The integral is: integral(2/(t^2-1),t,2,infinity)

The integral of two over the quantity t squared minus one, from 2 to infinity.

I used trig substitution and ended up in the end with:

2*[sqrt(t^2-1)]+c for the integral, which is not what my calculator gets.

2 times the limit as B goes to infinity of [ln((b^2-1)^(1/2)-ln((3)^(1/2))]

The answer should be ln(3), which is 2*ln((3)^(1/2)), so I thought maybe I'm on the right track, but isn't the limit of as b approaches infinity of

ln((b^2-1)^(1/2), inifinity?

Here are my steps.

t=secu

dt=secutanudu

t^2-1=tan^2u

2*integral((cscdu)/tan^2u)

=2*integral(ln(abs(sinu)/abs(cosu+1)))

sinu= sqrt(t^2-1)/t

cosu=1/t

...

Is there something I am doing wrong.