Let's compute the integral: $\int (x^3-4x+2)dx$

ⅰ $\int (x^3-4x+2)dx$ = $\int x^{3}dx-\int 4xdx+\int 2dx$

$\int x^{3}dx-\int 4xdx+\int 2dx$

ⓘ An antiderivative of x^n is x^(n+1)/(n+1), for n ≠ -1: $\int x^{3}dx$ = $\frac{x^4}{4}$■A constant factor stays in front of the integral. An antiderivative of x^n is x^(n+1)/(n+1), for n ≠ -1: $-\int 4xdx$ = $-4·\frac{x^2}{2}$■The derivative of $2x$ is $2$, so $2x$ is an antiderivative of $2$: $\int 2dx$ = $2x$

$\frac{x^4}{4}-4·\frac{x^2}{2}+2x$

ⓘ Multiplying a fraction by a number multiplies its numerator: $4·\frac{x^2}{2}$ = $\frac{4x^2}{2}$■A fraction is simplified by dividing the numerator and the denominator by the same number: $-\frac{4x^2}{2}$ = $-2x^2$

$\frac{x^4}{4}-2x^2+2x$

▶Indefinite integrals are defined up to an additive constant C, so this yields: $\frac{x^4}{4}$$-2x^2$$+2x$$+C$