Indefinite integral: ∫ (x^3 - 1)^2 dx?

Question

Help me understand this. I understand that indefinite intergrals are the antiderivative of the equation plus a constant.
For example, ∫ cos(x) dx = sin(x)+C where C is an unknown constant [because cos(x) is derivative of sin(x)]. However, I don't know where to start with this:
∫ (x^3 - 1)^2 dx
Please help me understand how to do this.

Answer 1

start by writing it out like

integral ( (x^3 - 1)^2 ) dx =
integral ( (x^6 - 2x^3 + 1) )dx =
x^7 / 7 - 1/2 x^4 + x + C

Answer 2

Expand the bracket first and then integrate each term:

∫ (x^3-1)^2 = ∫(x^6 - 2x^3+1)dx = x^7/7 - x^4/2 + x + C


Generally, ∫x^n dx = x^n+1/n+1
and also ∫1 dx = x

Answer 3

(x³ - 1).(x³ - 1) = x^(6) - 2x³ + 1
I = ∫ x^(6) - 2.x³ + 1 dx
I = x^(7) / 7 - 2.x^(4) / 4 + x + C
I = x^7 / 7 - (1/2).x^4 + x + C