Answer
$$\frac{{2\pi \sqrt 6 }}{3}$$
Work Step by Step
$$\eqalign{
& \int_0^\infty {\frac{4}{{\sqrt x \left( {x + 6} \right)}}} dx \cr
& \frac{4}{{\sqrt x \left( {x + 6} \right)}}{\text{ has an infinite discontinuity at }}x = 0,{\text{so we can write}} \cr
& \int_0^\infty {\frac{4}{{\sqrt x \left( {x + 6} \right)}}} dx = \mathop {\lim }\limits_{a \to {0^ + }} \int_a^1 {\frac{4}{{\sqrt x \left( {x + 6} \right)}}} dx + \mathop {\lim }\limits_{b \to \infty } \int_1^b {\frac{4}{{\sqrt x \left( {x + 6} \right)}}} dx \cr
& \cr
& *{\text{Integrating }}\int {\frac{4}{{\sqrt x \left( {x + 6} \right)}}dx} \cr
& {\text{Let }}{u^2} = x,{\text{ }}2udu = dx \cr
& \int {\frac{4}{{\sqrt x \left( {x + 6} \right)}}dx} = \int {\frac{4}{{\sqrt {{u^2}} \left( {{u^2} + 6} \right)}}\left( {2u} \right)du} = \int {\frac{8}{{{u^2} + 6}}du} \cr
& = \frac{8}{{\sqrt 6 }}{\tan ^{ - 1}}\left( {\frac{u}{{\sqrt 6 }}} \right) + C \cr
& = \frac{8}{{\sqrt 6 }}{\tan ^{ - 1}}\left( {\frac{{\sqrt x }}{{\sqrt 6 }}} \right) + C,{\text{ then}} \cr
& \cr
& \mathop {\lim }\limits_{a \to {0^ + }} \int_a^1 {\frac{4}{{\sqrt x \left( {x + 6} \right)}}} dx + \mathop {\lim }\limits_{b \to \infty } \int_1^b {\frac{4}{{\sqrt x \left( {x + 6} \right)}}} dx \cr
& = \mathop {\lim }\limits_{a \to {0^ + }} \left[ {\frac{8}{{\sqrt 6 }}{{\tan }^{ - 1}}\left( {\frac{{\sqrt x }}{{\sqrt 6 }}} \right)} \right]_a^1 + \mathop {\lim }\limits_{b \to \infty } \left[ {\frac{8}{{\sqrt 6 }}{{\tan }^{ - 1}}\left( {\frac{{\sqrt x }}{{\sqrt 6 }}} \right)} \right]_1^b \cr
& = \frac{8}{{\sqrt 6 }}\mathop {\lim }\limits_{a \to {0^ + }} \left[ {{{\tan }^{ - 1}}\left( {\frac{1}{{\sqrt 6 }}} \right) - {{\tan }^{ - 1}}\left( {\frac{a}{{\sqrt 6 }}} \right)} \right] \cr
& + \frac{8}{{\sqrt 6 }}\mathop {\lim }\limits_{b \to \infty } \left[ {{{\tan }^{ - 1}}\left( {\frac{{\sqrt b }}{{\sqrt 6 }}} \right) - {{\tan }^{ - 1}}\left( {\frac{1}{{\sqrt 6 }}} \right)} \right] \cr
& {\text{Evaluate the limits}} \cr
& = - \frac{8}{{\sqrt 6 }}{\tan ^{ - 1}}\left( {\frac{{{0^ + }}}{{\sqrt 6 }}} \right) + \frac{8}{{\sqrt 6 }}{\tan ^{ - 1}}\left( {\frac{{\sqrt \infty }}{{\sqrt 6 }}} \right) \cr
& = - \frac{8}{{\sqrt 6 }}\left( 0 \right) + \frac{8}{{\sqrt 6 }}\left( {\frac{\pi }{2}} \right) \cr
& = \frac{{4\pi }}{{\sqrt 6 }} \cr
& {\text{Rationalizing}} \cr
& = \frac{{4\pi \sqrt 6 }}{{\sqrt 6 \sqrt 6 }} \cr
& = \frac{{2\pi \sqrt 6 }}{3} \cr} $$
