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Math Help - Integration of Rational Functions by Partial Fractions

  1. #1
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    Integration of Rational Functions by Partial Fractions

    Hey guys i'm a bit confused and clueless when it comes to writing out the form of the partial fraction decomposition of the function.


    Maybe you guys can help me understand through examples.

    2 / (x^2 + 3x - 4)

    now I know I can factor the Q(x) to (x + 4)(x - 1)

    So the partial fractioning for this will be (A / (x + 4)) + (B / (x - 1)) right? I have no problems doing simple ones such as this.




    Now here is an example of one of which i'm totally clueless as where the extra factors are coming from.

    1 / (x^6 - x^3)

    now I can factor the Q(x) to (x^3)(x - 1)(x^2 +x + 1). Now how would I partial fraction this and the most important WHY? The "WHY" is what I really need most. Thanks
    Last edited by JonathanEyoon; June 19th 2008 at 09:30 AM. Reason: Had to fix + 4 to - 4
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    Quote Originally Posted by JonathanEyoon View Post
    Hey guys i'm a bit confused and clueless when it comes to writing out the form of the partial fraction decomposition of the function.


    Maybe you guys can help me understand through examples.

    2 / (x^2 + 3x +4)

    now I know I can factor the Q(x) to (x + 4)(x - 1)

    So the partial fractioning for this will be (A / (x + 4)) + (B / (x - 1)) right? I have no problems doing simple ones such as this.

    Now here is an example of one of which i'm totally clueless as where the extra factors are coming from.

    1 / (x^6 - x^3)

    now I can factor the Q(x) to (x^3)(x - 1)(x^2 +x + 1). Now how would I partial fraction this and the most important WHY? The "WHY" is what I really need most. Thanks
    When you have repeated factors you must include a term for each degree of the repeated factor. So for example in your problem

    \frac{1}{x^3(x-1)(x^2+x+1)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x^  3}+\frac{D}{x-1}+\frac{Ex+F}{x^2+x+1}

    The same is true for repeated quadratic factors for example

    \frac{3}{x^2(x^2+9)^2}=\frac{A}{x}+\frac{B}{x^2}+\  frac{Cx+D}{x^2+9}+\frac{Ex+F}{(x^2+9)^2}

    I hope this helps
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  3. #3
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    Hello =)

    Quote Originally Posted by JonathanEyoon View Post
    Hey guys i'm a bit confused and clueless when it comes to writing out the form of the partial fraction decomposition of the function.


    Maybe you guys can help me understand through examples.

    2 / (x^2 + 3x +4)

    now I know I can factor the Q(x) to (x + 4)(x - 1)

    So the partial fractioning for this will be (A / (x + 4)) + (B / (x - 1)) right? I have no problems doing simple ones such as this.

    Now here is an example of one of which i'm totally clueless as where the extra factors are coming from.

    1 / (x^6 - x^3)

    now I can factor the Q(x) to (x^3)(x - 1)(x^2 +x + 1). Now how would I partial fraction this and the most important WHY? The "WHY" is what I really need most. Thanks
    For the first one, I guess it is (x^2+3x-4), right ?


    When dealing with powers, for example [P(x)]^n, with P(x) a polynomial of degree 1 (that is to say x-4, 2x+3, etc...), we write this :

    \frac{\dots}{[P(x)]^n}=\frac{A}{P(x)}+\frac{B}{[P(x)]^2}+\dots+\frac{N}{[P(x)]^n}

    When you have an irreducible polynomial, such as x^2+x+1, then :

    \frac{\dots}{(x-1)(x^2+x+1)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+x+1}


    Good luck


    Edit : niaaah >.< too slcow !
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    For repeated factors, I must include a term and it goes the same for quadratics. Let me try one on here and tell me if it's correct.


    (2x + 1) / ((x + 1)^3)((x^2 + 4)^2)


    (A / (x + 1)) + B / ((x + 1)^2) + C / ((x + 1)^3) + Dx +E / (x^2 + 4) + Fx + G / ((x^2 +4)^2)


    what do you think?
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    Quote Originally Posted by JonathanEyoon View Post
    For repeated factors, I must include a term and it goes the same for quadratics. Let me try one on here and tell me if it's correct.


    (2x + 1) / ((x + 1)^3)((x^2 + 4)^2)


    (A / (x + 1)) + B / ((x + 1)^2) + C / ((x + 1)^3) + Dx +E / (x^2 + 4) + Fx + G / ((x^2 +4)^2)


    what do you think?
    I think it's correct


    Extra feature !!!! Here is what you should get :

    \frac{-(38*x-8)}{625*(x^2+4)}-\frac{21*x+19}{125*(x^2+4)^2}+\frac{38}{625*(x+1)}  +\frac{6}{125*(x+1)^2}-\frac{1}{25*(x+1)^3}
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  6. #6
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    I really appreciate the help guys! I have one more question


    How am I supposed to know when to Equate for quadratics, linear, and constants?


    Example. ( This is kinda ugly )

    Example : (-x^3 + 2x - x + 1) = A(x^4 + 2x^2 + 1) + B(x^4 + x^2) + C(x^3 + x) + Dx^2 + Ex.


    Thanks in advance
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  7. #7
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    Quote Originally Posted by JonathanEyoon View Post
    I really appreciate the help guys! I have one more question


    How am I supposed to know when to Equate for quadratics, linear, and constants?


    Example. ( This is kinda ugly )

    Example : (-x^3 + 2x - x + 1) = A(x^4 + 2x^2 + 1) + B(x^4 + x^2) + C(x^3 + x) + Dx^2 + Ex.


    Thanks in advance
    I think the 2nd term should be 2x^2

    First we multiply out the right hand side to get

    -x^3+2x^2-x+1=Ax^4+2Ax^2+A+Bx^4+Bx^2+Cx^3+Cx+Dx^2+Ex

    Now we group all terms of the same degree (same exponent) to get

    -x^3+2x^2-x+1=Ax^4+Bx^4+Cx^3+2Ax^2+Bx^2+Dx^2+Cx+Ex+A

    Now we can factor the right hand side to get

    -x^3+2x^2-x+1=(A+B)x^4+(C)x^3+(2A+B+D)x^2+(C+E)x+A

    Now we can equate the coeffients from both sides of the equation. For exapmle since there is no x^4 term on the left hand side its coeffeint must be zero so... 0=A+B now for the cubic term -1=C and so on 2=2A+B+D and -1=C+E finally 1=A

    Now we can solve this system to get A=1, B=-1,C=-1,D=1,E=0
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  8. #8
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    ah Ic. Thanks alot! I'm doing the homework for these types of problems at the moment. Will let you guys know if I have any questions.
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