# How do you find the general indefinite integral of (13x^2+12x^-2)dx?

Feb 15, 2015

The answer is: $\frac{13}{3} {x}^{3} - 12 {x}^{-} 1 + c$

Using this integral:

$\int {\left[f \left(x\right)\right]}^{n} f ' \left(x\right) \mathrm{dx} = {\left[f \left(x\right)\right]}^{n + 1} / \left(n + 1\right) + c$.

So:

$\int \left(13 {x}^{2} + 12 {x}^{-} 2\right) \mathrm{dx} = 13 {x}^{2 + 1} / \left(2 + 1\right) + 12 {x}^{- 2 + 1} / \left(- 2 + 1\right) + c =$

$= \frac{13}{3} {x}^{3} - 12 {x}^{-} 1 + c$.