calc

is the integral of 2x/(x^3-x)
=-1/2ln(x+1)+1/2ln(x-1)

I got this by doing this:
(\=integral sign)
2\1/(x^2-1)
2\1/(x+1)(x-1)
=A/(x+1)+B/(x-1)
=Ax-A+Bx+B=1
Ax+Bx=0
B-A=1
B=1+A
(1+A)x+Ax=0
x+Ax+Ax=0
1+AB=0
2A=-1
A=-1/2
B+1/2=1, B=1/2
etc.

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asked by sarah
  1. I differ by a factor of 2

    2[ A/(x+1) +B/(x-1) ]
    A = -1/2, B = +1/2
    so
    2 int [ -.5 dx/(x+1) + .5 dx /(x-1) ]

    1 int [ dx/(x+1) + dx/(x-1) ]

    ln (x+1) + ln (x-1)
    which is
    ln[ (x+1)/(x-1) ]

    1. 👍 0
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    posted by Damon
  2. To see if your answer is correct, take the derivative. Itis
    -(1/2)/(x+1) +(1/2)/(x-1)
    = [-(1/2)(x-1) + (1/2)(x+1)]/(x^2 -1)
    = 1/(x^2-1)
    Your riginal integrand is equal to 2/(x^2-1), so your answer seems to be off by a factor of 2.

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    posted by drwls
  3. Sorry
    ln (x+1)(x-1)
    which is
    ln(x^2-1)

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    posted by Damon
  4. Which I should have seen in the first place.

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    posted by Damon
  5. took ya long enuff

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    posted by sarah
  6. Yeah, but I am really old :)

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    posted by Damon

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