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Math Help - Calculus- Integration w/ partial fractions

  1. #1
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    Calculus- Integration w/ partial fractions

    Find the integral of (x^2+4)/(x^2-4) dx.


    The answer is x + 2 ln | x-2/x+2 |

    Here's what I did so far:
    (x^2+4)/(x^2-4) = (x^2+4)/[x+2][x-2].
    I'm not really sure how to split them apart. I tried making it Ax+B/x+2 + Cx+D/x-2 but when I worked it out I got: A+C = 1, C-B = 2..? I don't think that was right.

    Thank you for your help!!
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  2. #2
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    \frac{x^2 + 4}{x^2 - 4} = \frac{x^2 - 4 + 8}{x^2 - 4} = 1 + \frac{8}{x^2-4}

    partial fraction set up for last quotient ...

    \frac{8}{x^2-4} = \frac{A}{x+2} + \frac{B}{x-2}

    finish
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Just for fun: \frac 8{(x + 2)(x - 2)} = \frac {2x + 4 - (2x - 4)}{(x + 2)(x - 2)} =  \frac {2x + 4}{(x + 2)(x - 2)} - \frac {2x - 4}{(x + 2)(x - 2)} = \frac 2{x - 2} - \frac 2{x + 2}
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  4. #4
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    Quote Originally Posted by Jhevon View Post

    Just for fun: \frac 8{(x + 2)(x - 2)} = \frac {2x + 4 - (2x - 4)}{(x + 2)(x - 2)} =  \frac {2x + 4}{(x + 2)(x - 2)} - \frac {2x - 4}{(x + 2)(x - 2)} = \frac 2{x - 2} - \frac 2{x + 2}
    Not fun!! Actually better and faster!
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  5. #5
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    Quote Originally Posted by Jhevon View Post
    Just for fun: \frac 8{(x + 2)(x - 2)} = \frac {2x + 4 - (2x - 4)}{(x + 2)(x - 2)} =  \frac {2x + 4}{(x + 2)(x - 2)} - \frac {2x - 4}{(x + 2)(x - 2)} = \frac 2{x - 2} - \frac 2{x + 2}

    wow never would've thought of it
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