# How do you find the antiderivative of (4x^2 - 3x)e^x?

Mar 14, 2018

$4 {e}^{x} {x}^{2} - 11 {e}^{x} x + 11 {e}^{x}$

#### Explanation:

The antiderivative of a function is its integral. Here, we need to solve:

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

According to integration by parts, $\int f \left(x\right) g \left(x\right) \mathrm{dx} = f \left(x\right) \int g \left(x\right) \mathrm{dx} - \int f ' \left(x\right) \left(\int g \left(x\right) \mathrm{dx}\right) \mathrm{dx}$.

Here, $f \left(x\right) = 4 {x}^{2} - 3 x$ and $g \left(x\right) = {e}^{x}$.

But since $\int {e}^{x} \mathrm{dx} = {e}^{x}$, we can just write:

${e}^{x} \left(4 {x}^{2} - 3 x\right) - \int \left(\frac{d}{\mathrm{dx}} \left(4 {x}^{2} - 3 x\right)\right) {e}^{x} \mathrm{dx}$

${e}^{x} \left(4 {x}^{2} - 3 x\right) - \int \left(8 x - 3\right) {e}^{x} \mathrm{dx}$

Integrating by parts the integral, we get:

${e}^{x} \left(4 {x}^{2} - 3 x\right) - {e}^{x} \left(8 x - 3\right) + \int 8 {e}^{x} \mathrm{dx}$

${e}^{x} \left(4 {x}^{2} - 3 x\right) - {e}^{x} \left(8 x - 3\right) + 8 {e}^{x}$

${e}^{x} \left(\left(4 {x}^{2} - 3 x\right) - \left(8 x - 3\right) + 8\right)$

${e}^{x} \left(4 {x}^{2} - 3 x - 8 x + 3 + 8\right)$

${e}^{x} \left(4 {x}^{2} - 11 x + 11\right)$

$4 {e}^{x} {x}^{2} - 11 {e}^{x} x + 11 {e}^{x}$