Results 1 to 2 of 2

Math Help - Integration by Partial Fractions

  1. #1
    Member RedBarchetta's Avatar
    Joined
    Apr 2008
    From
    United States
    Posts
    114

    Integration by Partial Fractions

    <br />
\int {\frac{{2s + 2}}<br />
{{(s^2  + 1)(s - 1)^3 }}ds} <br />

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

    <br />
\begin{gathered}<br />
  \frac{{2s + 2}}<br />
{{(s^2  + 1)(s - 1)^3 }} = \frac{{As + B}}<br />
{{s^2  + 1}} + \frac{C}<br />
{{s - 1}} + \frac{D}<br />
{{(s - 1)^2 }} + \frac{E}<br />
{{(s - 1)^3 }} \hfill \\<br />
  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 \\ <br />
\end{gathered} <br />

    Plugging in s=1 I get E=2

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

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

    when s=2

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

    when s=-2

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

    s=-1

    <br />
\begin{gathered}<br />
  0 = (B - A)( - 2)^3  + C(2)( - 2)^2  + D(2)( - 2) + 2(2) \hfill \\<br />
  0 = 8A - 8B + 8C - 4D + 4 \hfill \\ <br />
\end{gathered} <br />

    ....and on and on. How do you even begin to solve these?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by RedBarchetta View Post
    <br />
\int {\frac{{2s + 2}}<br />
{{(s^2  + 1)(s - 1)^3 }}ds} <br />

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

    <br />
\begin{gathered}<br />
  \frac{{2s + 2}}<br />
{{(s^2  + 1)(s - 1)^3 }} = \frac{{As + B}}<br />
{{s^2  + 1}} + \frac{C}<br />
{{s - 1}} + \frac{D}<br />
{{(s - 1)^2 }} + \frac{E}<br />
{{(s - 1)^3 }} \hfill \\<br />
  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 \\ <br />
\end{gathered} <br />

    Plugging in s=1 I get E=2

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

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

    when s=2

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

    when s=-2

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

    s=-1

    <br />
\begin{gathered}<br />
  0 = (B - A)( - 2)^3  + C(2)( - 2)^2  + D(2)( - 2) + 2(2) \hfill \\<br />
  0 = 8A - 8B + 8C - 4D + 4 \hfill \\ <br />
\end{gathered} <br />

    ....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
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Partial fractions for integration
    Posted in the Calculus Forum
    Replies: 6
    Last Post: September 10th 2011, 10:22 PM
  2. Integration partial fractions
    Posted in the Calculus Forum
    Replies: 6
    Last Post: September 10th 2011, 02:58 AM
  3. Integration by partial fractions
    Posted in the Calculus Forum
    Replies: 2
    Last Post: October 1st 2009, 01:18 AM
  4. integration via partial fractions
    Posted in the Calculus Forum
    Replies: 5
    Last Post: March 27th 2009, 10:13 AM
  5. Integration by Partial Fractions
    Posted in the Calculus Forum
    Replies: 2
    Last Post: November 5th 2008, 09:00 AM

Search Tags


/mathhelpforum @mathhelpforum