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Math Help - Integration of exp(x^2)*(x^2-1)/x^3

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    Integration of exp(x^2)*(x^2-1)/x^3

    I am working on a double integral, and I am now stuck on this integral.
    exp(x^2)*(x^2-1)/x^3

    Wolfram returned this as an answer:
    exp(x^2)/2x

    Any help or suggestions will be appreciated!
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  2. #2
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    Re: Integration of exp(x^2)*(x^2-1)/x^3

    Multiply both numerator and denominator by x to make the integral
    \int \frac{e^{x^2}(x^2-1)}{x^4} xdx
    and use the substitution u= x^2 so that du= 2xdx, \frac{1}{2}du= xdx.
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    Re: Integration of exp(x^2)*(x^2-1)/x^3

    Thanks!
    I know the answer is exp(x)/2x, becuase I can confirm it using the quotient rule for differentiation, but could someone please explain how to show the integration process?
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  4. #4
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    Re: Integration of exp(x^2)*(x^2-1)/x^3

    Quote Originally Posted by dnftp View Post
    Thanks!
    I know the answer is exp(x)/2x, becuase I can confirm it using the quotient rule for differentiation, but could someone please explain how to show the integration process?
    HallsOfIvy JUST DID! Do the substitution he suggested. If you have any more problems, post your working and we can help you where you are stuck.
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    Re: Integration of exp(x^2)*(x^2-1)/x^3

    I realised that I had wrote the equation wrong...it should be
    \int \frac{e^{x^2}(x^2-1)+1}{x^4} xdx using the substitution suggested I got, \int \frac{e^{u}(u-1)+1}{2u^2} du=  \frac{e^{u}-1}{2u} .
    Substituting back to x,  \frac{e^{x^2}-1}{2x^2} . The limits on the integration were from 0 to 2.
     \frac{e^{4}-1}{8} - 0 = 6.6997...
    However, when I do this on wolfram, I get,  \frac{e^{4}-5}{8} = 6.1997...
    I can't see what I am doing wrong.
    Last edited by dnftp; October 7th 2012 at 09:08 PM.
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