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Partial Fractions

  1. Dec 30, 2007 #1
    Hi!

    There's this one problem that I'm having troubles with. I've tried using the decomposition method, but I've ended up getting a messy answer. If someone can give me tips or the solution to the problem, I'll appreciate it. Here's the problem: solve the integral of 1/ y^2-1 dx.
     
  2. jcsd
  3. Dec 30, 2007 #2

    nicksauce

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    I assume you mean [tex]\int\frac{1}{x^2-1}dx[/tex].

    So if you want to use partial fractions, you can decompose it as such:
    [tex]\frac{1}{x^2-1}=\frac{A}{x+1}+\frac{B}{x-1}[/tex]. How would you continue?
     
  4. Dec 30, 2007 #3
    I used that method however it came out pretty ugly.

    (1) A/ y+1 + B/ y-1

    (2) 1 = (y+1) (y-1) A/ y+1 + (y+1) (y-1) B/ y-1 -I multiplied each side by the LCD

    (3) 1 = A(y-1) +B(y+1) -Cancelled algebraically

    (4) 1 = y(A+B) + (-A+B)

    Since there's only one constant, am I assume that (A+B) =1 and similarly that -A+B =1? In order to identify A and B, I set up the system of equations as:

    A+B =1
    -A+B= 1

    I know there's something wrong here because for any subsitution I would get "0" Plus, I've never seen a partial decomp problem where A and B would equal to the same number.
     
    Last edited: Dec 30, 2007
  5. Dec 30, 2007 #4

    nicksauce

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    A little mistake at the end. Since there is no y term on the left side, you require A+B = 0, while -A + B = 1.
     
  6. Dec 30, 2007 #5
    Thanks Nicksauce,

    The only rational answer I can see why A+B can equal to 0 is because the costants on both sides of the equation are equal somehow

    Here's what I got from using that helpful tip

    (1) 1/ y^2-1 = A/ y-1 + B/ y+1

    (2) -1/2/ y-1 + 1/2/ y+1

    (3) integral 1/ y^2-1 dx = integral - dx/ 2(y-1) + dx/ 2(y+1)

    (4) -1/2 integral dx/ y-1 + 1/2 integral dx/ y+1

    = 1/2 ln (y+1) -1/2 ln (y-1) whew!

    = 1/2 ln ((y-1)/(y+1))

    Thanks for the help! :-)
     
    Last edited: Dec 30, 2007
  7. Dec 30, 2007 #6

    nicksauce

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    No problemo.
     
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