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

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Answer 1 · 2015-06-17T20:42:52

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.)

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#.