# calculus

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how would you do this improper integral

1/(x-1)
from 0 to 2

this is improper at one, so I split it up into two integrals
ln(x-1) from 0-1 and
ln(x-1) from 1-2

I then did for the first one the (lim t->1(-) of ln(t-1))-(ln(0-1))

and then the same thing for the second part
I didn't know if this was right though, or what the answer would be

• calculus -

You used a correct procedure, but both integrals you broke it up into involve the log of zero or a negative number, both of which are "improper". I don't see how one can get a finite integral this way.
Here is another way that might work:
Define the integral of the sum of two limits. One is the integral from 0 to 1-a and the other is the integral from 1+a to 2, as value of a goes to zero. Consider a as always positive. You may find that two two terms always cancel each other whatever a is, so that the limit will be zero.

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