Answer
$$2{\left( {x - 2} \right)^{1/2}}\left( {x + 4} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{3x}}{{\sqrt {x - 2} }}} dx \cr
& {\text{set }}u = x - 2{\text{ then }}\frac{{du}}{{dx}} = 1,\,\,\,\,\,du = dx \cr
& u = x - 2 \to x = u + 2 \cr
& {\text{write the integrand in terms of }}u \cr
& \int {\frac{{3x}}{{\sqrt {x - 2} }}} dx = \int {\frac{{3\left( {u + 2} \right)}}{{\sqrt u }}} du \cr
& = 3\int {\frac{{u + 2}}{{{u^{1/2}}}}} du \cr
& = 3\int {\left( {{u^{1/2}} + 2{u^{ - 1/2}}} \right)} du \cr
& {\text{using }}\int {{u^n}du = \frac{{{u^{n + 1}}}}{{n + 1}} + C} \cr
& = 3\left( {\frac{{{u^{3/2}}}}{{3/2}} + \frac{{2{u^{1/2}}}}{{1/2}}} \right) + C \cr
& = 3\left( {\frac{{2{u^{3/2}}}}{3} + 4{u^{1/2}}} \right) + C \cr
& = 2{u^{3/2}} + 12{u^{1/2}} + C \cr
& {\text{replace }}x - 2{\text{ for }}u \cr
& = 2{\left( {x - 2} \right)^{3/2}} + 12{\left( {x - 2} \right)^{1/2}} + C \cr
& {\text{factoring}} \cr
& = 2{\left( {x - 2} \right)^{1/2}}\left[ {x - 2 + 6} \right] + C \cr
& = 2{\left( {x - 2} \right)^{1/2}}\left( {x + 4} \right) + C \cr} $$
