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How do I do this simple looking integral?

  1. Dec 8, 2007 #1
    How do I do this simple looking integral?????

    Ok, we are supposed to evaluate

    int from -1 to 0 of x/(x+1) dx
    ---> my first step was to write (x+1) -1/(x+1) to form the two integrals: of 1 and 1/(x+1)


    --> is this the right way? The answer is "it diverges."
     
  2. jcsd
  3. Dec 8, 2007 #2

    EnumaElish

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    You are on the right path; what is the antiderivative of 1/(x+1)?
     
  4. Dec 8, 2007 #3
    If I get the antiderivatives:

    x] - [ln(u)] ? -- Am i supposed to substitute u for (x+1) before proceeding?
     
  5. Dec 8, 2007 #4
    you don't have to, this problem you can pretty much eyeball.

    but if you had like 2x+1 in the denominator, then you would need worry about constants.
     
  6. Dec 8, 2007 #5
    So, the lim n -->-1 of ln(n) is infinity?
     
  7. Dec 8, 2007 #6
    If we let u=x+1 then du=dx

    So then x=u-1 and dx=du

    It's a way similar to what you did, and since ln(0) is undef. it diverges.
     
  8. Dec 8, 2007 #7
    so, I do have to sub x+1 back I guess...
     
  9. Dec 8, 2007 #8

    HallsofIvy

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    Either that or change the limits of integration as you go. If you make the substitution u= x+1 and your limits of integration, with respect to x, were x= -1 and x= 0, then the limits of integration become u= -1+1= 0 and u= 0+ 1= 1.
     
  10. Dec 8, 2007 #9
    but then the integral wouldn't diverge, would it?
     
  11. Dec 9, 2007 #10
    ln(u), where u=0 will definitely diverge
     
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