Question

how do i evaluate the improper integral of (1/ x^2 - 6x +8) from 1 to 3. I know there is a discontinuity at x=2 so we split the integral. I just have problem evaluating the limit. Any help would be greatly appreciated :D

Answer 1

i don't think it is going to work

Answer 2

you can integrate via partial fractions, and you will get the log

Answer 3

\[\frac{1}{2}\int\frac{dx}{x-4}-\frac{1}{2}\int\frac{dx}{x-2}\]

Answer 4

anti derivative is the log, and
\[\lim_{x\to 2}\ln(x-2)\] does not exist, either from the left or the right

Answer 5

i.e. the integral does not converge

Answer 6

yea thanks a lot. and i guess u cant even evaluate ln(t-4) as t approaches 2 since it willl be ln (-2) which is undefined

Answer 7

This is such a common error.
I think we need an evangelist to go about preaching the Order of Operations.

(1/ x^2 - 6x +8) is NOT whaat you intend.

(1/ (x^2 - 6x +8)) The inner parentheses are NOT optional.

Answer 8

Sorry that is what I meant..So the integral is divergent since ln(-2) is undefined?

Answer 9

I know that is what you meant. Obviously, so did satellite73. But, it is NOT what you wrote. Please write what you mean. Learning to communicate clearly is quite a large portion of the battle - maybe not half, but it's a good piece.

ln(-2)? No. That appears nowhere in this problem. The integral fails to converge around x = 2.

Answer 10

What I mean is that when u evaluate the integral u get ln(x-4) as one of the terms as x approaches 2+. So is it undefined?

Answer 11

Sorry that is what I meant..So the integral is divergent since ln(-2) is undefined?

Answer 12

Still no. ln(-2) has nothing to do with anything, just like ln(0). These are meaningless terms. Don't use them.

This integral fails to converge around x = 2 and around x = 4. Why do you have to say anything else?