {\lim }\limits_{b \to \infty } \left[ {{{\tan }^{ - 1}}\left( {\frac{{\sqrt b }}{2}} \right) - {{\tan }^{ - 1}}\left( {\frac{{\sqrt 1 }}{2}} \right)} \right] \cr
& {\text{Evaluate the limits}} \cr
& = 2{\tan ^{ - 1}}\left( {\frac{1}{2}} \right) - {\tan ^{ - 1}}\left( {\frac{0}{2}} \right) + 2{\tan ^{ - 1}}\left( {\frac{\infty }{2}} \right) - 2{\tan ^{ - 1}}\left( {\frac{1}{2}} \right) \cr
& = 2{\tan ^{ - 1}}\left( \infty \right) \cr
& = 2\left( {\frac{\pi }{2}} \right) \cr
& = \pi \cr} $$
