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How do I do this simple looking integral?

  • Thread starter frasifrasi
  • Start date
  • #1
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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."
 

Answers and Replies

  • #2
EnumaElish
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You are on the right path; what is the antiderivative of 1/(x+1)?
 
  • #3
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If I get the antiderivatives:

x] - [ln(u)] ? -- Am i supposed to substitute u for (x+1) before proceeding?
 
  • #4
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If I get the antiderivatives:

x] - [ln(u)] ? -- Am i supposed to substitute u for (x+1) before proceeding?
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.
 
  • #5
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So, the lim n -->-1 of ln(n) is infinity?
 
  • #6
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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.
 
  • #7
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so, I do have to sub x+1 back I guess...
 
  • #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.
 
  • #9
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but then the integral wouldn't diverge, would it?
 
  • #10
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ln(u), where u=0 will definitely diverge
 

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