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Question about integral and natural log

  1. Sep 13, 2011 #1
    1. The problem statement, all variables and given/known data

    Find the integral to the following expression

    [itex]\int\frac{1}{x-1}dx[/itex]


    2. Relevant equations

    [itex]\int udv = uv - \int vdu[/itex]

    [itex]\int\frac{1}{x}dx = ln(x) [/itex]

    3. The attempt at a solution

    Given the information above, would a correct answer to this problem be ln(x-1) ? If not, what am I doing wrong ?

    thanks
     
  2. jcsd
  3. Sep 13, 2011 #2

    lanedance

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    how non need to use int by parts a simple substitution will do, ln(x-1) is correct

    if you don't show what you doing (your steps it woudl eb hard to tell what was going wrong as well ;)
     
  4. Sep 13, 2011 #3

    oh i just realised if you use substitution of variable u=x-1, it makes it easy to show that integral of 1/(x-1) is ln(x-1) :)
     
    Last edited: Sep 13, 2011
  5. Sep 13, 2011 #4

    lanedance

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    good guess, is that all you need?
     
  6. Sep 13, 2011 #5
    You can attack the problem as follows

    let F(x)=[itex]\int\frac{1}{x-1}dx[/itex]


    From then on make substitution x=x+1

    which yields

    F(x+1)=[itex]\int\frac{1}{x}dx[/itex]

    Find out this integral which is quite obvious ( do not forget to add a constant to it )

    since what you found is F(x+1) now re-substitute x=x-1
     
  7. Sep 13, 2011 #6

    yeah pretty much. i worked it out myself using the substituiton and it came out fine
     
  8. Sep 13, 2011 #7

    Hootenanny

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    :grumpy:
     
  9. Sep 13, 2011 #8

    lanedance

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    agree, would always use a different variable to represent the substitution eg. u = x-1, then du = dx
     
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