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Math Help - integral

  1. #1
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    integral

    integral((x^2)-1/((x^2)+6x+9)
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by lord12 View Post
    integral((x^2)-1/((x^2)+6x+9)
    I assume this is

    \int\frac{x^2-1}{x^2+6x+9}dx

    Well \frac{x^2-1}{x^2+6x+9}=1-\frac{6}{x+3}+\frac{8}{(x+3)^2}
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  3. #3
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    Edit: eep sorry again mathstud. Please disregeard this post

    I think you mean
    \int\frac{x^2-1}{x^2+6x+9}dx
    You have an extra open bracket that you haven't closed at the start of x^2 which makes thins confusing.

    First we need to reduce the numerator to a degree smaller than the denominator using long division. Here is an example of how to do it using 2 common notations. you should get
    \int 1+\frac{-6x-10}{x^2+6x+9}dx
    Do you know how to proceed from here?
    Last edited by badgerigar; June 25th 2008 at 06:11 PM. Reason: mathstud beat me to it
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by badgerigar View Post
    Edit: eep sorry again mathstud. Please disregeard this post

    I think you mean
    \int\frac{x^2-1}{x^2+6x+9}dx
    You have an extra open bracket that you haven't closed at the start of x^2 which makes thins confusing.

    First we need to reduce the numerator to a degree smaller than the denominator using long division. Here is an example of how to do it using 2 common notations. you should get
    \int 1+\frac{-6x-10}{x^2+6x+9}dx
    Do you know how to proceed from here?
    do not worry about it.
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