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

Oct 25, 2016

The integral is $\frac{{x}^{2} - 1}{2} {e}^{{x}^{2}} + C$

#### Explanation:

First we use the substitution $u = {x}^{2}$
so $\mathrm{du} = 2 x \mathrm{dx}$
So the integral becomes $\int {x}^{3} {e}^{{x}^{2}} \mathrm{dx}$
$= \frac{1}{2} \int u {e}^{u} \mathrm{du}$

This is the integration by parts
let $p = u$ then $p ' = 1$
and $v ' = {e}^{u}$ then $v = {e}^{u}$

$\int p v ' = p v - \int p ' v$

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

$= \frac{u - 1}{2} {e}^{u}$

Going back to x

$\int {x}^{3} {e}^{{x}^{2}} \mathrm{dx} = \frac{{x}^{2} - 1}{2} {e}^{{x}^{2}} + C$