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

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

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

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

$3·\frac{x^3}{3}+2·\frac{x^2}{2}$

ⓘ Multiplying a fraction by a number multiplies its numerator: $3·\frac{x^3}{3}$ = $\frac{3x^3}{3}$, $2·\frac{x^2}{2}$ = $\frac{2x^2}{2}$■A fraction is simplified by dividing the numerator and the denominator by the same number: $\frac{3x^3}{3}$ = $x^3$, $\frac{2x^2}{2}$ = $x^2$

$x^3+x^2$

▶Indefinite integrals are defined up to an additive constant C, so this yields: $x^3+x^2+C$