# How do you integrate int (4x^3+6x^2-1)dx?

Dec 8, 2016

We can use the power rule for integration all the way across that integrand.

$\int \left(4 {x}^{3} + 6 {x}^{2} - 1\right) \mathrm{dx}$

Keep in mind that:

$\int {x}^{n} \mathrm{dx} = {x}^{n + 1} / \left(n + 1\right) + c$

So, unsimplified:

$= 4 \cdot \frac{{x}^{3 + 1}}{3 + 1} + 6 \cdot \frac{{x}^{2 + 1}}{2 + 1} - 1 \left(x\right) + c$

$= {x}^{4} + 2 {x}^{3} - x + c$