Question

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

Answer 1

It can be written as: `1/3(x-3)^3+C` or as `1/3x^x-3x^2+9x +C`.
(The `C`'s are different.)

Explanation:

Expansion :

`(x-3)^2 = x^2-6x+9`

Antidifferentiate term by term to get:

`1/3x^3-1/2(6x^2)+9x+C` which simplifies to:

`1/3x^x-3x^2+9x +C`.

Substitution
Let `u=(x-3)`, so `(du)/dx = 1` and we have `u^2 (du)/dx`

and the antiderivative (by reversing the chain rule) of `u^2 (du)/dx`

is `1/3 u^3 +C`.

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

`1/3 (x-3)^3 +C`.