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

Aug 3, 2016

$\frac{1}{9} \cdot \left(3 x - 1\right) \cdot {e}^{3 x} + C$.

Explanation:

Let, $I = \int 2 x {e}^{3 x} \mathrm{dx} \Rightarrow I = 2 \int x {e}^{3 x} \mathrm{dx}$.

To find $I$, we will use the following Rule of Integration by Parts :

$\int u v \mathrm{dx} = u \int v \mathrm{dx} - \int \left\{\frac{\mathrm{du}}{\mathrm{dx}} \int v \mathrm{dx}\right\} \mathrm{dx}$.

We take, u=x, so, (du)/dx=1, &, v=e^(3x), so, intvdx=1/3e^(3x). So,

$I = x \cdot \frac{1}{3} {e}^{3 x} - \int \left\{1 \cdot \frac{1}{3} {e}^{3 x}\right\} \mathrm{dx}$

$= \frac{x}{3} {e}^{3 x} - \frac{1}{3} \int {e}^{3 x} \mathrm{dx}$

$= \frac{x}{3} {e}^{3 x} - \frac{1}{3} \cdot \frac{1}{3} {e}^{3 x}$

$\therefore I = \frac{1}{9} \cdot \left(3 x - 1\right) \cdot {e}^{3 x} + C$.