Answer
$$\frac{{2{x^2} - \left( {1/2} \right)x - 3/4}}{6}\sqrt {x - {x^2}} + \frac{1}{{16}}{\sin ^{ - 1}}\left( {2x - 1} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {x\sqrt {x - {x^2}} } dx \cr
& {\text{write in terms of }}u.{\text{ let }}u = x,\,\,\,du = dx \cr
& = \int {u\sqrt {u - {u^2}} } du \cr
& \cr
& {\text{Use the Endpaper Integral Table to evaluate the integral by the formula }} \cr
& \int {u\sqrt {2au - {u^2}} du} = \frac{{2{u^2} - au - 3{a^2}}}{6}\sqrt {2au - {u^2}} + \frac{{{a^3}}}{2}{\sin ^{ - 1}}\left( {\frac{{u - a}}{a}} \right) + C \cr
& {\text{we have }}u - {u^2} = 2au - {u^2},{\text{ then }}u = 2au,\,\,\,a = 1/2 \cr
& \int {u\sqrt {u - {u^2}} } du = \frac{{2{u^2} - \left( {1/2} \right)u - 3{{\left( {1/2} \right)}^2}}}{6}\sqrt {u - {u^2}} + \frac{{{{\left( {1/2} \right)}^3}}}{2}{\sin ^{ - 1}}\left( {\frac{{u - 1/2}}{{1/2}}} \right) + C \cr
& {\text{simplifying}} \cr
& \int {u\sqrt {u - {u^2}} } du = \frac{{2{u^2} - \left( {1/2} \right)u - 3/4}}{6}\sqrt {u - {u^2}} + \frac{1}{{16}}{\sin ^{ - 1}}\left( {2u - 1} \right) + C \cr
& \cr
& {\text{Write in terms of }}x{\text{, and replace }}x{\text{ for }}u \cr
& \int {x\sqrt {x - {x^2}} } dx = \frac{{2{x^2} - \left( {1/2} \right)x - 3/4}}{6}\sqrt {x - {x^2}} + \frac{1}{{16}}{\sin ^{ - 1}}\left( {2x - 1} \right) + C \cr} $$
