Use the formula

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\sqrt[\phantom{2}]{2au-u^2}du=\frac{u-a}{2}\sqrt[\phantom{2}]{2au-u^2}+\frac{a^2}{2}co{s}^{-1}(\frac{a-u}{a})+C$

to evaluate the following integral.

${\displaystyle \underset{0}{\overset{1}{}}}\sqrt[\phantom{2}]{x-x^2}dx$