Answer
$=x\arctan{2x}-\frac{1}{8}\ln(4x^{2}+1)+C$
Work Step by Step
$\int \arctan {2x}dx$
$=uv-\int u'vdx$
$u=\arctan{2x},dv=\int dx$
$u'=\frac{1}{1+4x^{2}}dx,v=x$
$=x\arctan{2x}-\int \frac{x}{4x^{2}+1}dx$
Use u-substitution. (I use z here to not get confused with the previous u)
$z=4x^{2}+1$
$\frac{dz}{dx}=8x$
$dx=\frac{1}{8x}dz$
$=x\arctan{2x}-\frac{1}{8}\int \frac{dz}{z}$
$=x\arctan{2x}-\frac{1}{8}\ln(4x^{2}+1)+C$
