- #1

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## 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.

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- Thread starter cscott
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- #1

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

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

- #2

G01

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

- #3

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## 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.

- #4

Avodyne

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- #5

CompuChip

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[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.

- #6

Avodyne

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

- #8

HallsofIvy

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Avodyne's suggestion is the best in my opinion.

- #9

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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.

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

Thanks for all the suggestions.

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- #10

dynamicsolo

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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) ...

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- #11

HallsofIvy

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Actually, you should get

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

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

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