# What is the integral of 2xe^x?

Feb 24, 2018

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

#### Explanation:

To find:

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

By constant rule

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

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

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

We can integrate using the parts rule

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

Here,

$u = x \to \mathrm{du} = \mathrm{dx}$

$\mathrm{dv} = {e}^{x} \mathrm{dx} \to v = {e}^{x}$

Substituting

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

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

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

$I = x {e}^{x} - {e}^{x}$

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

Feb 24, 2018

2xe^(x)-2e^(x)+C

#### Explanation:

We have: $\int \left(2 x {e}^{x}\right) \mathrm{dx}$

This integral can be evaluated using integration by parts.

Let $u = 2 x R i g h t a r r o w \frac{\mathrm{du}}{\mathrm{dx}} = 2$ and $\frac{\mathrm{dv}}{\mathrm{dx}} = {e}^{x} R i g h t a r r o w v = {e}^{x}$:

$R i g h t a r r o w \int \left(2 x {e}^{x}\right) \mathrm{dx} = 2 x {e}^{x} - \int \left(2 {e}^{x}\right) \mathrm{dx}$

$\therefore \int \left(2 x {e}^{x}\right) \mathrm{dx} = 2 x {e}^{x} - 2 {e}^{x} + C$