# How do you integrate int x^3*e^(x^2/2)  using integration by parts?

Nov 16, 2017

${x}^{3} \cdot {e}^{{x}^{2} / 2} \cdot \mathrm{dx} = \left({x}^{2} - 2\right) \cdot {e}^{{x}^{2} / 2} + C$

#### Explanation:

${x}^{3} \cdot {e}^{{x}^{2} / 2} \cdot \mathrm{dx}$

=${x}^{2} \cdot x {e}^{{x}^{2} / 2} \cdot \mathrm{dx}$

I used integration by parts with $u = {x}^{2}$ and $\mathrm{dv} = x {e}^{\left({x}^{2} / 2\right)} \cdot \mathrm{dx}$, so $\mathrm{du} = 2 x \cdot \mathrm{dx}$ and $v = {e}^{{x}^{2} / 2}$

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

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

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

=${x}^{2} \cdot {e}^{{x}^{2} / 2} - 2 {e}^{{x}^{2} / 2} + C$

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

Nov 16, 2017

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

#### Explanation:

We need

$\int p q ' = p q - \int p ' q$

Here,

Perform the substitution

$u = {x}^{2}$, $\implies$, $\mathrm{du} = 2 x \mathrm{dx}$, $\implies$, $\mathrm{dx} = \frac{1}{2 x} \mathrm{du}$

So,

$\int {x}^{3} {e}^{{x}^{2} / 2} \mathrm{dx} = \frac{1}{2} \int u {e}^{\frac{u}{2}} \mathrm{du}$

Perform the integration by parts

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

$q ' = {e}^{\frac{u}{2}}$, $\implies$, $q = 2 {e}^{\frac{u}{2}}$

Therefore,

$\int {x}^{3} {e}^{{x}^{2} / 2} \mathrm{dx} = \frac{1}{2} \int u {e}^{\frac{u}{2}} \mathrm{du}$

$= \frac{1}{2} \left(2 u {e}^{\frac{u}{2}} - 2 \int {e}^{\frac{u}{2}} \mathrm{du}\right)$

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

$= {x}^{2} {e}^{{x}^{2} / 2} - 2 {e}^{{x}^{2} / 2} + C$

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