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Thread: Partial fraction method on this problem

  1. #1
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    Partial fraction method on this problem

    I'm just not getting this at all, and I lost the page from class where he did it.

    I asked before and didn't get much further. I know there are a few methods for doing this.

    The initial problem is the integral of dx/(x^2-1)^2 , I'm supposed to integrate by using partial fractions. I get to the point where, if I label my unknowns A, B, C, and D, I can figure out B and D are 1/4. Then I get stuck.

    What's the easiest way to tackle this?

    Thanks!
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  2. #2
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    P.S, the steps he gave us were these:

    1For f(x)=p(x)/q(x)

    1. if degree P >= degree Q, divide it out.

    2. Factor the denominator completely

    3. Write the quotient as a sum.

    4. Set the original fraction equal to the sum of the fraction.

    5. Clear the equation of fractions.

    6. Combine like terms.

    7. Equate coefficients of like powers of X (? How do I do this?)

    8. Solve for coefficients
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  3. #3
    Super Member Quacky's Avatar
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    Quote Originally Posted by Wolvenmoon View Post
    P.S, the steps he gave us were these:

    1For f(x)=p(x)/q(x)

    1. if degree P >= degree Q, divide it out.

    2. Factor the denominator completely

    3. Write the quotient as a sum.

    4. Set the original fraction equal to the sum of the fraction.

    5. Clear the equation of fractions.

    6. Combine like terms.

    7. Equate coefficients of like powers of X (? How do I do this?)

    8. Solve for coefficients
    Ugh this took me forever!

    You get:

    $\displaystyle \displaystyle\frac{1}{(x^2-1)^2}=\frac{A}{x+1}+\frac{B}{(x+1)^2}+\frac{C}{x-1}+\frac{D}{(x-1)^2}$

    Which then gives:

    $\displaystyle A(x+1)(x-1)^2+B(x-1)^2+C(x+1)^2(x-1)+D(x+1)^2=1$
    So, you let $\displaystyle x=1$

    $\displaystyle 4D=1$

    $\displaystyle D=0.25 $

    So you're fine so far.

    You let $\displaystyle x=-1$

    $\displaystyle 4B=1$

    $\displaystyle B=0.25$

    Again, no problem.

    But there isn't a substitution that will work for C and A.

    So it gets a little more tricky.

    Substitute in the values for B and D:

    $\displaystyle A(x+1)(x-1)^2+\frac{1}{4}(x-1)^2+C(x+1)^2(x-1)+\frac{1}{4}(x+1)^2=1$

    Now, the left hand side has to be equal to the right hand side, obviously.Read this paragraph slowly! There is $\displaystyle 0x^3$ on the right hand side of the equation. So there has to be a total of $\displaystyle 0x^3$ on the left. So, expand the brackets, but only look at the $\displaystyle x^3$ terms:
    You'll get the following terms involving an $\displaystyle x^3$:

    $\displaystyle Ax^3 + Cx^3$
    So $\displaystyle Ax^3 + Cx^3 = 0x^3$ as there are no $\displaystyle x^3 $ terms on the right.
    Therefore 1) A + C = 0

    Now I don't want to look at the $\displaystyle x^2$ terms because they will be tricky. I'll skip to terms which have no $\displaystyle x$ coefficients.

    From the left hand side:

    $\displaystyle A+\frac{1}{4}-C+\frac{1}{4}=1$
    So 2) A-C=0.5

    Two equations, two unknowns with A and C left to find.

    Does that make sense?
    Last edited by Quacky; Feb 24th 2011 at 05:06 PM. Reason: Recorrected some wording.
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  4. #4
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    Thank you!
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