Answer
$$\frac{1}{3}{\left( {{x^2} + 12x} \right)^{3/2}} + C$$
Work Step by Step
$$\eqalign{
& \int {\left( {\sqrt {{x^2} + 12x} } \right)\left( {x + 6} \right)} dx \cr
& {\text{set }}u = {x^2} + 12x{\text{ then }}\frac{{du}}{{dx}} = 2x + 12,\,\,\,\,\,\frac{{du}}{{dx}} = 2\left( {x + 6} \right) \cr
& \to \left( {x + 6} \right)dx = \frac{1}{2}du \cr
& {\text{write the integrand in terms of }}u \cr
& \int {\left( {\sqrt {{x^2} + 12x} } \right)\left( {x + 6} \right)} dx = \int {\sqrt u } \left( {\frac{1}{2}du} \right) \cr
& = \frac{1}{2}\int {\sqrt u } du \cr
& {\text{write the radical as }}{u^{1/2}} \cr
& = \frac{1}{2}\int {{u^{1/2}}} du \cr
& {\text{using }}\int {{u^n}du = \frac{{{u^{n + 1}}}}{{n + 1}} + C} \cr
& = \frac{1}{2}\left( {\frac{{{u^{3/2}}}}{{3/2}}} \right) + C \cr
& = \frac{1}{3}{u^{3/2}} + C \cr
& {\text{replace }}{x^2} + 12x{\text{ for }}u \cr
& = \frac{1}{3}{\left( {{x^2} + 12x} \right)^{3/2}} + C \cr} $$
