Answer
$$\int\frac{\sqrt{2-x}}{\sqrt{x}}dx=\sqrt{x(2-x)}+2\sin^{-1}\sqrt{\frac{x}{2}}+C$$
Work Step by Step
$$A=\int\frac{\sqrt{2-x}}{\sqrt{x}}dx$$
We set $u=\sqrt x$, which means $$du=\frac{1}{2\sqrt x}dx$$ $$\frac{1}{\sqrt{x}}dx=2du$$
Also, we can rewrite $x=u^2$
Therefore, $$A=2\int\sqrt{2-u^2}du$$
Here apply Formula 45, which states that
$$\int\sqrt{a^2-x^2}dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\frac{x}{a}+C$$
for $a=\sqrt2$.
$$A=2\Big(\frac{u}{2}\sqrt{2-u^2}+\frac{2}{2}\sin^{-1}\frac{u}{\sqrt2}\Big)+C$$ $$A=u\sqrt{2-u^2}+2\sin^{-1}\frac{u}{\sqrt2}+C$$ $$A=\sqrt x\sqrt{2-x}+2\sin^{-1}\frac{\sqrt x}{\sqrt2}+C$$ $$A=\sqrt{x(2-x)}+2\sin^{-1}\sqrt{\frac{x}{2}}+C$$