# Thread: Simple integration question help

1. ## Simple integration question help

Hey guys, after differentiating $y=x^{2}e^{-x^{2}}$by the product rule to get $y=2xe^{-x^{2}}-2x^{3}e^{-x^{2}}
$

how do I use this result when integrating $\displaystyle\int x^{3}e^{-x^{2}}\,dx$

thanks a lot

2. $\displaystyle \frac{d}{dx}(x^2e^{-x^2}) = 2x\,e^{-x^2} - 2x^3e^{-x^2}$ is equivalent to

$\displaystyle x^2e^{-x^2} = \int{2x\,e^{-x^2} - 2x^3e^{-x^2}\,dx}$

$\displaystyle x^2e^{-x^2} = -e^{-x^2} - 2\int{x^3e^{-x^2}\,dx}$.

Now solve for your integral.

3. Originally Posted by Prove It
$\displaystyle \frac{d}{dx}(x^2e^{-x^2}) = 2x\,e^{-x^2} - 2x^3e^{-x^2}$ is equivalent to

$\displaystyle x^2e^{-x^2} = \int{2x\,e^{-x^2} - 2x^3e^{-x^2}\,dx}$

$\displaystyle x^2e^{-x^2} = -e^{-x^2} - 2\int{x^3e^{-x^2}\,dx}$.

Now solve for your integral.
so this is right?
$\displaystyle\int x^3e^{-x^2}\,dx=\frac {-(x^{2}e^{-x^2}+e^{-x^2})}{2}$

4. Originally Posted by aonin
so this is right?
$\displaystyle\int x^3e^{-x^2}\,dx=\frac {-(x^{2}e^{-x^2}+e^{-x^2})}{2}$
Yes