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Integration of Rational Functions by Partial Fractions

  1. Aug 25, 2008 #1
    1. The problem statement, all variables and given/known data


    ∫ 10/(x-1)(x^2+9)

    would i change this into 10/ (x-1) (x+3) (x+3)

    then= A/ x-1 + B/ X+3 + C/ x+3
     
  2. jcsd
  3. Aug 25, 2008 #2
    You have a sum of squares. (x^2 + 9) =/= (x+3)(x+3) = x^2 +6x + 9.
     
  4. Aug 25, 2008 #3
    woops.
     
  5. Aug 25, 2008 #4
    do i just leave it like that x^2+9
     
  6. Aug 25, 2008 #5
    this one is strange
     
  7. Aug 25, 2008 #6
    How about this 3x (1/3x+3)

    yea or nah
     
  8. Aug 25, 2008 #7
    if you have a second power the partial fraction you use is Bx+C/D

    for ex.

    a/(Ax+B)(Cx^2+D)

    if (Cx^2+D) cannot be factored then your partial fraction would be

    a/(Ax+B)(Cx^2+D) = E/(Ax+B) + Fx+G/(Cx^2+D)


    where, a,A,B,C,D,E,F,G are all constants
     
  9. Aug 25, 2008 #8

    how does 3x(1/3x +3) = (x^2+9)?

    it = (x^2 + 9x)
     
  10. Aug 25, 2008 #9
    Oright I'll use your technique. Thanks man.
     
  11. Aug 25, 2008 #10
    So it would be A/x-1 + Bx+C/ x^2+9
     
  12. Aug 25, 2008 #11
    correct
     
  13. Aug 25, 2008 #12
    ok here's what i did 10= A(x^2+9) + (Bx+C) (x-1)

    10= Ax^2 + 9A + Bx^2- Bx + Cx -C
    10= (A+B) x^2 + (9A-C) -(B+C)x
    ok so A+B=0, 9A-C = 10, B+C =0

    is that correct
     
  14. Aug 25, 2008 #13
    be careful

    10= Ax^2 + 9A + Bx^2- Bx + Cx -C


    you have

    10= (A+B) x^2 + (9A-C) -(B+C)x = (A+B) x^2 + (9A-C) -Bx-Cx = remember that negative is distributive.
     
  15. Aug 25, 2008 #14
    Ok this is where I need help. Bx+Cx,
     
  16. Aug 25, 2008 #15
    ok i have to go to sleep now lol, hopefully PF wont kill me for telling its easy enough..:(

    Just make it (C-B)x

    then solve

    ur equations are right A+B=0, 9A-C = 10, B+C =0 but the last one should be changed i think you now..
     
  17. Aug 25, 2008 #16
    oh ok, so would it be (B-C)x
     
  18. Aug 25, 2008 #17
    Thanks a lot man.
     
  19. Aug 25, 2008 #18
    oh shoot another thing, by the looks of it not that im to judge but u I THINK you may need help later in the actual integration so i will give u the formula,


    you will need EVENTUALLY need to integrate

    A/x^2+B^2
    =
    (1/a)tan-1(x/a)

    just do the integration and you will see one place u will be like how do i integrate

    -1/x^2+9
    then use the above formula

    ok going now, hopefully someone else will help if u get stuck integrating
     
  20. Aug 25, 2008 #19
    I was wondering how the answer in the back of the book had that. Thanks.
     
  21. Aug 25, 2008 #20
    so A=1, C=-1, and B=1
     
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