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Partial Fractions

  1. Mar 26, 2008 #1
    [SOLVED] Partial Fractions

    1. Evaluate:

    [tex]\int \frac{dx}{x^{2} -1}[/tex]



    Attempt:

    [tex]\int \frac{dx}{x^{2} -1}[/tex]

    [tex]= \int \frac{dx}{(x+1)(x-1)}[/tex]

    [tex]= \frac{A}{(x+1)} + \frac{B}{(x-1)}[/tex]

    [tex]= \frac{Ax - A + Bx + B}{(x+1)(x-1)}[/tex]

    Where do I got from here? Thanks
     
  2. jcsd
  3. Mar 26, 2008 #2

    tiny-tim

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    Hi LadiesMan! :smile:

    (erm … what happened to the integral sign, and dx … ?)

    Easy! You just solve Ax - A + Bx + B = 1.

    So A = … , B = … ? :smile:
     
  4. Mar 26, 2008 #3
    so A = Ax + Bx + B -1
    and B = -Ax + A - Bx -1??
     
  5. Mar 26, 2008 #4

    rock.freak667

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    Ax-A+Bx+B=1 for all values of x

    meaning for any value of x you put in there, it will always be equal to 1
    Choose some suitable values for x to get A and B
     
  6. Mar 26, 2008 #5
    You could do A(x-1) + B(x+1) = 1

    Then to get rid of the A, substitute x = 1 and find B.

    To get rid of the B, substitute x = -1 and find A.
     
  7. Mar 26, 2008 #6
    That means:

    (1/2)ln(x-1) - (1/2)ln(x+1) + C

    Therefore Since: (1/2)ln(x-1) = ln(x-1)^1/2 and (1/2)ln(x +1)= (1/2)ln(x +1)^1/2

    ln(x-1)^1/2 - (1/2)ln(x +1)^1/2 + C

    Natural log rule ln a - ln b = ln (a/b)

    Thank you =)
     
  8. Mar 27, 2008 #7

    tiny-tim

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    … never substitute …

    No! :frown:

    It's a polynomial, and you just put all the coefficients equal to 0.

    It's (A+B)x + (B-A-1) = 0,

    so both the brackets must be 0. :smile:

    Never substitute - it makes it look as if you don't understand what a polynomial is!

    (I agree that, technically, that's the same as substituting x = 0 and ∞; but that requires an extra line, and it doesn't work if you have x^2 or higher.)
     
  9. Mar 27, 2008 #8

    HallsofIvy

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    Tiny Tim, since that equation is true for all x, you certainly could get substitute any two values of x to get 2 equations to solve for a and b. Of course, it is true that if a polynomial is 0 for all x, then its coefficients must all be 0- that's typically simpler than substituting specific values for x.

    Ladies Man, a somewhat simpler method for partial fraction is this:
    You want to find A, B such that
    [tex]\frac{A}{x+1}+ \frac{B}{x-1}= \frac{1}{x^2- 1}[/tex]
    Multiply both sides by x2-1 to eliminate the denominators:
    [tex]A(x-1)+ B(x+ 1)= 1[/itex]
    Now substitute x= 1 and x= -1 to find A and B.
     
  10. Mar 27, 2008 #9
    Thanks everyone! =P
     
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