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Thread: Integration by Partial Fractions

  1. #1
    Member RedBarchetta's Avatar
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    Integration by Partial Fractions

    $\displaystyle
    \int {\frac{{2s + 2}}
    {{(s^2 + 1)(s - 1)^3 }}ds}
    $

    Alright, so I split it up the integrand into partial fractions and try to find the coefficients:

    $\displaystyle
    \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 \\
    \end{gathered}
    $

    Plugging in s=1 I get E=2

    $\displaystyle
    \begin{gathered}
    4 = (As + B)(0)^3 + C(s^2 + 1)(0)^2 + D(s^2 + 1)(0) + E(2) \hfill \\
    4 = E(2) \Rightarrow E = 2 \hfill \\
    \end{gathered}
    $

    But solving for the other variables gets tricky.....

    when s=2

    $\displaystyle
    \begin{gathered}
    6 = (As + B)(1)^3 + C(5)(1)^2 + D(5)(1) + E(5) \hfill \\
    2A + B + 5C + 5D = 0 \hfill \\
    \end{gathered}
    $

    when s=-2

    $\displaystyle
    \begin{gathered}
    - 2 = (B - 2a)( - 27) + C(5)( - 3)^2 + D(5)( - 3) + 2(5) \hfill \\
    54A + 45C - 15D - 27B + 12 = 0 \hfill \\
    \end{gathered}
    $

    s=-1

    $\displaystyle
    \begin{gathered}
    0 = (B - A)( - 2)^3 + C(2)( - 2)^2 + D(2)( - 2) + 2(2) \hfill \\
    0 = 8A - 8B + 8C - 4D + 4 \hfill \\
    \end{gathered}
    $

    ....and on and on. How do you even begin to solve these?
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by RedBarchetta View Post
    $\displaystyle
    \int {\frac{{2s + 2}}
    {{(s^2 + 1)(s - 1)^3 }}ds}
    $

    Alright, so I split it up the integrand into partial fractions and try to find the coefficients:

    $\displaystyle
    \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 \\
    \end{gathered}
    $

    Plugging in s=1 I get E=2

    $\displaystyle
    \begin{gathered}
    4 = (As + B)(0)^3 + C(s^2 + 1)(0)^2 + D(s^2 + 1)(0) + E(2) \hfill \\
    4 = E(2) \Rightarrow E = 2 \hfill \\
    \end{gathered}
    $

    But solving for the other variables gets tricky.....

    when s=2

    $\displaystyle
    \begin{gathered}
    6 = (As + B)(1)^3 + C(5)(1)^2 + D(5)(1) + E(5) \hfill \\
    2A + B + 5C + 5D = 0 \hfill \\
    \end{gathered}
    $

    when s=-2

    $\displaystyle
    \begin{gathered}
    - 2 = (B - 2a)( - 27) + C(5)( - 3)^2 + D(5)( - 3) + 2(5) \hfill \\
    54A + 45C - 15D - 27B + 12 = 0 \hfill \\
    \end{gathered}
    $

    s=-1

    $\displaystyle
    \begin{gathered}
    0 = (B - A)( - 2)^3 + C(2)( - 2)^2 + D(2)( - 2) + 2(2) \hfill \\
    0 = 8A - 8B + 8C - 4D + 4 \hfill \\
    \end{gathered}
    $

    ....and on and on. How do you even begin to solve these?
    it's a bit messy. but i think the more straightforward way is to expand everything and equate coefficients.

    but to continue here, you have 4 unknowns left, you need four equations.

    plug in s = 0 to get another equation. then you can use the elimination method to solve the system. similar to how you use the elimination method to solve a system of two equations and two unknowns
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