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Homework Help: Partial Fractions Help - Calc II Integration

  1. Sep 20, 2008 #1
    1. The problem statement, all variables and given/known data
    [tex]\int[/tex]1/(s[tex]^{2}[/tex](s-1)[tex]^{2}[/tex]) ds

    2. Relevant equations
    Partial Fractions


    3. The attempt at a solution
    = [tex]\frac{A}{s^{2}}[/tex]+[tex]\frac{B}{s-1}[/tex]+[tex]\frac{C}{(s-1)^{2}}[/tex]

    Setting numerators equal to each other:
    1 = A(s-1)(s-1)[tex]^{2}[/tex] + Bs[tex]^{2}[/tex](s-1)[tex]^{2}[/tex]+Cs[tex]^{2}[/tex](s-1)
    1=(As-A)(s[tex]^{2}[/tex]-2s+1)+Bs[tex]^{2}[/tex](s[tex]^{2}[/tex]-2s+1)+Cs[tex]^{3}[/tex]+Cs[tex]^{2}[/tex]
    1=As[tex]^{3}[/tex]-2As[tex]^{2}[/tex]+As-As[tex]^{2}[/tex]+2As-A+Bs[tex]^{4}[/tex]-2Bs[tex]^{3}[/tex]+Bs[tex]^{2}[/tex]+Cs[tex]^{3}[/tex]-Cs[tex]^{3}[/tex]+Cs[tex]^{2}[/tex]
    1=Bs[tex]^{4}[/tex]+(A-2B+C)s[tex]^{3}[/tex]+(-3A+B-C)s[tex]^{2}[/tex]+3As-A

    Using coefficients to solve for variables (here's where something is up!)
    s[tex]^{4}[/tex]: 0=B
    s[tex]^{3}[/tex]: 0=A-2B+C
    s[tex]^{2}[/tex]: 0=-3A+B-C
    s: 0=3A
    #: 1=-A

    They don't add up! Am I doing something wrong?
    Thank you for everyone's time and help!
     
  2. jcsd
  3. Sep 20, 2008 #2

    Dick

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    Homework Helper

    Don't you need a D/s term as well? If you can't solve it you are missing a variable. (s+1)^2 is a repeated factor and you have two variables for it. Why not for s^2 as well?
     
  4. Sep 20, 2008 #3
    Ohh okay! I'll give it a try! :)

    EDIT: Adding a D/s term only seemed to make the situation worse. Unless I made a slight error, there is now no term A, B, C, or D that stands unattached to an "s" and thus there is no variable to equate to the constant.

    Ideas??
     
    Last edited: Sep 20, 2008
  5. Sep 21, 2008 #4
    I'm not sure if you really have to do it by partial fractions, but a simpler way to do this problem is by trigonometric substitution.
     
  6. Sep 21, 2008 #5

    HallsofIvy

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    [tex]\frac{1}{s^2(s-1)^2}= \frac{A}{s}+ \frac{B}{s^2}+ \frac{C}{s-1}+ \frac{D}{(s-1)^2}[/tex]
    [tex]1= As(s-1)^2+ B(s-1)^2+ Cs^2(s-1)+ Ds^2[/tex]
    Taking s= 0, B= 1.
    Taking s= 1, D= 1.
    Taking s= -1, 1= -4A+ 4B- 2C+ D= -4A+ 4- 2C+ 1 so 4A+ 2C= 4.
    Taking s= 2, 1= 2A+ B+ 4C+ 4D= 2A+ 1+ 4C+ 4 so 2A+ 4C= -3
     
  7. Sep 21, 2008 #6
    Thank you so much, Hall of Ivy! I had a feeling I was doing way too much work. Now I understand :)
     
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