# What is the antiderivative of  (x-3)^2?

Jun 17, 2015

It can be written as: $\frac{1}{3} {\left(x - 3\right)}^{3} + C$ or as $\frac{1}{3} {x}^{x} - 3 {x}^{2} + 9 x + C$.
(The $C$'s are different.)

#### Explanation:

Expansion :

${\left(x - 3\right)}^{2} = {x}^{2} - 6 x + 9$

Antidifferentiate term by term to get:

$\frac{1}{3} {x}^{3} - \frac{1}{2} \left(6 {x}^{2}\right) + 9 x + C$ which simplifies to:

$\frac{1}{3} {x}^{x} - 3 {x}^{2} + 9 x + C$.

Substitution
Let $u = \left(x - 3\right)$, so $\frac{\mathrm{du}}{\mathrm{dx}} = 1$ and we have ${u}^{2} \frac{\mathrm{du}}{\mathrm{dx}}$

and the antiderivative (by reversing the chain rule) of ${u}^{2} \frac{\mathrm{du}}{\mathrm{dx}}$

is $\frac{1}{3} {u}^{3} + C$.

Replacing $u$ by $x - 3$ gets us:

$\frac{1}{3} {\left(x - 3\right)}^{3} + C$.