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Math Help - Problem with integral

  1. #1
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    Problem with integral

    This is probably really simple but I cant see it. Where does the 4x come from in this integral?



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  2. #2
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    This problem is looking like a partial fractions decomposition. However, in order to do that, the degree of the numerator must be strictly less than the degree of the denominator. Hence, the first step would be to perform polynomial long division. Apparently, you get 1 + remainder/denominator. Thus, when you integrate 1, you get the x in the answer. Make sense?

    Incidentally, this post belongs in the Calculus forum, even if you're not attending a university.
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  3. #3
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    Thanks, that makes sense.
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  4. #4
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    Quote Originally Posted by joe1990 View Post
    This is probably really simple but I cant see it. Where does the 4x come from in this integral?



    \displaystyle\huge\frac{x}{x+1}+\frac{x-2}{x+3}=\frac{2x(x-2)-2}{(x+1)(x-3)}

    so

    \displaystyle\huge\ 2\left[\frac{x(x-2)}{(x+1)(x-3)}\right]-\frac{2}{(x+1)(x-3)}=\frac{x}{x+1}+\frac{x-2}{x-3}

    which gives

    \displaystyle\huge\frac{x(x-2)}{(x+1)(x-3)}=\left[\frac{x}{x+1}+\frac{x-2}{x-3}+\frac{2}{(x+1)(x-3)}\right]\frac{1}{2}

    Splitting the rightmost term into partial fractions leads to your result.

    Maybe simplest is

    \displaystyle\huge\frac{x(x-2)}{(x+1)(x-3)}=\frac{x^2-2x}{x^2-2x-3}=\frac{\left(x^2-2x-3\right)+3}{x^2-2x-3}=1+\frac{3}{(x+1)(x-3)}

    Then when you integrate that, you get "x" and the logs.
    Last edited by Archie Meade; August 20th 2010 at 03:13 AM.
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    You're welcome. Have a good one!
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