Answer
$(y+1)\tan^{-1}\sqrt y-\sqrt y+C$
Work Step by Step
Use Formula:
$\int x^n\tan^{-1} \space axdx=\dfrac{x^{n+1}}{n+1}\tan^{-1}ax-\dfrac{a}{n+1}\int\frac{x^{n+1}dx}{1+a^2x^2}$
Let $I=\int\tan^{-1}\sqrt y \space dy$
Suppose $u=\sqrt y \implies du=\dfrac{dy}{2\sqrt y} \implies du=\dfrac{dy}{2u}$
Therefore, $I=2\int u\tan^{-1}(u) \space du\\=2(\dfrac{u^2}{2}\tan^{-1}(u)-\dfrac{1}{2}\int\dfrac{u^2}{1+u^2}du)\\= u^2\tan^{-1} (u)-(u-\tan^{-1}(u))+C\\=(u^2+1)\tan^{-1}(u)-u+C\\=(y+1)\tan^{-1}\sqrt y-\sqrt y+C$
