Question

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

Answer 1

We are looking for a function whose derivative is $(ax+b{)}^2$
$\frac{d}{dx}(ax+b{)}^3=3a(ax+b{)}^2$
$\Rightarrow (ax+b{)}^2=\frac{1}{3a}\frac{d}{dx}(ax+b{)}^3$
$\Rightarrow (ax+b{)}^2=\frac{d}{dx}\left[\frac{1}{3a}\right(ax+b{)}^3]$
Hence, the anti-derivative of $(ax+b{)}^2$ is $\frac{1}{3a}(ax+b{)}^3$