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Math Help - fraction decomposition of a rational function

  1. #1
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    fraction decomposition of a rational function

    The form of the partial fraction decomposition of a rational function is given below.



    A=
    B=
    C=

    Now evaluate the indefinite integral.
    i need a b and c and the integral
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  2. #2
    Senior Member DeMath's Avatar
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    Quote Originally Posted by dat1611 View Post
    The form of the partial fraction decomposition of a rational function is given below.



    A=
    B=
    C=

    Now evaluate the indefinite integral.
    i need a b and c and the integral
    \frac{{2{x^2} + 3x + 30}}<br />
{{\left( {x + 4} \right)\left( {{x^2} + 9} \right)}} = \frac{A}<br />
{{x + 4}} + \frac{{Bx + C}}<br />
{{{x^2} + 9}} \Leftrightarrow

    \Leftrightarrow 2{x^2} + 3x + 30 = \left( {{x^2} + 9} \right)A + \left( {x + 4} \right)\left( {Bx + C} \right) =

    = A{x^2} + 9A + B{x^2} + Cx + 4Bx + 4C =

    = \left( {A + B} \right){x^2} + \left( {4B + C} \right)x + 9A + 4C \Rightarrow

    \Rightarrow \left\{ \begin{gathered}<br />
  A + B = 2, \hfill \\<br />
  4B + C = 3, \hfill \\<br />
  9A + 4C = 30; \hfill \\ <br />
\end{gathered}  \right. \Leftrightarrow \left\{ \begin{gathered}A = 2 - B, \hfill \\<br />
  C = 3 - 4B, \hfill \\<br />
  9\left( {2 - B} \right) + 4\left( {3 - 4B} \right) = 30; \hfill \\ <br />
\end{gathered}  \right. \Leftrightarrow

    \Leftrightarrow \left\{ \begin{gathered}A = 2 - B, \hfill \\C = 3 - 4B, \hfill \\30 - 25B = 30; \hfill \\ \end{gathered}  \right. \Leftrightarrow \left\{ \begin{gathered}A = 2, \hfill \\B = 0, \hfill \\C = 3. \hfill \\ \end{gathered}  \right.

    \frac{{2{x^2} + 3x + 30}}<br />
{{\left( {x + 4} \right)\left( {{x^2} + 9} \right)}} = \frac{2}<br />
{{x + 4}} + \frac{3}<br />
{{{x^2} + 9}}.
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  3. #3
    Senior Member Danneedshelp's Avatar
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    Here is how I would do it...

    1) Multiply both sides by x+4 and take the limit \lim_{x\rightarrow{-4}}x+4. This will automatically give you A=2

    2) Multiply both sides by x and take the limit as x goes to infinity. You will find B=0.

    3) Simply, let x=0 and you will find C=3 right away.

    So, A=2, B=0, and C=3.

    Thats it!
    Last edited by Danneedshelp; July 30th 2009 at 05:16 PM.
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  4. #4
    Senior Member Danneedshelp's Avatar
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    Sorry about the double post!
    Last edited by Danneedshelp; July 30th 2009 at 05:11 PM.
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  5. #5
    Senior Member DeMath's Avatar
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    Quote Originally Posted by dat1611 View Post

    Now evaluate the indefinite integral.
    i need a b and c and the integral
    \int {\frac{{2{x^2} + 3x + 30}}{{\left( {x + 4} \right)\left( {{x^2} + 9} \right)}}dx}  = 2\int {\frac{{dx}}{{x + 4}}}  + 3\int {\frac{{dx}}<br />
{{{x^2} + 9}}} .

    2\int {\frac{{dx}}{{x + 4}}}  = 2\int {\frac{{d\left( {x + 4} \right)}}<br />
{{x + 4}}}  = 2\ln \left| {x + 4} \right| + C.

    3\int {\frac{{dx}}{{{x^2} + 9}}}  = \frac{1}{3}\int {\frac{{dx}}{{{{\left( {x/3} \right)}^2} + 1}}}  = \left\{ \begin{gathered}\frac{x}{3} = u, \hfill \\\frac{1}{3}dx = du \hfill \\ \end{gathered}  \right\} =

    = \int {\frac{{dx}}{{{u^2} + 1}}}  = \arctan u + C = \arctan \frac{x}{3} + C.

    \boxed{\int {\frac{{2{x^2} + 3x + 30}}{{\left( {x + 4} \right)\left( {{x^2} + 9} \right)}}dx}  = 2\ln \left| {x + 4} \right| + \arctan \frac{x}<br />
{3} + C}
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