Question

How do you integrate `f(x)=(x^2+x+1)(x-1)` using the product rule?

Answer 1

There is no product rule for integration.

Answer 2

Jim H. is correct; there is no product rule for integration but there is an integration method called, Integration by Parts , that is so closely related to the product rule that the product rule can be used to derive the method. The reference contains the derivation and I will not repeat it here but I will use the method to perform the integration.

NOTE: Normally, I would not use integration by parts; I would, perform the multiplication, simplify, and then integrate each term but you requested the use of the product rule, therefore, I will use the closest possible relative to the product rule.

The integration by parts formula is:

`int udv = uv - intvdu`

To apply the formula to the given factors, we let: `u = (x-1)` and `dv = (x^2+x+1)dx` this implies that `du = 1 dx` and `v = (x^3/3+x^2/2+x)`

Substituting these 4 equations into the formula:

`int (x-1)(x^2+x+1)dx = (x-1)(x^3/3+x^2/2+x) - int(x^3/3+x^2/2+x)dx`

The remaining integral is done "term-by-term", using the power rule, `int x^n dx = 1/(n+1)x^(n+1) + C`:

`int (x-1)(x^2+x+1)dx = (x-1)(x^3/3+x^2/2+x) - (x^4/12+x^3/6+x^2/2)+ C`

Please compare the above answer with the answer that one receives, if one just performs the multiplication:

`int (x-1)(x^2+x+1)dx = int x(x^2+x+1)-(x^2+x+1)dx`

`int (x-1)(x^2+x+1)dx = int x^3+x^2+x-x^2-x-1dx`

`int (x-1)(x^2+x+1)dx = int x^3-1dx`

`int (x-1)(x^2+x+1)dx = x^4/4-x+ C`

This method does not require the tedious simplification that the other method requires.