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Math Help - Integration by partial fractions

  1. #1
    Junior Member
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    Integration by partial fractions

    It was easier to just make a scan, so here it is:
    http://img195.imageshack.us/img195/9144/scan0001tv.jpg

    The answer I'm getting is wrong. It's an easy problem, but I'm sure I'm doing something stupid. Please advise.

    Thanks for your time.
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  2. #2
    MHF Contributor
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    Did you consider splitting the numerator in the first place! 1 - 9/(x^3 - 3x^2) should be a bit easier...
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  3. #3
    Super Member
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     \frac{ x^3 - 3x^2 - 9}{ x^3 - 3x^2 }

     = 1 - \frac{9}{x^2(x-3)}

     = 1 - [  \frac{1}{x-3} - \frac{x+3}{x^2} ]

     = 1 + \frac{1}{x} + \frac{3}{x^2} - \frac{1}{x-3}

    Therefore , its integral is

     x + \ln(x) - \frac{3}{x} - \ln(x-3) + C
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