Question

find the Anti derivatives or integral of function (ax  + b)² the method of inspection. 

Answer 1

HELLO DEAR,

the anti derivatives of (ax + b)² is a function of x whose derivative is (ax + b)².

we know:- $\bold{\frac{d}{dx}(ax + b)^3 = 3a(ax + b)^2}$

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

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

Hence, Anti derivatives of (ax + b)² is 1/3a(ax + b)².

I HOPE ITS HELP YOU DEAR,
THANKS