# Evaluate the integral? :  int x^3e^(x^2) dx

Aug 4, 2017

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

#### Explanation:

We need the integration by parts

$\int p ' q \mathrm{dx} = p q - \int p q ' \mathrm{dx}$

We perform the substitution

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

Therefore,

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

We apply the integration by parts

$p ' \left(u\right) = {e}^{u}$, $\implies$, $p \left(u\right) = {e}^{u}$

$q \left(u\right) = u$, $\implies$, $q ' \left(u\right) = 1$

Therefore,

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

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

Aug 4, 2017

We seek:

$\int \setminus {x}^{3} {e}^{{x}^{2}} \setminus \mathrm{dx} = \frac{1}{2} \left({x}^{2} - 1\right) {e}^{{x}^{2}} + C$

#### Explanation:

We seek:

$I = \int \setminus {x}^{3} {e}^{{x}^{2}} \setminus \mathrm{dx}$

Note as a helper that:

$\frac{d}{\mathrm{dx}} \left({e}^{{x}^{2}}\right) = 2 x {e}^{{x}^{2}} \iff \int \setminus 2 x {e}^{{x}^{2}} \setminus \mathrm{dx} = {e}^{{x}^{2}}$

So we can write the integral as:

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

We can now use the formula for Integration By Parts (IBP):

$\int \setminus u \frac{\mathrm{dv}}{\mathrm{dx}} \setminus \mathrm{dx} = u v - \int \setminus v \frac{\mathrm{du}}{\mathrm{dx}} \setminus \mathrm{dx}$, or less formally
$\text{ } \int \setminus u \setminus \mathrm{dv} = u v - \int \setminus v \setminus \mathrm{du}$

I was taught to remember the less formal rule in word; "The integral of udv equals uv minus the integral of vdu". If you struggle to remember the rule, then it may help to see that it comes a s a direct consequence of integrating the Product Rule for differentiation.

Let $\left\{\begin{matrix}u & = {x}^{2} & \implies \frac{\mathrm{du}}{\mathrm{dx}} & = 2 x \\ \frac{\mathrm{dv}}{\mathrm{dx}} & = 2 x {e}^{{x}^{2}} & \implies v & = {e}^{{x}^{2}}\end{matrix}\right.$

Then plugging into the IBP formula:

$\int \setminus \left(u\right) \left(\frac{\mathrm{dv}}{\mathrm{dx}}\right) \setminus \mathrm{dx} = \left(u\right) \left(v\right) - \int \setminus \left(v\right) \left(\frac{\mathrm{du}}{\mathrm{dx}}\right) \setminus \mathrm{dx}$

gives us

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

$\therefore 2 I = {x}^{2} {e}^{{x}^{2}} - \int \setminus 2 x {e}^{{x}^{2}} \setminus \mathrm{dx} + A$
$\text{ } = {x}^{2} {e}^{{x}^{2}} - {e}^{{x}^{2}} + A$
$\text{ } = \left({x}^{2} - 1\right) {e}^{{x}^{2}} + A$

Hence

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