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

Sep 1, 2016

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

#### Explanation:

throughout , remember that $\frac{d}{\mathrm{dx}} \left({e}^{{x}^{2}}\right) = 2 x {e}^{{x}^{2}}$

So:
$\int {x}^{3} {e}^{{x}^{2}} \mathrm{dx}$

setting it up for IBP
$= \int \textcolor{red}{{x}^{2}} \frac{d}{\mathrm{dx}} \left(\textcolor{b l u e}{\frac{1}{2} {e}^{{x}^{2}}}\right) \mathrm{dx}$

so by IBP
$= \textcolor{red}{{x}^{2}} \left(\textcolor{b l u e}{\frac{1}{2} {e}^{{x}^{2}}}\right) - \int \frac{d}{\mathrm{dx}} \left(\textcolor{red}{{x}^{2}}\right) \left(\textcolor{b l u e}{\frac{1}{2} {e}^{{x}^{2}}}\right) \mathrm{dx}$

$= \frac{1}{2} {x}^{2} {e}^{{x}^{2}} - \int 2 x \cdot \frac{1}{2} {e}^{{x}^{2}} \mathrm{dx}$

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

$= \frac{1}{2} {x}^{2} {e}^{{x}^{2}} - \int \frac{1}{2} \frac{d}{\mathrm{dx}} \left({e}^{{x}^{2}}\right) \mathrm{dx}$

$= \frac{1}{2} {x}^{2} {e}^{{x}^{2}} - \frac{1}{2} {e}^{{x}^{2}} + C$

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