# How do you evaluate the integral int x^3e^(x^2)"d"x?

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

#### Explanation:

Let ${x}^{2} = t \setminus \implies 2 x = \mathrm{dt} \setminus \mathmr{and} \setminus x \mathrm{dx} = \setminus \frac{\mathrm{dt}}{2}$
$\setminus \therefore \setminus \int {x}^{3} {e}^{{x}^{2}} \setminus \mathrm{dx}$
$= \setminus \int {x}^{2} {e}^{{x}^{2}} \left(x \mathrm{dx}\right)$
$= \setminus \int t {e}^{t} \setminus \frac{\mathrm{dt}}{2}$
$= \setminus \frac{1}{2} \setminus \int t {e}^{t} \setminus \mathrm{dt}$
$= \setminus \frac{1}{2} \left(t \setminus \int {e}^{t} \setminus \mathrm{dt} - \setminus \int \left(\setminus \frac{d}{\mathrm{dt}} \left(t\right) \setminus \cdot \setminus \int {e}^{t} \setminus \mathrm{dt}\right) \mathrm{dt}\right)$
$= \setminus \frac{1}{2} \left(t {e}^{t} - \setminus \int \left(1 \setminus \cdot {e}^{t}\right) \mathrm{dt}\right)$
$= \setminus \frac{1}{2} \left(t {e}^{t} - \setminus \int {e}^{t} \mathrm{dt}\right)$
$= \setminus \frac{1}{2} \left(t {e}^{t} - {e}^{t}\right) + C$
$= \setminus \frac{1}{2} \left(t - 1\right) {e}^{t} + C$
$= \frac{1}{2} \left({x}^{2} - 1\right) {e}^{{x}^{2}} + C$