Question

Find an anti derivative (or integral) of the given function by the method of inspection.
$(ax+b{)}^2$

Answer 1

We know that,
$\frac{d}{dx}(ax+b{)}^3=3a(ax+b{)}^2$
$\Rightarrow (ax+b{)}^2=\frac{1}{3a}\frac{d}{dx}(ax+b{)}^3$
$\therefore (ax+b{)}^2=\frac{d}{dx}\left(\frac{1}{3a}\right(ax+b{)}^3)$
Therefore, the anti derivative of $(ax+b{)}^2$ is $\frac{1}{3a}(ax+b{)}^3$.