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Math Help - partial fractions problem

  1. #1
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    partial fractions problem

    Hi all ...

    Here is an integration which we use partial fractions
    I coulnd get the four constants .. any help?

     \int \frac{4x^2+3}{(x^2+x+1)^2}dx =
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  2. #2
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     <br />
x^2+x+1 = \left(x+\frac{1}{2}+\frac{\sqrt{3}}{2}i\right)\lef  t(x+\frac{1}{2}-\frac{\sqrt{3}}{2}i\right)<br />
    Last edited by pickslides; December 30th 2009 at 01:37 PM. Reason: bad latex
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  3. #3
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    Hello, Abbas!

    Here is an integration which we use partial fractions.
    I couldn't get the four constants .. any help?

     \int \frac{4x^2+3}{(x^2+x+1)^2}\,dx

    We have: . \frac{4x^2 + 3}{(x^2+x+1+1)^2} \;=\;\frac{Ax}{x^2+x+1} + \frac{B}{x^2+x+1} + \frac{Cx}{(x^2+x+1)^2} + \frac{D}{(x^2+x+1)^2}

    Then: . 4x^2+3 \;=\;Ax(x^2+x+1) + B(x^2+x+1) + Cx + D


    x =1:\quad \;\:7 \:=\: A(1)(3) + B(3) + C(1) + D

    x = \text{-}1: \quad \text{-}1 \:=\: A(-1)(1) + B(1) + C(-1) + D

    x=2:\quad\, 35 \:=\: A(2)(7) + B(7) + C(2) + D

    x=0:\quad\;\: 3 \:=\: A(0) + B(1) + C(0) + D


    Solve the system: . \begin{array}{ccc}3A + 3B + C + D &=& 7 \\<br />
\text{-}A + B - C + D &=& \text{-}1 \\<br />
14A + 7B + 2C + D &=& 36 \\<br />
B + D &=& 3 \end{array}


    You should get: . \begin{Bmatrix}A &=& 4 \\ B &=&\text{-}4 \\ C &=& 0 \\ D &=& 7 \end{Bmatrix}

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  4. #4
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    . \frac{4x^2 + 3}{(x^2+x+1+1)^2} \;=\;\frac{Ax}{x^2+x+1} + \frac{B}{x^2+x+1} + \frac{Cx}{(x^2+x+1)^2} + \frac{D}{(x^2+x+1)^2}

    4x^2+3 \;=\;Ax(x^2+x+1) + B(x^2+x+1) + Cx + D

    4x^2 + 3 = Ax^3 + (A+B)x^2 + (A+B+C)x + (B+D)

    equating coefficients ...

    A = 0

    B = 4

    C = -4

    D = -1






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  5. #5
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    Quote Originally Posted by Abbas View Post
    Hi all ...

    Here is an integration which we use partial fractions
    I coulnd get the four constants .. any help?

     \int \frac{4x^2+3}{(x^2+x+1)^2}dx =
    I'll compute \int{\frac{x^{2}}{\left( x^{2}+x+1 \right)^{2}}\,dx}. (Note that when computing these you'll realize how to do the other integral, they're the same.)

    We have t^{2}+t+1=\frac{4t^{2}+4t+4}{4}=\frac{(2t+1)^{2}+3  }{4}. Now in your integral put x=\frac1t and get -\int{\frac{dt}{\left( t^{2}+t+1 \right)^{2}}}=-16\int{\frac{dt}{\left( (2t+1)^{2}+3 \right)^{2}}}, now put 2t+1=\sqrt3u and the integral becomes -\frac{8}{3\sqrt{3}}\int{\frac{du}{\left( u^{2}+1 \right)^{2}}}.

    As for solving the last integral, write the numerator as (1+u^2)-u^2, and split the integral into two ones, integrate by parts the second one and we're done.
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  6. #6
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    Here is another method ,


    I write  4x^2 + 3 in the form of

     A(2x+1) + B(x^2-1) +C(x^2+ x+1)

    I obtained  A= -7/3 , B=-2/3 ,C= 14/3


    the integral

     \int \frac{2x+1}{ (x^2 + x + 1 )^2}~dx = - \frac{1}{ 1 + x+x^2}


     \int \frac{x^2-1}{ (x^2 + x + 1 )^2}~dx

     = \int \frac{ 1 - 1/x^2}{  ( x +1/x + 1)^2 }~dx

    sub.  x + 1/x = t

    you will be able to continue then ,

    the last integral , you just need to make a trigo. substitution . ..
    Last edited by simplependulum; December 30th 2009 at 10:18 PM.
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