# How do you integrate xe^(2x)dx?

Mar 7, 2015

I would use Integration by Parts: Jun 7, 2017

$\int \setminus x {e}^{2 x} \setminus \mathrm{dx} = \frac{1}{4} {e}^{2 x} \left(2 x - 1\right) + C$

#### Explanation:

We can use the formula for Integration By Parts (IBP):

$\int \setminus u \frac{\mathrm{dv}}{\mathrm{dx}} \setminus \mathrm{dx} = u v - \int \setminus v \frac{\mathrm{du}}{\mathrm{dx}} \setminus \mathrm{dx}$, or less formally
$\text{ } \int \setminus u \setminus \mathrm{dv} = u v - \int \setminus v \setminus \mathrm{du}$

I was taught to remember the less formal rule in word; "The integral of udv equals uv minus the integral of vdu". If you struggle to remember the rule, then it may help to see that it comes a s a direct consequence of integrating the Product Rule for differentiation.

Essentially we would like to identify one function that simplifies when differentiated, and identify one that simplifies when integrated (or is at least is integrable).

So for the integrand $x {e}^{2 x}$, hopefully you can see that $x$ simplifies when differentiated and ${e}^{2 x}$ effectively remains unchanged.

Let $\left\{\begin{matrix}u & = x & \implies & \frac{\mathrm{du}}{\mathrm{dx}} & = 1 \\ \frac{\mathrm{dv}}{\mathrm{dx}} & = {e}^{2 x} & \implies & v & = \frac{1}{2} {e}^{2 x}\end{matrix}\right.$

Then plugging into the IBP formula:

$\int \setminus \left(u\right) \left(\frac{\mathrm{dv}}{\mathrm{dx}}\right) \setminus \mathrm{dx} = \left(u\right) \left(v\right) - \int \setminus \left(v\right) \left(\frac{\mathrm{du}}{\mathrm{dx}}\right) \setminus \mathrm{dx}$

gives us

$\int \setminus \left(x\right) \left({e}^{2 x}\right) \setminus \mathrm{dx} = \left(x\right) \left(\frac{1}{2} {e}^{2 x}\right) - \int \setminus \left(\frac{1}{2} {e}^{2 x}\right) \left(1\right) \setminus \mathrm{dx}$
$\int \setminus x {e}^{2 x} \setminus \mathrm{dx} = \frac{1}{2} x {e}^{2 x} - \frac{1}{2} \int \setminus {e}^{2 x} \setminus \mathrm{dx}$
$\text{ } = \frac{1}{2} x {e}^{2 x} - \frac{1}{4} {e}^{2 x} + C$
$\text{ } = \frac{1}{4} {e}^{2 x} \left(2 x - 1\right) + C$