Answer
$\sqrt{x-x^2}+\dfrac{1}{2}\sin^{-1} (2x-1 )+C$
Work Step by Step
Apply formula : $\int \frac{\sqrt{2ax-x^2}}{x}dx=\sqrt{2ax-x^2}+a\sin^{-1} (\dfrac{x-a}{a})+C$
Consider $I=\int\dfrac{\sqrt{x-x^2}}{x}dx =\sqrt{x-x^2}+\dfrac{1}{2}\sin^{-1} (\dfrac{x-(1/2)}{\dfrac{1}{2}})+C \\=\sqrt{x-x^2}+\dfrac{1}{2}\sin^{-1} (2x-1 )+C$
