Answer
$$\int\frac{\cos^{-1}\sqrt x}{\sqrt x}dx=2\sqrt x\cos^{-1}\sqrt x-2\sqrt{1-x}+C$$
Work Step by Step
$$A=\int\frac{\cos^{-1}\sqrt x}{\sqrt x}dx$$
We set $u=\sqrt x$, which means $$du=\frac{1}{2\sqrt x}dx$$ $$\frac{1}{\sqrt x}dx=2du$$
Therefore, $$A=2\int\cos^{-1}udu$$
Now apply Formula 104, which states that
$$\int \cos^{-1}axdx=x\cos^{-1}ax-\frac{1}{a}\sqrt{1-a^2x^2}+C$$
for $a=1$.
$$A=2\Big(u\cos^{-1}u-\frac{1}{1}\sqrt{1-u^2}+C\Big)$$ $$A=2\sqrt x\cos^{-1}\sqrt x-2\sqrt{1-x}+C$$
