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Math Help - Partial fractions...stuck on one part

  1. #1
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    Cool Partial fractions...stuck on one part

    ∫ dx/(2x-1)

    Here's what I have so far:

    1 = A(2x-1) + B(2x-1) + C

    [solve for c]
    x=1/2
    1= A(0) + B(0) + C
    1=C

    But in order to solve for A or B, I'd have to use x=1/2, but that makes BOTH terms 0.

    Am I supposed to plug in a random # for x and then get a system of equations, or is there an easier way I'm missing?
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  2. #2
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    Substitution would be better for this question

    Let u = 2x-1

    \dfrac{du}{dx} = 2 \: \rightarrow \: dx = \dfrac{1}{2}du


    \int \dfrac{\frac{1}{2}du}{u^3} = \dfrac{1}{2} \int \dfrac{du}{u^3} = \dfrac{1}{2} \int u^{-3}\, du
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  3. #3
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    Quote Originally Posted by janedoe View Post
    ∫ dx/(2x-1)

    Here's what I have so far:

    1 = A(2x-1) + B(2x-1) + C

    [solve for c]
    x=1/2
    1= A(0) + B(0) + C
    1=C

    But in order to solve for A or B, I'd have to use x=1/2, but that makes BOTH terms 0.

    Am I supposed to plug in a random # for x and then get a system of equations, or is there an easier way I'm missing?
    why partial fractions?

    let u = 2x-1

    du = 2\, dx

    \displaystyle \frac{1}{2} \int \frac{2}{(2x-1)^3} \, dx

    \displaystyle \frac{1}{2} \int \frac{du}{u^3}

    \displaystyle -\frac{1}{4u^2} + C

    \displaystyle -\frac{1}{4(2x-1)^2} + C
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  4. #4
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    wow i totally over-thought that, thanks guys!
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  5. #5
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    Quote Originally Posted by janedoe View Post
    Here's what I have so far:

    1 = A(2x-1) + B(2x-1) + C
    When you set-up the partial fractions as \displaystyle \frac{1}{(2x-1)^3} = \frac{a}{2x-1}+\frac{b}{(2x-1)^2}+\frac{c}{(2x-1)^3},
    you can easily see that you will have:

    \displaystyle \int \frac{1}{(2x-1)^3}\;{dx} = \int\left\{\frac{a}{2x-1}+\frac{b}{(2x-1)^2}+\frac{c}{(2x-1)^3}\right\}\;{dx}

     \displaystyle= a\int\frac{1}{2x-1}\;{dx}+b\int\frac{1}{(2x-1)^2}\;{dx}+c\int\frac{1}{(2x-1)^3}\;{dx}

    Now, the last integral is the one which we have started with (times a constant).
    So our fraction was more desirable in its original form.

    Am I supposed to plug in a random # for x and then get a system of equations, or is there an easier way I'm missing?
    That's correct. It's an identity so any random number will do. Having found [LaTeX ERROR: Convert failed] , putting x = 1
    and x = -1 you will find that [LaTeX ERROR: Convert failed] , so in full circle you will end-up with:

    [LaTeX ERROR: Convert failed]

    which is your original integral.

    Morale: don't 'partial fractionalise' a partial fraction.
    Last edited by TheCoffeeMachine; October 17th 2010 at 05:58 PM.
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