# How do you integrate int x^2e^(4x) by integration by parts method?

Mar 21, 2017

The answer is $= {e}^{4 x} \left({x}^{2} / 4 - \frac{x}{8} + \frac{1}{32}\right) + C$

#### Explanation:

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

Here, we have

$u = {x}^{2}$, $\implies$, $u ' = 2 x$

$v ' = {e}^{4 x}$, $\implies$, $v = {e}^{4 x} / 4$

Therefore,

$\int {x}^{2} {e}^{4 x} \mathrm{dx} = \frac{{x}^{2} {e}^{4 x}}{4} - \int 2 x \cdot {e}^{4 x} \frac{\mathrm{dx}}{4}$

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

We apply the integration by parts a second time to find $\int x \cdot {e}^{4 x} \mathrm{dx}$

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

$v ' = {e}^{4 x}$, $\implies$, $v = {e}^{4 x} / 4$

Therefore,

$\int x \cdot {e}^{4 x} \mathrm{dx} = \frac{x}{4} {e}^{4 x} - \int {e}^{4 x} \frac{\mathrm{dx}}{4}$

$= \frac{x}{4} {e}^{4 x} - {e}^{4 x} / 16$

Putting the results together,

$\int {x}^{2} {e}^{4 x} \mathrm{dx} = \frac{{x}^{2} {e}^{4 x}}{4} - \frac{1}{2} \left(\frac{x}{4} {e}^{4 x} - {e}^{4 x} / 16\right)$

$= {e}^{4 x} \left({x}^{2} / 4 - \frac{x}{8} + \frac{1}{32}\right) + C$

Mar 30, 2017

There is a faster method

#### Explanation:

Introduce $t$ and write
$I = \int {e}^{4 t x} \mathrm{dx}$
Now differentiate with respect to t
$\frac{\mathrm{dI}}{\mathrm{dt}} = \int 4 x {e}^{4 t x} \mathrm{dx}$
Once more
$\frac{{d}^{2} I}{\mathrm{dt}} ^ 2 = \int 16 {x}^{2} {e}^{4 t x} \mathrm{dx}$
$I = {e}^{4 t x} / \left(4 t\right)$
$\frac{\mathrm{dI}}{\mathrm{dt}} = \frac{x}{t} {e}^{4 t x} - {e}^{4 t x} / \left(4 {t}^{2}\right)$
and again
$\frac{{d}^{2} I}{{\mathrm{dt}}^{2}} = \left(4 {x}^{2} / t - \frac{x}{{t}^{2}} - \frac{x}{t} ^ 2 + \frac{2}{t} ^ 3 \frac{1}{4}\right) {e}^{4 t x}$
$\frac{{d}^{2} I}{\mathrm{dt}} ^ 2 = \left(4 {x}^{2} - 2 x + 2\right) {e}^{4 x}$
Our integral is 1/16 I
$J = \frac{I}{16} = \left({x}^{2} - \frac{x}{4} + \frac{1}{8}\right) {e}^{4 x}$