How do you integrate f(x)=(-3+3x+3x^2-3x^3)(3-3x-x^2+3x^3+2x^4) using the product rule?

Jan 1, 2018

$- \frac{3}{4} {x}^{8} - \frac{3}{7} {x}^{7} + 3 {x}^{6} + \frac{9}{5} {x}^{5} - \frac{15}{2} {x}^{4} + {x}^{3} + 9 {x}^{2} - 9 x + C$

Explanation:

There is no product rule for integrals - the product rule only applies to derivatives.

In this case, we have to just expand the entire thing:
$\left(- 3 + 3 x + 3 {x}^{2} - 3 {x}^{3}\right) \left(3 - 3 x - {x}^{2} + 3 {x}^{3} + 2 {x}^{4}\right) =$

$= - 9 + 9 x + 3 {x}^{2} - 9 {x}^{3} - 6 {x}^{4} + 9 x - 9 {x}^{2} - 3 {x}^{3} + 9 {x}^{4} + 6 {x}^{5} + 9 {x}^{2} - 9 {x}^{3} - 3 {x}^{4} + 9 {x}^{5} + 6 {x}^{6} - 9 {x}^{3} + 9 {x}^{4} + 3 {x}^{5} - 9 {x}^{6} - 6 {x}^{7}$

Combining like terms, we get:
$- 6 {x}^{7} - 3 {x}^{6} + 18 {x}^{5} + 9 {x}^{4} - 30 {x}^{3} + 3 {x}^{2} + 18 x - 9$

We can then integrate using the reverse power rule, which says:
$\int \setminus {x}^{n} \setminus \mathrm{dx} = {x}^{n + 1} / \left(n + 1\right) + C$

In our case, we get:
$\int \setminus - 6 {x}^{7} - 3 {x}^{6} + 18 {x}^{5} + 9 {x}^{4} - 30 {x}^{3} + 3 {x}^{2} + 18 x - 9 \setminus \mathrm{dx} =$

$= - \frac{6}{8} {x}^{8} - \frac{3}{7} {x}^{7} + \frac{18}{6} {x}^{6} + \frac{9}{5} {x}^{5} - \frac{30}{4} {x}^{4} + \frac{3}{3} {x}^{3} + \frac{18}{2} {x}^{2} - 9 x + C =$

Simplifying, we get:
$= - \frac{3}{4} {x}^{8} - \frac{3}{7} {x}^{7} + 3 {x}^{6} + \frac{9}{5} {x}^{5} - \frac{15}{2} {x}^{4} + {x}^{3} + 9 {x}^{2} - 9 x + C$