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

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

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

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

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

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

    $\displaystyle \frac{{2{x^2} + 3x + 30}}
    {{\left( {x + 4} \right)\left( {{x^2} + 9} \right)}} = \frac{2}
    {{x + 4}} + \frac{3}
    {{{x^2} + 9}}.$
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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 $\displaystyle x+4$ and take the limit $\displaystyle \lim_{x\rightarrow{-4}}x+4$. This will automatically give you $\displaystyle A=2$

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

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

    So, $\displaystyle A=2, B=0,$ and $\displaystyle C=3$.

    Thats it!
    Last edited by Danneedshelp; Jul 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; Jul 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
    $\displaystyle \int {\frac{{2{x^2} + 3x + 30}}{{\left( {x + 4} \right)\left( {{x^2} + 9} \right)}}dx} = 2\int {\frac{{dx}}{{x + 4}}} + 3\int {\frac{{dx}}
    {{{x^2} + 9}}} .$

    $\displaystyle 2\int {\frac{{dx}}{{x + 4}}} = 2\int {\frac{{d\left( {x + 4} \right)}}
    {{x + 4}}} = 2\ln \left| {x + 4} \right| + C.$

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

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

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