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Math Help - Partial Fractions

  1. #1
    Junior Member
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    Partial Fractions

    I cant solve this problem:

    INTEGRAL OF: (x^3-4x^2-16) / (x^3-4x^2)

    I change the denominator into (x^2)(x-4) and setup the following partial fraction:

    A/x + (Bx+C)/x^2 + D/x-4

    When you multiply the numerators by the other two denominators though you don't get a term without an x, thus your left with nothing to set equal to 16.
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  2. #2
    Member RedBarchetta's Avatar
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    For partial fractions to work the degree of the numerator must be less than the degree of the denominator. Since both the numerator and denominator have a degree of three, you must do polynomial long division.

    <br />
\frac{{x^3  - 4x^2  - 16}}<br />
{{x^3  - 4x^2 }} = x^3  - 4x^2 \left){\vphantom{1{x^3  - 4x^2  - 16}}}\right.<br />
\!\!\!\!\overline{\,\,\,\vphantom 1{{x^3  - 4x^2  - 16}}}<br />

    After this, take the answer from the long division plus the remainder and then perform the partial fraction decomposition.
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  3. #3
    Math Engineering Student
    Krizalid's Avatar
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    \frac{x^{3}-4x^{2}-16}{x^{3}-4x^{2}}=1-\frac{16}{x^{3}-4x^{2}}.

    The remaining challenge is to descompose \frac{16}{x^{3}-4x^{2}}, and we can tackle it like this:

    \frac{16}{x^{2}(x-4)}=\frac{x^{2}-(x-4)(x+4)}{x^{2}(x-4)}=\frac{1}{x-4}-\frac{x+4}{x^{2}}=\frac{1}{x-4}-\frac{1}{x}-\frac{4}{x^{2}}.
    Last edited by Krizalid; September 26th 2008 at 03:18 AM.
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