# How do you integrate (x^3+2x^2-x) / x dx?

Jul 1, 2018

${x}^{3} / 3 + {x}^{2} - x + C$

#### Explanation:

At first we simplify the integrand

$\frac{{x}^{3} + 2 {x}^{2} - x}{x} = {x}^{3} / x + 2 {x}^{2} / x - \frac{x}{x} = {x}^{2} + 2 x - 1$
Now we can integrate the result:

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

we have used that

$\int {x}^{n} \mathrm{dx} = {x}^{n + 1} / \left(n + 1\right) + C$ if $n \ne - 1$