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April 24, 2014

April 24, 2014

Posted by **sarah** on Sunday, February 10, 2008 at 10:54pm.

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 -
**drwls**, Sunday, February 10, 2008 at 11:27pmDid you read my previous answer?

- calculus -
**sarah**, Sunday, February 10, 2008 at 11:39pmyes, but I wasn't sure of the final answer from your answer.

- calculus -
**drwls**, Monday, February 11, 2008 at 2:26amI left the final calculation up to you. If you define the integral as the sum of two parts that each approach within a distance a of x=1, and let a gapproach zero, than the answer will be zero because the two parts will always cancel, not matter how small a is.

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