Question

Find an anti derivative (or integral) of the following function by the method of inspection: (ax + b)2

Answer 1

The anti derivative of (ax + b)2 is the function of x whose derivative is (ax + b)2 .
It is known that,

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

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

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

Therefore, the anti derivative of (ax + b)2 is $\frac{1}{3a}$(ax + b)3 .