Answer
$$\int\tan^{-1}\sqrt ydy=(y+1)\tan^{-1}\sqrt y-\sqrt y+C$$
Work Step by Step
$$A=\int\tan^{-1}\sqrt ydy$$
We set $u=\sqrt y$, which means $$du=\frac{dy}{2\sqrt y}=\frac{dy}{2u}$$ $$dy=2udu$$
Therefore, $$A=2\int u\tan^{-1}udu$$
Use Formula 108, which states that
$$\int x^n\tan^{-1}axdx=\frac{x^{n+1}}{n+1}\tan^{-1}ax-\frac{a}{n+1}\int\frac{x^{n+1}dx}{1+a^2x^2}$$
for $a=1$ and $n=1$.
$$A=2\Big(\frac{u^2}{2}\tan^{-1}u-\frac{1}{2}\int\frac{u^2}{1+u^2}du\Big)$$ $$A=u^2\tan^{-1}u-\int\frac{u^2}{1+u^2}du$$ $$A=u^2\tan^{-1}u-\Big(\int1du-\int\frac{1}{1+u^2}du\Big)$$ $$A=u^2\tan^{-1}u-\Big(u-\tan^{-1}u\Big)+C$$ $$A=u^2\tan^{-1}u+\tan^{-1}u-u+C$$ $$A=(u^2+1)\tan^{-1}u-u+C$$ $$A=(y+1)\tan^{-1}\sqrt y-\sqrt y+C$$
