Answer
$$\frac{2}{{15}}{\left( {x - 2} \right)^{3/2}}\left( {3x + 4} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {x\sqrt {x - 2} } dx \cr
&\text{We find}
& \\ u = {\left( {x + a} \right)^{1/n}} \cr
& {\text{Then}}{\text{,}} \cr
& {\text{Take }}\,\,\,u = {\left( {x - 2} \right)^{1/2}} \to \,\,\,\,\,\,\,\,{u^2} = x - 2,\,\,\,\,\,\,2udu = dx \cr
& \cr
& {\text{write the integrand in terms of }}u \cr
& \int {x\sqrt {x - 2} } dx = \int {\left( {{u^2} + 2} \right)u} \left( {2udu} \right) \cr
& = \int {\left( {{u^2} + 2} \right)} \left( {2{u^2}du} \right) \cr
& = \int {\left( {2{u^4} + 4{u^2}} \right)} du \cr
& \cr
& {\text{integrating}} \cr
& = \frac{{2{u^5}}}{5} + \frac{{4{u^3}}}{3} + C \cr
& {\text{factoring}} \cr
& = \frac{2}{{15}}{u^3}\left( {3{u^2} + 10} \right) + C \cr
& \cr
& {\text{write the integrand in terms of }}x{\text{; replace }}u = {\left( {x - 2} \right)^{1/2}} \cr
& = \frac{2}{{15}}{\left( {{{\left( {x - 2} \right)}^{1/2}}} \right)^3}\left( {3{{\left( {{{\left( {x - 2} \right)}^{1/2}}} \right)}^2} + 10} \right) + C \cr
& = \frac{2}{{15}}{\left( {x - 2} \right)^{3/2}}\left( {3x - 6 + 10} \right) + C \cr
& = \frac{2}{{15}}{\left( {x - 2} \right)^{3/2}}\left( {3x + 4} \right) + C \cr} $$
