I am trying to evaluate $\int \arctan4t \, dt$
I make $u = \arctan4t$, $du = \frac{4}{1-4t}\,dt$
and $dv = dt$ and $v = t$
I then make it the form of $uv = \int v \, du$
$t\arctan4t - \int \frac{4t}{1-16t^2} \, dt$
Now to get the integral of $\int \frac{4t}{1-16t^2} \, dt$ I just pull out the 4 and use the identity I have memorized.
But I just now realised that this is not possible since I have a $16t^2$ so I have no idea how to advance from here.