# Integrate int xe^(2x)dx ?

Jul 10, 2018

This will require a single application of integration by parts.

We let $\mathrm{dv} = {e}^{2 x} \mathrm{dx}$ and $u = x$. Then $v = \frac{1}{2} {e}^{2 x}$ and $\mathrm{du} = \mathrm{dx}$.

By integration by parts we have

$\int u \mathrm{dv} = u v - \int v \mathrm{du}$

$\int x {e}^{2 x} \mathrm{dx} = \frac{1}{2} x {e}^{2 x} - \int \frac{1}{2} {e}^{2 x} \mathrm{dx}$

$\int x {e}^{2 x} \mathrm{dx} = \frac{1}{2} x {e}^{2 x} - \frac{1}{4} {e}^{2 x} + C$

Hopefully this helps!

Jul 10, 2018

The answer is $= \left(2 x - 1\right) {e}^{2 x} / 4 + C$

#### Explanation:

Perform an integration by parts

$\int u v ' \mathrm{dx} = u v - \int u ' v \mathrm{dx}$

Here,

The integral is

$I = \int x {e}^{2 x} \mathrm{dx}$

$u = x$, $\implies$, $u ' = 1$

$v ' = {e}^{2 x}$, $\implies$, $v = {e}^{2 x} / 2$

Therefore, the integral is

$I = \frac{x {e}^{2 x}}{2} - \frac{1}{2} \int {e}^{2 x} \mathrm{dx}$

$= \frac{x {e}^{2 x}}{2} - \frac{1}{4} {e}^{2 x} + C$

$= \left(2 x - 1\right) {e}^{2 x} / 4 + C$