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Help with integration via partial fractions

  1. Apr 23, 2006 #1
    I'm supposed to integrate this using partial fractions:

    [tex]\int\frac{1}{(x-1)^2(x+1)} \ dx [/tex]

    I've started to split the integrand into more readily integrated fractions by stating...

    [tex] \frac{A}{(x-1)}+\frac{B}{(x-1)^2}+\frac{C}{(x+1)} = \frac{1}{(x-1)^2(x+1)}[/tex]

    combining the fractions via addition, and making the numerators equal to eachother (because we have common denominators on each side), I get

    [tex]x^3(A+C)+X^2(-A+B-3C)+X(-A+3C)+A-B-C=1[/tex]

    From this equation, I can deduce the following:

    1. [tex] A+C=0 [/tex]

    2. [tex] -A+B-3C=0 [/tex]

    3. [tex] -A+3C=0 [/tex]

    4. [tex] A-B-C=1 [/tex]

    The problem is, the results I get from these equations contradict eachother. ie, if add equations 1 and 3 together, I get that C=0, but if I add equations 2 and 3 together, I get that [tex]C=\frac{-1}{4}[/tex]

    What in god's name am I doing wrong? I've done the calculations a number of times, and am pretty sure that my algebra isn't off...any help?

    I have a hunch that my problem comes from the way I split up the fraction into three parts..
     
    Last edited: Apr 23, 2006
  2. jcsd
  3. Apr 23, 2006 #2

    nrqed

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    There is no way to get C=-1/4 using only eqs 2 and 3. if you *add* them, C drops out. If you subtract them, B remains there
     
  4. Apr 23, 2006 #3

    dav2008

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    Well the problem off the bat is that you have 4 equations and 3 unknowns.

    The next (and directly related) problem is that I'm not quite sure where you're getting your x3 term from...

    Looking at your second line: [tex] \frac{A}{(x-1)}+\frac{B}{(x-1)^2}+\frac{C}{(x+1)} = \frac{1}{(x-1)^2(x+1)}[/tex]I'm not sure why you have B divided by (x-1)2.
     
    Last edited: Apr 23, 2006
  5. Apr 23, 2006 #4

    nrqed

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    Your mistake is that you multiplied A by (x-1)^2 (x+1). You must only multiply it by (x-1)*(x+1)

    EDIT: you made the same mistake with the others. You multiply by too many factors. For example, B must get multiplied by (x+1) only
     
    Last edited: Apr 23, 2006
  6. Apr 23, 2006 #5

    nrqed

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    If I recall, in partial fractions if there is a term of higher power, one must include terms all the way to the highest power hence A/(x-1) +B/(x-1)^2
     
  7. Apr 23, 2006 #6
    Well, my reasoning for splitting up the fractions the way I did, was because my teacher and my textbook both told me that when the denominator is comprised of linear factors, some of which are repeated, I should go about the problem the following manner:

    [tex]\frac{6x+7}{(x+2)^2} = \frac{A}{x+2} + \frac{B}{(X+2)^2}[/tex]

    the above is an example straight out of the textbook.

    The truth is, I was given very little justification for having to split up the fraction in this way, and therefore have a very muddled understanding of the whole process in general. My attempt at trying to solve the initial question basically represents me trying to piece all the rules I got for partial fractions in a disjointed manner.

    How do you guys suggest I go about tacking the original problem posted in my first post? Trigonometric substitution is out of question right now, since I haven't covered it yet. Perhaps I should try and use integration by parts? I don't know. I'm honestly pretty stumped with this one. I think all I need is a nudge in the right direction here.
     
  8. Apr 23, 2006 #7

    dav2008

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    [tex]\int\frac{1}{(x-1)^2(x+1)} \ dx [/tex] is the same as
    [tex]\int\frac{1}{(x-1)(x-1)(x+1)} \ dx [/tex]

    I guess you're right that it doesn't really matter what the denominator of your fractions are as long as you multiply A, B, and C by the proper terms.

    In your example of [tex]\frac{6x+7}{(x+2)^2} = \frac{A}{x+2} + \frac{B}{(X+2)^2}[/tex] you would have your equation as [tex]A(x+2)+B=6x+7[/tex]
     
  9. Apr 23, 2006 #8

    nrqed

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    No, keep going! You did write the correct decomposition. It is just when you put them over a common denominator that you multiplied by two many factors.

    For example [itex] {A \over (x-1)} = {A (x-1)(x+1) \over (x-1)^2 (x+1) } [/itex] and so on. You see? The idea is to bring all your terms over the common factor (x-1)^2(x+1) so that now you can set all this equal to your initial expression. Your mistake was to multiply A, B and C all by too many factors.

    Pat
     
  10. Apr 23, 2006 #9
    !!!

    Good lord, I feel like a moron :rofl:


    thanks!!
     
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