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

Toolbox:
• $\frac{d}{dx}[F(x)+c]=f(x),$
• $\int f(x)\;dx=F(x)+c$
We know that $\frac{d}{dx}(ax+b)^3=3a(ax+b)^2$

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

$\qquad\;\;=\frac{d}{dx}\begin{bmatrix}\frac{1}{3a}(ax+b)^3\end{bmatrix}.$

Hence $\frac{1}{3a}(ax+b)^3$ is an anti derivative of $(ax+b)^2$.