Answer
$$\frac{2}{3}\sqrt {x - 2} \left( {x + 4} \right) + C $$
Work Step by Step
$$\eqalign{
& \int {\frac{{xdx}}{{\sqrt {x - 2} }}} \cr
& = \int {x{{\left( {x - 2} \right)}^{ - 1/2}}} dx \cr
& {\text{use the table of integrals from the book}} \cr
& {\text{by the formula 2}}2:\,\,\,\,\,\,\,\int {x{{\left( {ax + b} \right)}^n}dx = \frac{{{{\left( {ax + b} \right)}^{n + 1}}}}{{{a^2}}}\left[ {\frac{{ax + b}}{{n + 2}} - \frac{b}{{n + 1}}} \right]} + C \cr
& {\text{setting }}a = 1{\text{ and }}b = - 2{\text{ then}} \cr
& \int {x{{\left( {x - 2} \right)}^{ - 1/2}}} dx = \frac{{{{\left( {x - 2} \right)}^{ - 1/2 + 1}}}}{{{{\left( 1 \right)}^2}}}\left[ {\frac{{x - 2}}{{ - 1/2 + 2}} - \frac{{ - 2}}{{ - 1/2 + 1}}} \right] + C \cr
& {\text{simplifying}} \cr
& = {\left( {x - 2} \right)^{1/2}}\left[ {\frac{{x - 2}}{{3/2}} + \frac{2}{{1/2}}} \right] + C \cr
& = {\left( {x - 2} \right)^{1/2}}\left( {\frac{2}{3}x + \frac{8}{3}} \right) + C \cr
& = \frac{2}{3}\sqrt {x - 2} \left( {x + 4} \right) + C \cr} $$
