Substitution and integration by parts

1. Sep 27, 2007

cscott

1. The problem statement, all variables and given/known data

Can anybody help me integrate $$x^3 e^{x^2}$$

3. The attempt at a solution
I can't see how to do it by substitution or integration by parts.

2. Sep 27, 2007

G01

Here's a hint. Did you try doing one and THEN the other?

3. Sep 27, 2007

l46kok

Integration by parts should work. Set u equal to the term which will diminish eventually as you keep taking the derivative of and dv equal to the other term.

4. Sep 27, 2007

Avodyne

It's hard to integrate e^(x^2). It's easy to integrate e^x. Does this suggest a substitution you might consider?

5. Sep 27, 2007

CompuChip

Actually exp(x^2) is a sort of standard integral. It's called a Gaussian integral. You can find it in about any physics book, vector calculus book or at the front of any integral table, sometimes with proof. The idea is basically to calculate
$$\left( \int e^{x^2} \, dx \right)^2 = \left( \int e^{x^2} \, dx \right) \left( \int e^{y^2} \, dy \right) = \left( \int e^{x^2 + y^2} \, dx \, dy \right)$$
and evaluate it using polar coordinates.

6. Sep 27, 2007

Avodyne

That only works if you're doing a definite integral over an infinite range. The question asks for the indefinite integral.

7. Sep 27, 2007

dextercioby

Not to mention that you need a "-" sign in the exponent, else it won't converge...

8. Sep 27, 2007

HallsofIvy

Staff Emeritus
Avodyne's suggestion is the best in my opinion.

9. Sep 27, 2007

cscott

I think I see where I messed up in my substitution, which made me think it'd be useless

So with a u substitution I should get $$\frac{1}{2}u \cdot e^{u}$$ ?

Thanks for all the suggestions.

Last edited: Sep 27, 2007
10. Sep 27, 2007

dynamicsolo

Yes, that does work. (I was still editing my reply while you posted this.) You should get

(1/2)[ (x^2) - 1 ] · exp(x^2) + C .

It's one of those oddball integrals that looks like one of the exponents went the wrong way, but when you check your result, differentiating the exp(x^2) provides the factor that gets you the (x^3) ...

Last edited: Sep 27, 2007
11. Sep 27, 2007

HallsofIvy

Staff Emeritus
Actually, you should get
$$\frac{1}{2}u e^u du$$