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Integration of Rational Functions

  1. Feb 28, 2012 #1
    1. The problem statement, all variables and given/known data
    ∫(7x^2+22x-54)/((x-2)(x+4)(x-1)) dx


    2. Relevant equations

    Partial functions

    3. The attempt at a solution

    ∫(7x^2+22x-54)/((x-2)(x+4)(x-1)) dx = ∫(Ax)/(x2+2x-8) + B/(x-1)

    = ∫A(x-1)dx + ∫B(x^2+2x-8)dx

    Or is it:
    = ∫(AX+B)/(x^2-2x-8) dx + ∫ C/(x-1)dx

    Have I done the seperation of the numerator and denominators right? I'm stuck right about here..I don't know what Bx2 equals to. 7? That doesn't seem to make sense, I must have done something wrong earlier.
    Please help, thanks!
     
    Last edited: Feb 28, 2012
  2. jcsd
  3. Feb 28, 2012 #2

    SammyS

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    Do a complete partial fraction decomposition.

    [itex]\displaystyle \frac{7x^2+22x-54}{(x-2)(x+4)(x-1)}=\frac{A}{x-2}+\frac{B}{x+4}+\frac{C}{x-1}\,.[/itex]
     
  4. Feb 28, 2012 #3
    Okay, so I did that.

    = ∫ [itex]\frac{A(x+4)(x-1) + B(x-2)(x-1) + C(x-2)(x+4)}{(x-2)(x+4)(x-1)}[/itex]

    and this is what I get after multiplying everything out. Three equations. Now I just need to solve for A,B,C...

    -4A+B-8C=-54

    3A-3B+2C=22

    A+B+C =7
     
    Last edited: Feb 28, 2012
  5. Feb 28, 2012 #4

    SammyS

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    Do you know how to do partial fraction decomposition?
     
  6. Feb 28, 2012 #5

    SammyS

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    [itex]\displaystyle \frac{7x^2+22x-54}{(x-2)(x+4)(x-1)}=\frac{A(x+4)(x-1) + B(x-2)(x-1) + C(x-2)(x+4)}{(x-2)(x+4)(x-1)} [/itex]

    So that   [itex]\displaystyle 7x^2+22x-54=A(x+4)(x-1) + B(x-2)(x-1) + C(x-2)(x+4)\,. [/itex]

    There's a neat trick to finishing this.

    Let x=2 to find A.

    Let x=-4 to find B.

    Let x=1 to find C.
     
  7. Feb 28, 2012 #6
    ∫ [itex]\frac{A(x+4)(x-1) + B(x-2)(x-1) + C(x-2)(x+4)}{(x-2)(x+4)(x-1)}[/itex] = ∫ [itex]\frac{7x^2+22x-54}{(x-2)(x+4)(x-1)}[/itex]dx ?


    The values for X are the zeroes, could you explain to me how they are used to find A,B,C? (the trick you typed above)
     
    Last edited: Feb 28, 2012
  8. Feb 28, 2012 #7

    SammyS

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    See post #5.
     
  9. Feb 28, 2012 #8

    [STRIKE]Okay,

    I multiplied everything out: Ax^2+3AX-4A+Bx^2-3Bx+2B+Cx^2+2Cx-8C

    and ...I'm back to what I did in post #3 solving for A,B,C ?

    Sorry, it's getting late and I'm getting a little frustrated, maybe I'm not thinking.[/STRIKE]


    nevermind, I got it now :)

    A = 3
    B = -1
    C = 5

    So the answer should be ∫f(x)dx = 3ln|x-2|-ln|x+4|+5ln|x-1| + C

    I made a calculation error in post #3.

    Thanks for your help Sammy! :)
     
    Last edited: Feb 28, 2012
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