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Math Help - Problems with Partial Fractions

  1. #1
    DCC
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    Problems with Partial Fractions

    Hi, I've been working on these integral evaluations and I'm stuck. Any help on any of them would be greatly appreciated, thanks in advance!

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

    My attempt:

    2x^3-3x+2=\frac{A}{x^2-1}+\frac{B}{x-1}+\frac{C}{x} (Tried breaking it down, couldn't get it to work )

    2. \int\frac{x^4+1}{x^3+x}dx

    My attempt:

    long division to get \int xdx-\int\frac{x^2+1}{x^3+x}dx (not sure how to show that in Latex)

    Then x^2+1=\frac{A}{X^2+1}+\frac{B}{x+1}+\frac{C}{x}

    Not sure where to go from there...I imagine that I'm missing the same idea in both of these, so any help would be appreciated. Thanks again!
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  2. #2
    Senior Member Pinkk's Avatar
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    For #1:

    x^{3}-x=x(x^{2}-1)=x(x+1)(x-1)

    \frac{2x^{3}-2x+2}{x^{3}-x}=\frac{A}{x}+\frac{B}{x+1}+\frac{C}{x-1}

    Multiply both sides of the equation by x(x+1)(x-1) and go on from there.

    For #2:

    x^{3}+x=x(x^{2}+1)

    \frac{x^{4}+1}{x^{3}+x}=\frac{A}{x}+\frac{Bx+C}{x^  {2}+1}

    Multiply both sides of the equation by x(x^{2}+1) and go on from there.



    Edit: Don't forget to divide beforehead as in Plato's post; I forgot that part.
    Last edited by Pinkk; May 15th 2009 at 03:02 PM.
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  3. #3
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    Here is a solution.
    Attached Thumbnails Attached Thumbnails Problems with Partial Fractions-pf5.gif  
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  4. #4
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    Quote Originally Posted by Pinkk View Post

    For #2:

    x^{3}+x=x(x^{2}+1)

    \frac{x^{4}+1}{x^{3}+x}=\frac{A}{x}+\frac{Bx+C}{x^  {2}+1}

    Multiply both sides of the equation by x(x^{2}+1) and go on from there.
    Your response for number two is clearly incorrect as multiplication on both sides by the cubic will equate a quartic on the left to a quadratic on the right. The way it should be done is as follows:

    \frac{x^4+1}{x(x^2+1)} \equiv x + \frac{1-x^2}{x(x^2+1)}

    then

    \frac{1-x^2}{x(x^2+1)} \equiv \frac{A}{x} + \frac{Bx +C}{x^2 +1}

    yielding

    \frac{x^4+1}{x(x^2+1)} \equiv x + \frac{1}{x} - \frac{2x}{x^2 +1}<br />
.
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  5. #5
    Senior Member Pinkk's Avatar
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    Yes, that's before I realized the highest degree of the numerator was larger than the highest degree of the denominator.
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  6. #6
    DCC
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    I think I've got them both, thanks all!
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  7. #7
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    Quote Originally Posted by Pinkk View Post
    Yes, that's before I realized the highest degree of the numerator was larger than the highest degree of the denominator.
    My apologies for not noticing your edit.
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