Answer
$\sqrt{x(2-x)}+2\sin^{-1} (\sqrt{\dfrac{x}{2}})+C$
Work Step by Step
Suppose $u=\sqrt x \implies du=\dfrac{1}{2\sqrt x}dx \implies \dfrac{1}{\sqrt{x}}dx=2 \space du$
and Let $x=u^2$
Therefore, $I=2\int\sqrt{2-u^2}du\\=2(\dfrac{u}{2}\sqrt{2-u^2}+\dfrac{2}{2}\sin^{-1}\dfrac{u}{\sqrt2} )+C\\=u\sqrt{2-u^2}+2\sin^{-1}\dfrac{u}{\sqrt2}+C \\=\sqrt x\sqrt{2-x}+2\sin^{-1} (\dfrac{\sqrt x}{\sqrt2}) +C \\=\sqrt{x(2-x)}+2\sin^{-1} (\sqrt{\dfrac{x}{2}})+C$