1. Integral

I'm desperately stuck on the following integral:

Integral: x^5*e^(-1/2x^2)

Thanks

2. Parts reduces the x^5 by a power of 2 each time. So using it twice leaves $xe^{\frac{-x^2}{2}}$ which hopefully you can just integrate.

3. confused

Thanks for the reply but I'm still a little confused.

How does by parts reduce x^5 by 2 each time? Doesn't it only reduce it by one?

4. $I=\int x^5 e^{\frac{-x^2}{2}}dx$

Take $u=x^5, dv=e^{\frac{-x^2}{2}}$

So $du=5x^4, v=\frac{e^{\frac{-x^2}{2}}}{-x}$
We want I = uv - int(v)du

So $I=-x^4 e^{\frac{-x^2}{2}} + \int 5x^3 e^{\frac{-x^2}{2}}$

etc...

5. Originally Posted by Thomas154321
$I=\int x^5 e^{\frac{-x^2}{2}}dx$

Take $u=x^5, dv=e^{\frac{-x^2}{2}}$

So $du=5x^4, v=\frac{e^{\frac{-x^2}{2}}}{-x}$ (**)
We want I = uv - int(v)du

So $I=-x^4 e^{\frac{-x^2}{2}} + \int 5x^3 e^{\frac{-x^2}{2}}$

etc...
The (**) step is simply not true! The best thing to do first is a substitution $u = - \frac{x^2}{2}$ giving

$-4 \int u^2 e^u\,du$ the integration by parts twice.