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How do you integrate (x)/(x-1)

  1. Aug 29, 2011 #1
    1. The problem statement, all variables and given/known data
    How do you integrate S x/(x-1) dx


    2. Relevant equations

    IBP

    3. The attempt at a solution
    I tried using Integration by Parts with u= x/(x-1) and dv = dx and that did not work out. Then I tried using u= 1/(x-1) and dv = xdx but that did not work either.

    A google search said "Just use the substitution u = x+1, then replace dx with du and you get u+1/u = 1 + 1/u which you can integrate to give u + ln u, thus = x-1 + ln(x-1)". But I am pretty sure that you can't just add +1 to the numerator like that.

    I also know that it can be done using tables but is there a way not to use tables? Did I do my integration wrong?

    Thanks
     
  2. jcsd
  3. Aug 29, 2011 #2

    micromass

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    You don't add +1 to the numerator. We do the substitution u=x-1. What happens if you do that substitution??
     
  4. Aug 29, 2011 #3
    Hint Hint:
    [itex]\frac{1}{x-1}[/itex]+[itex]\frac{x-1}{x-1}[/itex]=[itex]\frac{x}{x-1}[/itex]
     
  5. Aug 29, 2011 #4
    You can even do it without substitution by changing the equation to (x)/(x-1) = (x-1+1)/(x-1) = 1+1/(x-1), then integrating to get x+ln(x-1)+C.

    Wow! All 3 posts within a minute!
     
  6. Aug 29, 2011 #5
    Perhaps this will make it clearer,

    Suppose [itex]u=x-1[/itex]. Then clearly [itex]du=dx[/itex]. But notice also that [itex]u+1=x[/itex] by the first equation. Therefore, the integral becomes,

    [itex]\int \frac{u+1}{u}du[/itex]

    Now, it is trivial. :)
     
  7. Aug 29, 2011 #6
    We're all just so stoked about integration, we can't resist. If only all students shared our passion for integration.... ;)
     
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