Answer
$$\frac{{x - 1}}{2}\sqrt {2x - {x^2}} + \frac{1}{2}{\sin ^{ - 1}}\left( {x - 1} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\sqrt {2x - {x^2}} } dx \cr
& {\text{Complete the square to the radicand }}2x - {x^2} \cr
& 2x - {x^2} = - \left( {{x^2} - 2x + 1} \right) + 1 \cr
& 2x - {x^2} = 1 - {\left( {x - 1} \right)^2} \cr
& \cr
& \int {\sqrt {2x - {x^2}} } dx = \int {\sqrt {1 - {{\left( {x - 1} \right)}^2}} } dx \cr
& \cr
& {\text{Let }}u = x - 1,\,\,\,\,du = dx \cr
& \,\,\,\,\,\,\,\,\int {\sqrt {1 - {{\left( {x - 1} \right)}^2}} } dx = \int {\sqrt {1 - {u^2}} } du \cr
& \cr
& {\text{Integrate by tables using }}\int {\sqrt {{a^2} - {u^2}} du = \frac{u}{2}\sqrt {{a^2} - {u^2}} + \frac{{{a^2}}}{2}{{\sin }^{ - 1}}\left( {\frac{u}{a}} \right) + C} \cr
& \int {\sqrt {1 - {u^2}} } du = \frac{u}{2}\sqrt {1 - {u^2}} + \frac{1}{2}{\sin ^{ - 1}}\left( u \right) + C \cr
& \cr
& {\text{write in terms of }}x,{\text{ replacing }}x - 1{\text{ for }}u \cr
& \int {\sqrt {2x - {x^2}} } dx = \frac{{x - 1}}{2}\sqrt {2x - {x^2}} + \frac{1}{2}{\sin ^{ - 1}}\left( {x - 1} \right) + C \cr} $$
