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Integral question

  1. Apr 21, 2008 #1
    Why is the anti derivative of 1/(x-3) equal to -ln[tex]\left|x+3\right|[/tex]. Why isn't it just ln[tex]\left|x-3\right|[/tex]. Does it have something to do with the absolute value? Thanks.
     
  2. jcsd
  3. Apr 21, 2008 #2
    for negative x?

    using just log |x-3| defines the differentiation function for pos x only
     
  4. Apr 21, 2008 #3

    exk

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    it's not -ln|x+3| and is in fact ln|x-3| . Take the derivative of your first answer and second and see which one works.
     
  5. Apr 21, 2008 #4
    Alright so I'm guessing there was a typo. Thanks for your help.
     
  6. Apr 21, 2008 #5
    d/dx (log |x-3|) != 1/(x-3)
    for x element of (-inf, inf) for sure! edit: wrong

    ooops.. my mistake. I was ignoring the absolute part
     
    Last edited: Apr 21, 2008
  7. Apr 21, 2008 #6

    rock.freak667

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    Homework Helper

    For [itex]x \neq 3[/itex]
     
  8. Apr 21, 2008 #7

    exk

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    the original function is undefined for x=3 also btw.
     
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