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Integral of 1/(x-1) has two answers?

  1. Jan 12, 2013 #1
    1. The problem statement, all variables and given/known data

    Hi so I'm integrating 1/(x-1) but I think it may have two answers and I'm not sure if I'm right or wrong

    2. Relevant equations

    y = 1/(x-1)

    3. The attempt at a solution

    well if you integrate this function you would get

    y = ln(x-1) (using u substituition)
    I think it makes sense, you take the derivative and you get the same thing back.

    now... I think there maybe another answer? it looks like this

    y = ln(1-x)

    if you take the derivative of this function you would get

    y = 1/(1-x) χ -1
    which would simplify to
    y = 1/(x-1)

    so yeah... i'm not sure which is the answer y = ln(x-1) or y = ln(1-x)

    I believe I am missing something, thanks for the help :)
     
  2. jcsd
  3. Jan 12, 2013 #2

    tiny-tim

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    Hi RadiantL! :smile:

    ln(x) isn't defined if x < 0

    ∫ 1/x dx = ln(|x|) +C, not ln(x) + C :wink:
     
  4. Jan 12, 2013 #3
    its ln|x-1| + C and as the previous replier said don't forget to put the absolute value symbol around x-1 because the absolute value is here f(y)=f(-y) which means that ln|x-1|=ln|1-x|
     
  5. Jan 12, 2013 #4
    Oh I see, that makes sense haha. Thanks tiny-tim and Dalek1099!
     
  6. Jan 12, 2013 #5

    Ray Vickson

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    [tex] \int \frac{1}{1-x} \, dx = - \ln(1-x),[/tex] because when you defferentiate the right-hand-side you get back the integrand.
     
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