Answer
$$\int \frac{dy}{\sqrt{\tan^{-1}y}(1+y^2)}=2\sqrt{\tan^{-1}y}+C$$
Work Step by Step
$$A=\int \frac{dy}{\sqrt{\tan^{-1}y}(1+y^2)}$$
We set $a=\sqrt{\tan^{-1}y}$, which means $$da=\frac{(\tan^{-1}y)'dy}{2\sqrt{\tan^{-1}y}}=\frac{dy}{2\sqrt{\tan^{-1}y}(1+y^2)}$$ $$\frac{dy}{\sqrt{\tan^{-1}y}(1+y^2)}=2da$$
Therefore, $$A=\int 2da$$ $$A=2a+C$$ $$A=2\sqrt{\tan^{-1}y}+C$$
