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Math Help - Partial Fractions

  1. #1
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    Partial Fractions

    hi again!

    i have to make partialfractions out of the following integral:

    \int \frac{1-x}{x^3-2x^2+x-2} dx

    guessing a critical point for the denominator gives me x_{0}=2

    then polynomial division gives me

    (x^3-2x^2+x-2) / (x-2) = x^2+1

     \rightarrow \frac {1-x}{(x-2)(x^2+1)} = \frac {A}{x-2} + \frac{B}{x^2+1}

    is this correct ? if so, i dont seem to get an equation system that is solvable...

    could you give me a quick hint to do partial fractions in general ? like the different cases etc..., my professors script is a mess
    thanks
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  2. #2
    MHF Contributor alexmahone's Avatar
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    \frac {1-x}{(x-2)(x^2+1)} = \frac {A}{x-2} + \frac{B}{x^2+1}
    Multiply throughout by (x-2)
    \frac {1-x}{x^2+1} = A + B\frac{x-2}{x^2+1}
    Sub x=2
    -\frac{1}{5}=A
    A=-\frac{1}{5}
    Last edited by alexmahone; June 11th 2009 at 09:42 AM.
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  3. #3
    AMI
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    Quote Originally Posted by coobe View Post
     \rightarrow \frac {1-x}{(x-2)(x^2+1)} = \frac {A}{x-2} + \frac{B}{x^2+1}
    Actually, it is \frac {1-x}{(x-2)(x^2+1)} = \frac {A}{x-2} + \frac{Bx+C}{x^2+1}
    \Rightarrow 1-x=A(x^2+1)+(Bx+C)(x-2) \Rightarrow A+B=0,-2B+C=-1,A-2C=1 \Rightarrow A=-\frac{1}{5},B=\frac{1}{5},C=-\frac{3}{5}.

    Did you look here: Partial fraction - Wikipedia, the free encyclopedia ?! It seems pretty well written..
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  4. #4
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    sorry i dont get it, why you need a C when there are only 2 factors in the denominator.... and i dont get the x with the B neither
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  5. #5
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    Quote Originally Posted by coobe View Post
    sorry i dont get it, why you need a C when there are only 2 factors in the denominator.... and i dont get the x with the B neither
    What do you mean by "two factors in the denominator"? The reason we need Bx+ C over x^2+ 1 is because it can't be factored!
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  6. #6
    AMI
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    Quote Originally Posted by coobe View Post
    sorry i dont get it, why you need a C when there are only 2 factors in the denominator.... and i dont get the x with the B neither
    Did you look at the link?? It's the case where you have an irreducible second degree factor in the denominator, just like HallsofIvy said..
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