Question

Find an anti derivative (or integral) of the following function by the method of inspection.

(axe + b)²

Answer 1

We know that,

${\displaystyle \frac{d}{dx}}$ (ax + b)² = 3a (ax + b)²

${\displaystyle \Rightarrow {(ax+b)}^2=\frac{1}{3a}\frac{d}{dx}{(ax+b)}^3}$

or (ax + b)² = ${\displaystyle \frac{d}{dx}\left[\frac{1}{3a}{(axee+b)}^3\right]}$ 

Hence, the antiderivative of (ax + b)² is ${\displaystyle \frac{1}{3a}}$(ax + b)³.