How do I do this simple looking integral?

  • Thread starter frasifrasi
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In summary, to evaluate the given integral, we first write it in the form of two integrals: one for 1 and one for 1/(x+1). Then, we can find the antiderivative for 1/(x+1) by using the substitution u=x+1. However, since ln(0) is undefined, the integral diverges. We can also make the substitution u=x+1 in the limits of integration to simplify the integration process. Overall, the integral diverges due to the undefined value of ln(0).
  • #1
frasifrasi
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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."
 
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  • #2
You are on the right path; what is the antiderivative of 1/(x+1)?
 
  • #3
If I get the antiderivatives:

x] - [ln(u)] ? -- Am i supposed to substitute u for (x+1) before proceeding?
 
  • #4
frasifrasi said:
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
So, the lim n -->-1 of ln(n) is infinity?
 
  • #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.
 
  • #7
so, I do have to sub x+1 back I guess...
 
  • #8
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
but then the integral wouldn't diverge, would it?
 
  • #10
ln(u), where u=0 will definitely diverge
 

FAQ: How do I do this simple looking integral?

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