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Integration by Partial Fractions - Long Problem

  1. Sep 21, 2008 #1
    1. The problem statement, all variables and given/known data
    [tex]
    \int {\frac{{2s + 2}}
    {{(s^2 + 1)(s - 1)^3 }}ds}
    [/tex]

    3. The attempt at a solution

    This is a long one....First, I split the integrand into partial fractions and find the coefficients:

    [tex]
    \begin{gathered}
    \frac{{2s + 2}}
    {{(s^2 + 1)(s - 1)^3 }} = \frac{{As + B}}
    {{s^2 + 1}} + \frac{C}
    {{s - 1}} + \frac{D}
    {{(s - 1)^2 }} + \frac{E}
    {{(s - 1)^3 }} \hfill \\
    2s + 2 = (As + B)(s - 1)^3 + C(s^2 + 1)(s - 1)^2 + D(s^2 + 1)(s - 1) + E(s^2 + 1) \hfill \\
    2s + 2 = (As + B)(s^3 - 3s^2 + 3s - 1) + C(s^4 - 2s^3 + 2s^2 - 2s + 1) + D(s^3 - s^2 + s - 1) + E(s^2 + 1) \hfill \\
    \end{gathered}
    [/tex]
     
  2. jcsd
  3. Sep 21, 2008 #2
    Part 2:

    Now find the coefficients. I suppose you can use a calculator if you want to check for these. :rofl:

    [tex]
    \begin{gathered}
    s^4 :A + C = 0 \hfill \\
    s^3 :B - 3A - 2C + D = 0 \hfill \\
    s^2 :3A - 3B + 2C - D = 0 \hfill \\
    s^1 :3B - A - 2C + D = 2 \hfill \\
    s^0 :C - B - D + E = 2 \hfill \\
    \end{gathered}
    [/tex]

    After solving....

    [tex]
    \begin{gathered}
    A: - 1/2 \hfill \\
    B:1/2 \hfill \\
    C:1/2 \hfill \\
    D: - 1 \hfill \\
    E:1 \hfill \\
    \end{gathered}
    [/tex]
     
  4. Sep 21, 2008 #3
    Part 3:

    [tex]
    \begin{gathered}
    \int {\frac{{2s + 2}}
    {{(s^2 + 1)(s - 1)^3 }}} = \int {\left[ {\frac{{As + B}}
    {{s^2 + 1}} + \frac{C}
    {{s - 1}} + \frac{D}
    {{(s - 1)^2 }} + \frac{E}
    {{(s - 1)^3 }}} \right]} ds \hfill \\
    \int {\frac{{2s + 2}}
    {{(s^2 + 1)(s - 1)^3 }}} = \int {\left[ {\frac{{( - 1/2)s + (1/2)}}
    {{s^2 + 1}} + \frac{{1/2}}
    {{s - 1}} + \frac{{ - 1}}
    {{(s - 1)^2 }} + \frac{1}
    {{(s - 1)^3 }}} \right]} ds \hfill \\
    \end{gathered}
    [/tex]
     
  5. Sep 21, 2008 #4
    Part 4:

    Alright, now just solving the integral:

    [tex]
    \begin{gathered}
    \int {\frac{{2s + 2}}
    {{(s^2 + 1)(s - 1)^3 }}} = \frac{1}
    {2}\int {\frac{{ds}}
    {{s^2 + 1}} - \frac{1}
    {2}\int {\frac{{s ds}}
    {{s^2 + 1}}} + \int {\frac{{1/2}}
    {{s - 1}}ds - } } \int {\frac{{ds}}
    {{(s - 1)^2 }} + \int {\frac{{ds}}
    {{(s - 1)^3 }}} } \hfill \\
    \int {\frac{{2s + 2}}
    {{(s^2 + 1)(s - 1)^3 }}} = \frac{1}
    {2}\tan ^{ - 1} s - \frac{1}
    {4}\ln (s^2 + 1) + \frac{1}
    {2}\ln |s - 1| + \frac{1}
    {{s - 1}} - \frac{1}
    {{2(s - 1)^2 }} + C \hfill \\
    \end{gathered}
    [/tex]

    Here's what the book says:

    [tex]
    \int {\frac{{2s + 2}}
    {{(s^2 + 1)(s - 1)^3 }}} = \frac{{ - 1}}
    {{(s - 1)^2 }} + \frac{1}
    {{(s - 1)}} + \tan ^{ - 1} s + C
    [/tex]

    So i'm not quite sure where I went wrong....Any input will be appreciated! :cool:
     
    Last edited: Sep 21, 2008
  6. Sep 21, 2008 #5

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    While equating coefficients works it is seldom the simplest way to find the coefficients.
    from
    [tex]2s + 2 = (As + B)(s - 1)^3 + C(s^2 + 1)(s - 1)^2 + D(s^2 + 1)(s - 1) + E(s^2 + 1)[/tex]
    If we let s= 1, so that s-1= 0, we get [itex]2+ 2= 4 = B(0)+ C(0)+ D(0)+ E(2)[/itex] so E= 2, not 1.

    Unfortunately, no other value of s makes everything collapse so easily but letting s= 0, -1, 2, and -2 should give fairly simple equations for A, B, C, and D.
     
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