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Math Help - Partial fraction question

  1. #1
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    Partial fraction question

    I need help solving:

    integral of (20x-30) / (x(x-2)(x^2-2x+5))

    If I try to decompose it;

    I will have

    LHS from above = (A/x)+(B/x-2) + ?

    Stuck here.
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  2. #2
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    You can use the third factor:
    <br />
\frac{20x-30}{x(x-2)(x^2-2x+5)} = \frac{A}{x} + \frac{B}{x-2} + \frac{C}{x^2 -2x+5}<br />
    C will a function of x.
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  3. #3
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    <br />
\frac{20x-30}{x(x-2)(x^2-2x+5)} = \frac{A}{x} + \frac{B}{x-2} + \frac{Cx+D}{x^2 -2x+5}<br />
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  4. #4
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    ohh I see, but I didnt think the polynomial could be there under Cx+D because its not factored?

    So, now I am suppose to multiply both sides by (x(x-2)(x^2-2x+5))?
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  5. #5
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    Quote Originally Posted by shadow85 View Post
    ohh I see, but I didnt think the polynomial could be there under Cx+D because its not factored?

    So, now I am suppose to multiply both sides by (x(x-2)(x^2-2x+5))?
    Yes
    and find the values of A,B,C and D
    substitute and evaluate the resulting integrals.
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  6. #6
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     \frac{20x-30}{x(x-2)(x^2-2x+5)}

     = \frac{20x-30}{(x^2-2x)((x^2-2x)+5)}

    If we let  u = x^2 - 2x

    and we know  \frac{1}{u(u+5)}= \frac{1}{5} \left ( \frac{1}{u} - \frac{1}{ u + 5 } \right )


    The original fraction can be expressed in the form of :

     \frac{20x-30}{5} \left (\frac{1}{u} - \frac{1}{ u + 5 } \right )

     = (4x - 6) \left (  \frac{ 1}{ x (x-2)} - \frac{1 }{ x^2 - 2x + 5 } \right )

    we can see the last term  \frac{4x-6}{ x^2 -2x +5 }

    the numerator is a polynomial with degree 1 , that means

    we do not need to break it down into pieces anymore .

    If we continue , we can obtain

     = \frac{2(2x-3)}{x(x-2)} - \frac{2(2x-3)}{ x^2 -2x +5 }

     = \frac{ 2[(3/2)(x-2) + (1/2)x]  }{x (x-2)}  - \frac{2(2x-3)}{ x^2 -2x +5 }

     =  \frac{3}{x} + \frac{1}{x-2}  - \frac{2(2x-3)}{x^2 -2x + 5}
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  7. #7
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    umm wasnt the second post, enough to solve this?
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  8. #8
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    Not quite, because you to solve you would have to have a polynomial in the third fraction. I did say that C was a polynomial, but I didn't say of what order.
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