Answer
$$2{\left( {x + 4} \right)^{3/2}} - 24{\left( {x + 4} \right)^{1/2}} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{3x}}{{\sqrt {x + 4} }}} dx \cr
& {\text{substitute }}u = x + 4,{\text{ }}x = u - 4,{\text{ }}dx = du \cr
& \int {\frac{{3x}}{{\sqrt {x + 4} }}} dx = \int {\frac{{3\left( {u - 4} \right)}}{{\sqrt u }}} du \cr
& = \int {\frac{{3u - 12}}{{\sqrt u }}} du \cr
& = \int {\left( {\frac{{3u}}{{\sqrt u }} - \frac{{12}}{{\sqrt u }}} \right)du} \cr
& = \int {\left( {3{u^{1/2}} - 12{u^{ - 1/2}}} \right)} du \cr
& {\text{find the antiderivative}} \cr
& = 3\left( {\frac{{{u^{3/2}}}}{{3/2}}} \right) - 12\left( {\frac{{{u^{1/2}}}}{{1/2}}} \right) + C \cr
& = 2{u^{3/2}} - 24{u^{1/2}} + C \cr
& {\text{ replacing }}u = x + 4 \cr
& = 2{\left( {x + 4} \right)^{3/2}} - 24{\left( {x + 4} \right)^{1/2}} + C \cr} $$
