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Math Help - [SOLVED] Integration by part problem with two different answers

  1. #1
    Senior Member x3bnm's Avatar
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    [SOLVED] Integration by part problem with two different answers

    I want to find the integration of following integral:

    \int x^3.e^{x^2}\, dx

    So i apply integration by part:

     u.v - \int v \,du\,\,

    <br />
Let \,\, u = x^3, du = 3x^2\,\,<br />
 and \,\, dv = e^{x^2}, v = 2x.e^{x^2}<br /> <br />

    So now th result is:

    \int x^3.e^{x^2}\, dx = 2x^4.e^{x^2} - \int\,\,6x^3.e^{x^2}\,\, dx

    7\!\!\int x^3.e^{x^2}\, dx = 2x^4.e^{x^2}

    \int x^3.e^{x^2}\, dx = \frac{2x^4.e^{x^2}}{7} + C

    Am i right? Because the answer to this problem is different back of the book.
    Answer is \frac{(x^2 -1).e^{x^2}}{2} + C

    The way i did it i can't find anything wrong with it. Can anyone kindly tell what
    is wrong with my way of solving this problem?
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by x3bnm View Post
    I want to find the integration of following integral:

    \int x^3.e^{x^2}\, dx

    So i apply integration by part:

     u.v - \int v \,du\,\,

    <br />
Let \,\, u = x^3, du = 3x^2\,\,<br />
and \,\, dv = e^{x^2}, v = 2x.e^{x^2}<br /> <br />

    So now th result is:

    \int x^3.e^{x^2}\, dx = 2x^4.e^{x^2} - \int\,\,6x^3.e^{x^2}\,\, dx

    7\!\!\int x^3.e^{x^2}\, dx = 2x^4.e^{x^2}

    \int x^3.e^{x^2}\, dx = \frac{2x^4.e^{x^2}}{7} + C

    Am i right? Because the answer to this problem is different back of the book.
    Answer is \frac{(x^2 -1).e^{x^2}}{2} + C

    The way i did it i can't find anything wrong with it. Can anyone kindly tell what
    is wrong with my way of solving this problem?
    I think that we can clear some of the "clutter". Let z=x^2 so dz=2x and this becomes \frac{1}{2}\int z\cdot e^{z}dz. Which is easily done. Does that help?

    EDIT: Ahh, did you differentiate when you should have integrated? For v
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  3. #3
    Senior Member x3bnm's Avatar
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    >EDIT: Ahh, did you differentiate when you should have integrated? For

    Yes i did. That's the mistake i made. Thanks for finding it.
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