# How do you integrate (x^5)*(e^(x^2/2))?

Sep 1, 2015

$\left(4 {x}^{4} - 4 {x}^{2} - 8\right) {e}^{{x}^{2} / 2} + c$

#### Explanation:

Let's assume ${x}^{2} / 2 = t$
$\implies \frac{2 x \mathrm{dx}}{2} = t , x \mathrm{dx} = \mathrm{dt}$

$= \int \left({x}^{5}\right) \cdot \left({e}^{{x}^{2} / 2}\right) \mathrm{dx}$ can be written as,

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

Substituting with $t$ we get

$= \int \left(2 t\right) \left(2 t\right) \left({e}^{t}\right) \mathrm{dt}$

$= 4 \int {t}^{2} \left({e}^{t}\right) \mathrm{dt}$

Using Integration by Parts,

∫(I)(II)dx=(I)∫(II)dx−∫((I)'∫(II)dx)dx

where $\left(I\right)$ and $\left(I I\right)$ are functions of $x$, and $\left(I\right)$ represents which will be differentiated and $\left(I I\right)$ will be integrated subsequently in the above formula

Similarly following for the problem,

$= 4 \left({t}^{2}\right) \int {e}^{t} \mathrm{dt} - 4 \int \left(\left({t}^{2}\right) ' \int {e}^{t} \mathrm{dt}\right) \mathrm{dt} + c$

$= 4 {t}^{2} {e}^{t} - 4 \int 2 t {e}^{t} \mathrm{dt} + c$

$= 4 {t}^{2} {e}^{t} - 8 \int t {e}^{t} \mathrm{dt} + c$

Applying again integration by parts in second term, we get

$= 4 {t}^{2} {e}^{t} - 8 \left(t {e}^{t} - \int {e}^{t} \mathrm{dt}\right) + c$

$= 4 {t}^{2} {e}^{t} - 8 \left(t {e}^{t} - {e}^{t}\right) + c$

$= 4 {t}^{2} {e}^{t} - 8 t {e}^{t} - 8 {e}^{t} + c$

Substituting $t$ back,

$= 4 {\left({x}^{2} / 2\right)}^{2} {e}^{{x}^{2} / 2} - 8 \left({x}^{2} / 2\right) {e}^{{x}^{2} / 2} - 8 {e}^{{x}^{2} / 2} + c$, where $c$ is a constant

$= \left(4 {x}^{4} - 4 {x}^{2} - 8\right) {e}^{{x}^{2} / 2} + c$