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Math Help - Partial fractions and integrals

  1. #1
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    Lightbulb Partial fractions and integrals

    Use the method of partial fractions to evaluate the following integral:
    \int (x^3 + 4x^2 - 6x)/(x^2 - 2x + 1)dx

    I have also solved for the denominator that (x^2 - 2x + 1) gives (x-1)^2

    Can anybody assist me to find A + B, for the numerators? Given the form:
    A/(x-1) + B/(x-1)


    Thank you and any help is much appreciated,


    Dranalion
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  2. #2
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    you can't apply it yet, you need to do a polynomial long division first, since the \deg p(x)>\deg q(x) given p(x) and q(x) polynomials for numerator and denominator respectively.
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  3. #3
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    Applying polynomial long division, I got the result:
     \int x + 6 + \frac{(5x - 6)}{(x-1)(x-1)} dx

    Given that earlier, I found that x^2 - 2x + 1 = (x-1)(x-1)

    Now, how do I make the \frac{(5x - 6)}{(x-1)(x-1)} part into the format:

    \frac{(A + B)}{(x-1)(x-1)}, where I need to find A and B

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  4. #4
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    no actually, the descomposition is \frac{5x-6}{(x-1)(x-1)}=\frac{a}{x-1}+\frac{b}{(x-1)^{2}}.
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  5. #5
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    How do I find A + B, either way?
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  6. #6
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    Quote Originally Posted by Dranalion View Post
    How do I find A + B, either way?
    The right hand side can be written \frac{a(x-1) + b}{(x - 1)^2}. Compare the numerator of the right hand side to the numerator of the left hand side.
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  7. #7
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    I'm lost.

    5x - 6 = Ax - (A+B)

    => 5 = A ...(1)
    => -6 = -A + B ...(2)

    Substituting A = 5 into (2) gives:

    => -6 = -5 + B

    Which gives B = -1

    Using my graphics calculator to check, I have somehow managed to get two different curves? As in x + 6 + \frac{5}{x-1} - \frac{1}{x-1}  \frac{(x^3 + 4x^2 - 6x)}{(x^2 - 2x + 1)}

    Help, please?
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  8. #8
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    Quote Originally Posted by Dranalion View Post
    I'm lost.

    5x - 6 = Ax - (A+B)

    => 5 = A ...(1)
    => -6 = -A + B ...(2)

    Substituting A = 5 into (2) gives:

    => -6 = -5 + B

    Which gives B = -1

    Using my graphics calculator to check, I have somehow managed to get two different curves? As in x + 6 + \frac{5}{x-1} - \frac{1}{x-1}  \frac{(x^3 + 4x^2 - 6x)}{(x^2 - 2x + 1)}

    Help, please?
    Of course they're not equal. Negligence begets error.

    Substitute your values for A and B into the expression given to you in post #4.
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