# How do you integrate int (x^3-4x+2)dx?

Feb 5, 2017

${x}^{4} / 4 - 2 {x}^{2} + 2 x + C .$

#### Explanation:

Recall that $\int {x}^{n} \mathrm{dx} = {x}^{n + 1} / \left(n + 1\right) + {c}_{1} , n \ne - 1 ,$and, for a constant $k$,

$\int \left[k f \left(x\right) \pm g \left(x\right)\right] \mathrm{dx} = k \int f \left(x\right) \mathrm{dx} \pm \int g \left(x\right) \mathrm{dx} + {c}_{2.}$

Hence, the reqd. Integral$= \int {x}^{3} \mathrm{dx} - 4 \int {x}^{1} \mathrm{dx} + 2 \int {x}^{0} \mathrm{dx}$

$= {x}^{3 + 1} / \left(3 + 1\right) - 4 \left({x}^{2} / 2\right) + 2 x$

$= {x}^{4} / 4 - 2 {x}^{2} + 2 x + C .$