Prove

The anti derivative of $(a x+b)^{2}$ is the function of $x$ whose derivative is $(a x+b)^{2}$.

It is known that,

$\frac{d}{d x}(a x+b)^{3}=3 a(a x+b)^{2}$

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

$\therefore(a x+b)^{2}=\frac{d}{d x}\left(\frac{1}{3 a}(a x+b)^{3}\right)$

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