#int 1/(t^2-9)^(1/2)dt#
to use the ID #tan^2 + 1 = sec^2#
let #t^2 = 9 sec^2 theta, t = 3 sec theta, dt = 3 sec theta tan theta d theta qquad triangle#
so the integration becomes
#int 1/(9 sec^2 theta -9)^(1/2) \ 3 sec theta tan theta d theta#
#= int 1/(3 tan theta) \ 3 sec theta tan theta d theta#
#= int sec theta d theta#
which is a standard integral
#= ln ( sec theta + tan theta) + C qquad star#
as in
# int \ d/(d theta) (ln ( sec theta + tan theta) + C) \ d theta#
#= int \ (sec theta tan theta + sec^2 theta)/( sec theta + tan theta) \ d theta#
#= int \ sec theta \ d theta#
reversing the sub in #triangle#, #star# becomes
#= ln ( t/3 + sqrt ((t/3)^2 - 1)) + C#
