Answer
$$\int\tan^{-1}ydy=y\tan^{-1}y-\frac{1}{2}\ln(1+y^2)+C$$
Work Step by Step
$$A=\int\tan^{-1}ydy$$
Take $u=\tan^{-1}y$ and $dv=dy$
We then have $du=\frac{1}{1+y^2}dy$ and $v=y$
Apply the formula $\int udv= uv-\int vdu$, we have $$A=y\tan^{-1}y-\int\frac{y}{1+y^2}dy$$
Take $z=1+y^2$, we then have $dz=2ydy$, meaning that $ydy=\frac{1}{2}dz$
That means $$A=y\tan^{-1}y-\int\frac{\frac{1}{2}dz}{z}$$ $$A=y\tan^{-1}y-\frac{1}{2}\int\frac{1}{z}dz$$ $$A=y\tan^{-1}y-\frac{1}{2}(\ln |z|)+C$$
Replace $z$ with $1+y^2$ back: $$A=y\tan^{-1}y-\frac{1}{2}\ln|1+y^2|+C$$
However, since $y^2+1\ge1\gt0$ for all $y$, we have $|1+y^2|=1+y^2$
$$A=y\tan^{-1}y-\frac{1}{2}\ln(1+y^2)+C$$
