Answer
$$2\left( {\frac{1}{3}{{\left( {\sqrt {2 - x} } \right)}^3} - 2\sqrt {2 - x} } \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{xdx}}{{\sqrt {2 - x} }}} \cr
& {\text{Integrate by the substitution method }} \cr
& \,\,{\text{Let }}u = 2 - x,\,\,\,\,\,\,\,x = 2 - u,\,\,\,\,\,\,dx = - du \cr
& \,\,\int {\frac{{xdx}}{{\sqrt {2 - x} }}} = \int {\frac{{2 - u}}{{\sqrt u }}} \left( { - du} \right) \cr
& \,\, = \int {\frac{{u - 2}}{{{u^{1/2}}}}} du \cr
& \cr
& {\text{Use the distributive property}} \cr
& = \int {\left( {\frac{u}{{{u^{1/2}}}} - \frac{2}{{{u^{1/2}}}}} \right)} du \cr
& = \int {\left( {{u^{1/2}} - 2{u^{ - 1/2}}} \right)} du \cr
& \cr
& {\text{Use the power rule for integration}} \cr
& = \frac{{{u^{3/2}}}}{{3/2}} - \frac{{2{u^{1/2}}}}{{1/2}} + C \cr
& = \frac{2}{3}{u^{3/2}} - 4{u^{1/2}} + C \cr
& \cr
& {\text{Write in terms of }}x,{\text{ replace }}2 - x{\text{ for }}u \cr
& = \frac{2}{3}{\left( {2 - x} \right)^{3/2}} - 4{\left( {2 - x} \right)^{1/2}} + C \cr
& {\text{Factoring}} \cr
& = 2\left( {\frac{1}{3}{{\left( {{{\left( {2 - x} \right)}^{1/2}}} \right)}^3} - 2{{\left( {2 - x} \right)}^{1/2}}} \right) + C \cr
& = 2\left( {\frac{1}{3}{{\left( {\sqrt {2 - x} } \right)}^3} - 2\sqrt {2 - x} } \right) + C \cr} $$
