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C4 Integration of e - Ahh watch

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    Find by recognition: (chain rule backwards)

    \int 3xe^{x^2} dx

    I'm completely stuck with this! A worked solution would be fantastic
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    What do you get when you differentiate e^{x^2} (with respect to x)?
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    IF you call the (e^x^2) (e^u) instead, then find du in terms of dx, you should be able to work it out more easily like that.
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    I think I was having problems differentiating  e^{x^2} like calcium said.

    If its  \frac{1}{2x} e^{x^2}

    =  \frac{3}{2} e^{x^2}

    But shouldnt the power be raised, as in normal integrations like  \int x(3x^2 + 1)^4 + c =  \frac{1}{30} (3x^2 + 1)^5 + c
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    Yep, the answer is \frac{3}{2}e^{x^2} (+c)

    Think about integrating e^x or e^3x. You don't change the exponent when integrating.
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    Why is e^x lonely?
    Because when it tries to integrate it just ends up with itself.

    That joke helps me everytime.
    (Add the constant of integration though :P)
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    (Original post by Earl Grey)
    I think I was having problems differentiating  e^{x^2} like calcium said.

    If its  \frac{1}{2x} e^{x^2}
    Noo. You can't integrate e^(x^2). You can differentiate it, and you can integrate xe^(x^2).
 
 
 
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Updated: March 16, 2008

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