How do I do this simple looking integral?

1. Dec 8, 2007

frasifrasi

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. Dec 8, 2007

EnumaElish

You are on the right path; what is the antiderivative of 1/(x+1)?

3. Dec 8, 2007

frasifrasi

If I get the antiderivatives:

x] - [ln(u)] ? -- Am i supposed to substitute u for (x+1) before proceeding?

4. Dec 8, 2007

rocomath

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. Dec 8, 2007

frasifrasi

So, the lim n -->-1 of ln(n) is infinity?

6. Dec 8, 2007

Feldoh

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. Dec 8, 2007

frasifrasi

so, I do have to sub x+1 back I guess...

8. Dec 8, 2007

HallsofIvy

Staff Emeritus
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. Dec 8, 2007

frasifrasi

but then the integral wouldn't diverge, would it?

10. Dec 9, 2007

Feldoh

ln(u), where u=0 will definitely diverge