I am working on a integration by parts problem.

$\displaystyle \int tan^{-1}(4t)dt$

Setting $\displaystyle u=tan^{-1}(4t)$, then $\displaystyle dv=dt$, $\displaystyle dv=\frac{1}{1+(4t)^2}$ and $\displaystyle v=t$

Plugging into the equation, we get:

$\displaystyle tan^{-1}(4t)(t)-\int \frac{4t}{1+(4t)^2}dt$

I am good with all of that, except the 4 in the numerator of the integrand. I don't see where that comes from, since v=t. With the 4, it does lead to the correct answer.

Can someone explain?