Answer
$y \tan^{-1}2y - \frac{1}{4} \ln(1+4y^2) + C$
Work Step by Step
Let $u = \tan^{-1} 2y, dv = dy$ and $du = \frac{2}{1+4y^2} dy, v = y$
Then
$\int \tan^{-1}2y dy$
$= y \tan^{-1} 2y - \int \frac{2y}{1+4y^2} dy$
$= y \tan^{-1} 2y - \int \frac{1}{t} (\frac{1}{4} dt)$
$= y \tan^{-1} 2y - \frac{1}{4}\ln |t| + C$
$= y \tan^{-1} 2y - \frac{1}{4}\ln (1+4y^2) + C$