Let's integrate: $\int (\frac{x^2}{4}-1)dx$

ⅰ $\int (\frac{x^2}{4}-1)dx$ = $\int \frac{x^2}{4}dx-\int 1dx$

$\int \frac{x^2}{4}dx-\int 1dx$

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

$\frac{1}{4}·\frac{x^3}{3}-x$

ⓘ To multiply fractions, multiply the numerators together and the denominators together: $\frac{1}{4}·\frac{x^3}{3}$ = $\frac{1·x^3}{4·3}$ = $\frac{x^3}{12}$

$\frac{x^3}{12}-x$

▶Indefinite integrals are defined up to an additive constant C, so this yields: $\frac{x^3}{12}-x+C$