Let's compute the integral: $\int (1+3x^2)dx$

ⅰ $\int (1+3x^2)dx$ = $\int 1dx+\int 3x^{2}dx$

$\int 1dx+\int 3x^{2}dx$

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

$x+3·\frac{x^3}{3}$

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

$x+x^3$

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