# How do you find the indefinite integral of int (3x^-2-4x^-3)dx?

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