evaluate integral or state that it is diverges

Question

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

-----------------------------------
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?

Answer 1

Hi 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).

Answer 2

Using the difference quotient of (x+h)and Evaluate f(x)=(x+y)^3