1. Definite Integral

I am taking the integral of a function that contains a discontinuity. So in order to do that, I first take the integral from 0 to 1, and from 1 to 2 and add the two integrals. Here's my problem: the function, once integrated, is ln|(x-1)|+x. So I have [ ln|(1-1)|+1 ] - [ ln|(0-1)|+0 .

ln(0) is - infinity? How do I evaluate the definite integral?

2. Originally Posted by onlinemathstudent
I am taking the integral of a function that contains a discontinuity. So in order to do that, I first take the integral from 0 to 1, and from 1 to 2 and add the two integrals. Here's my problem: the function, once integrated, is ln|(x-1)|+x. So I have [ ln|(1-1)|+1 ] - [ ln|(0-1)|+0 .

ln(0) is - infinity? How do I evaluate the definite integral?
What is the integral you are trying to integrate?

3. f(x) = (x^2+x) / (x^2 -1)

4. Originally Posted by onlinemathstudent
I am taking the integral of a function that contains a discontinuity. So in order to do that, I first take the integral from 0 to 1, and from 1 to 2 and add the two integrals. Here's my problem: the function, once integrated, is ln|(x-1)|+x. So I have [ ln|(1-1)|+1 ] - [ ln|(0-1)|+0 .

ln(0) is - infinity? How do I evaluate the definite integral?
is this the original integral?

$\displaystyle \displaystyle \int_0^1 \frac{1}{x-1} + 1 \, dx$

if so, it is an improper integral that diverges ... in other words, the following limit does not exist.

$\displaystyle \displaystyle \lim_{b \to 1^-} \int_0^b \frac{1}{x-1} + 1 \, dx$

5. Here's the whole problem:

6. Ok. After a little research into the matter, I see that there's no way to do this. It is an INFINITE discontinuity and therefore cannot be integrated. Had it been a JUMP discontinuity, then I cou;d have done the two integrals.