# calc

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

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1. You need to include absolute value signs in the argument of the logarithm:

ln|x-1| from 0-1 and
ln|x-1| from 1-2

What you find is that both the limits diverge logarithmically, so the integral doesn't exist. However, if you add both the terms together and take the limit in one go, then the divergent terms cancel. This is called the Cauchy principal value of the integral.

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