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March 5, 2015

March 5, 2015

Posted by **Anonymous** on Monday, February 11, 2008 at 3:41pm.

integral -oo, -2 [2/(x^2-1)] dx

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integral -oo, -2 [2/(x^2-1)] dx

Through partial fractions, I came up with lim [ln(x-1)-ln(x+1)] b, -2

b->-oo

I get (ln(3)-0)-(oo-oo)). The answer in the back of the book is ln(3). What am I doing wrong?

- calculus -
**H**, Monday, February 11, 2008 at 4:25pmHi there,

lim [ln(x-1)-ln(x+1)] b, -2 cannot get (ln(3)-0)-(oo-oo)) because ln(-00) has no definition. We should make it ln(x-1)-ln(x+1)=ln[(x-1)/(x+1)],then lim[ln[(x-1)/(x+1)]]b,-2 b->-00 is ln(3).

- calculus -
**Keith**, Monday, February 11, 2008 at 5:38pmUsing the difference quotient of (x+h)and Evaluate f(x)=(x+y)^3

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