## University Calculus: Early Transcendentals (3rd Edition)

$$\int\frac{\sqrt{x-x^2}}{x}dx=\sqrt{x-x^2}+\frac{1}{2}\sin^{-1}\Big(2x-1\Big)+C$$
$$A=\int\frac{\sqrt{x-x^2}}{x}dx$$ Use the Formula 125, which states that $$\int \frac{\sqrt{2ax-x^2}}{x}dx=\sqrt{2ax-x^2}+a\sin^{-1}\Big(\frac{x-a}{a}\Big)+C$$ for $a=1/2$ here: $$A=\sqrt{x-x^2}+\frac{1}{2}\sin^{-1}\Big(\frac{x-\frac{1}{2}}{\frac{1}{2}}\Big)+C$$ $$A=\sqrt{x-x^2}+\frac{1}{2}\sin^{-1}\Big(2x-1\Big)+C$$