# How do you integrate int x^2/e^(2x) by integration by parts method?

Feb 18, 2017

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

#### Explanation:

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

We must calculate $\int {x}^{2} / {e}^{2 x} \mathrm{dx} = \int {x}^{2} {e}^{- 2 x} \mathrm{dx}$

Let $u ' = {e}^{- 2 x}$, $\implies$, $u = {e}^{- 2 x} / \left(- 2\right) = - \frac{1}{2} {e}^{- 2 x}$

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

Therefore,

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

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

In order to calculate $\int x {e}^{- 2 x} \mathrm{dx}$, we do the integration by parts a second time

Let $u ' = {e}^{- 2 x}$, $\implies$, $u = {e}^{- 2 x} / \left(- 2\right) = - \frac{1}{2} {e}^{- 2 x}$

and $v = x$, $\implies$, $v ' = 1$

So,

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

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

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

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

Putting it all together

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

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