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

Mar 3, 2018

There is no product rule for integration.

Mar 3, 2018

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:

$\int u \mathrm{dv} = u v - \int v \mathrm{du}$

To apply the formula to the given factors, we let: $u = \left(x - 1\right)$ and $\mathrm{dv} = \left({x}^{2} + x + 1\right) \mathrm{dx}$ this implies that $\mathrm{du} = 1 \mathrm{dx}$ and $v = \left({x}^{3} / 3 + {x}^{2} / 2 + x\right)$

Substituting these 4 equations into the formula:

$\int \left(x - 1\right) \left({x}^{2} + x + 1\right) \mathrm{dx} = \left(x - 1\right) \left({x}^{3} / 3 + {x}^{2} / 2 + x\right) - \int \left({x}^{3} / 3 + {x}^{2} / 2 + x\right) \mathrm{dx}$

The remaining integral is done "term-by-term", using the power rule, $\int {x}^{n} \mathrm{dx} = \frac{1}{n + 1} {x}^{n + 1} + C$:

$\int \left(x - 1\right) \left({x}^{2} + x + 1\right) \mathrm{dx} = \left(x - 1\right) \left({x}^{3} / 3 + {x}^{2} / 2 + x\right) - \left({x}^{4} / 12 + {x}^{3} / 6 + {x}^{2} / 2\right) + C$

$\int \left(x - 1\right) \left({x}^{2} + x + 1\right) \mathrm{dx} = \int x \left({x}^{2} + x + 1\right) - \left({x}^{2} + x + 1\right) \mathrm{dx}$
$\int \left(x - 1\right) \left({x}^{2} + x + 1\right) \mathrm{dx} = \int {x}^{3} + {x}^{2} + x - {x}^{2} - x - 1 \mathrm{dx}$
$\int \left(x - 1\right) \left({x}^{2} + x + 1\right) \mathrm{dx} = \int {x}^{3} - 1 \mathrm{dx}$
$\int \left(x - 1\right) \left({x}^{2} + x + 1\right) \mathrm{dx} = {x}^{4} / 4 - x + C$