Integral question

  • Thread starter elitespart
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  • #1
elitespart
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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.
 

Answers and Replies

  • #2
rootX
465
4
for negative x?

using just log |x-3| defines the differentiation function for pos x only
 
  • #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.
 
  • #4
elitespart
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Alright so I'm guessing there was a typo. Thanks for your help.
 
  • #5
rootX
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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:
  • #6
rock.freak667
Homework Helper
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d/dx (log |x-3|) != 1/(x-3)
for x element of (-inf, inf) for sure!

For [itex]x \neq 3[/itex]
 
  • #7
exk
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the original function is undefined for x=3 also btw.
 

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