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Math Help - Antiderivatives Problem

  1. #1
    Junior Member
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    Unhappy Antiderivatives Problem

    Hi, I am trying to find the antiderivative of this problem, but it is not the correct answer.

    1/3x^2 dx

    I go from that to

    3x^-1/-1
    what am I suppose to do with this problem?

    In addition to that,
    I had a question when dividing with /x
    so if a problem is 3x^2/x do simply make it 3x^1 is that correct?
    And for fractions with roots at the bottom, how do I go to solving that? ex: -1/sqrt(y)


    Thanks for any help, especially for the problem at the top.
    Coke
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  2. #2
    Newbie
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    Hey there

    \int (x^2)/3 .dx = (x^3)/9 + C

    Yes, 3x^2/x = 3x

    Not sure what your last question means..?
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  3. #3
    MHF Contributor

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    Quote Originally Posted by cokeclassic View Post
    Hi, I am trying to find the antiderivative of this problem, but it is not the correct answer.

    1/3x^2 dx

    I go from that to

    3x^-1/-1
    what am I suppose to do with this problem?

    In addition to that,
    I had a question when dividing with /x
    so if a problem is 3x^2/x do simply make it 3x^1 is that correct?
    As long as x is not equal to 0, yes.

    And for fractions with roots at the bottom, how do I go to solving that? ex: -1/sqrt(y)
    What do you mean by "solve" that? What do you want to do with it? If you just want a different way to write it, perhaps to integrate, you could try -y^{1/2}.


    Thanks for any help, especially for the problem at the top.
    Coke
    Last edited by mr fantastic; November 26th 2009 at 04:02 AM. Reason: Fixed latex tags and a quote.
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  4. #4
    Junior Member
    Joined
    Sep 2008
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    Hi, I am not sure what you did with the first problem? Could you show me some steps please?
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