# How do you calculate the antiderivative of (-2x^3+14x^9)/(x^-2)?

Mar 25, 2018

$- \frac{1}{3} {x}^{6} + \frac{7}{6} {x}^{12} + C$

#### Explanation:

Recall that $\frac{a + b}{c} = \frac{a}{c} + \frac{b}{c}$

Thus, we can rewrite $\frac{- 2 {x}^{3} + 14 {x}^{9}}{x} ^ - 2$ as $\frac{- 2 {x}^{3}}{x} ^ - 2 + \frac{14 {x}^{9}}{x} ^ - 2 = - 2 {x}^{3 - \left(- 2\right)} + 14 {x}^{9 - \left(- 2\right)} = - 2 {x}^{5} + 14 {x}^{11}$

So, we want

$\int \left(- 2 {x}^{5} + 14 {x}^{11}\right) \mathrm{dx}$

Split this up, as we can split up sums or differences when integrating:

$\int - 2 {x}^{5} \mathrm{dx} + \int 14 {x}^{11} \mathrm{dx}$

Factor out the constants:

$- 2 \int {x}^{5} \mathrm{dx} + 14 \int {x}^{11} \mathrm{dx}$

Now, recall $\int {x}^{a} \mathrm{dx} = {x}^{a + 1} / \left(a + 1\right) + C$ where $C$ is the constant of integration, just an arbitrary constant.

So, integrating, we get

$\frac{- 2 {x}^{6}}{6} + \frac{14 {x}^{12}}{12} = - \frac{1}{3} {x}^{6} + \frac{7}{6} {x}^{12} + C$

Yes, we would technically have two constants of integration as we had two integrals, but we absorb them all into one constant.