Substitution and integration by parts

[cscott]

Homework Statement

Can anybody help me integrate [tex]x^3 e^{x^2}[/tex]

The Attempt at a Solution

I can't see how to do it by substitution or integration by parts.

 

[G01]

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

 

[asd1249jf]

Homework Statement

Can anybody help me integrate [tex]x^3 e^{x^2}[/tex]

The Attempt at a Solution

I can't see how to do it by substitution or integration by parts.

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.

 

[Avodyne]

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

 

[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
[tex]\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)[/tex]
and evaluate it using polar coordinates.

 

[Avodyne]

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

 

[dextercioby]

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

 

[HallsofIvy]

Avodyne's suggestion is the best in my opinion.

 

[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 [tex]\frac{1}{2}u \cdot e^{u}[/tex] ?

Thanks for all the suggestions.

 

[dynamicsolo]

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

So with a u substitution I should get [tex]\frac{1}{2}u \cdot e^{u}[/tex] ?

Thanks for all the suggestions.

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) ...

 

[HallsofIvy]

Actually, you should get
[tex]\frac{1}{2}u e^u du[/tex]