# How do you integrate (6x^5 -2x^4 + 3x^3 + x^2 - x-2)/x^3?

May 29, 2018

$= 2 {x}^{3} - {x}^{2} + 3 x + \ln | x | + \frac{1}{x} + \frac{1}{x} ^ 2 + C$

#### Explanation:

Since we're dividing by a single term, ${x}^{3} ,$ we may simplify the integrand as follows:

$\frac{6 {x}^{5} - 2 {x}^{4} + 3 {x}^{3} + {x}^{2} - x - 2}{x} ^ 3 = \frac{6 {x}^{5}}{x} ^ 3 - \frac{2 {x}^{4}}{x} ^ 3 + \frac{3 {x}^{3}}{x} ^ 3 + {x}^{2} / {x}^{3} - \frac{x}{x} ^ 3 - \frac{2}{x} ^ 3$

$= 6 {x}^{2} - 2 x + 3 + \frac{1}{x} - \frac{1}{x} ^ 2 - \frac{2}{x} ^ 3$

Thus, our integral becomes

$\int \left(6 {x}^{2} - 2 x + 3 + \frac{1}{x} - \frac{1}{x} ^ 2 - \frac{2}{x} ^ 3\right) \mathrm{dx}$

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

Integrate, recalling that

$\int {x}^{a} \mathrm{dx} = {x}^{a + 1} / \left(a + 1\right) + C , a \ne - 1 , \int \frac{1}{x} \mathrm{dx} = \ln | x | + C$

$= \int \left(6 {x}^{2} - 2 x + 3 + \frac{1}{x} - {x}^{-} 2 - 2 {x}^{-} 3\right) \mathrm{dx} = \frac{6}{3} {x}^{3} - \frac{2}{2} {x}^{2} + 3 x + \ln | x | + {x}^{-} 1 + \frac{2}{2} {x}^{-} 2 + C$

$= 2 {x}^{3} - {x}^{2} + 3 x + \ln | x | + \frac{1}{x} + \frac{1}{x} ^ 2 + C$