#=int1/sqrt((t-3)^2+4)dt#
which upon substituting #t-3=x# is the standard integral #int1/sqrt(x^2+a^2)dx = sinh^-1(x/a)# with #a=2#.
If the standard integral is not accessible, substitute #x=2 sinh u#, #dx=2 cosh u du# and use #cosh^2u - sinh^2u=1#:
#=int (2 cosh u)/sqrt(4 sinh^2u+4)du#
#=int 1du#
#=u+C#
#=sinh^-1(x/2)+C#
#=sinh^-1((t-3)/2)+C#.
If the #+4# had been #-4# you would have substituted #x= 2cosh u#.
The final answer can also be expressed using logarithms instead of inverse hyperbolic functions, because #sinh^-1x=ln(x+sqrt(x^2+1))#, not forgetting also that #ln (x/2+sqrt(x^2/4+1))# differs from #ln(x+sqrt(x^2+4))# by a constant which can be merged with the arbitrary constant of integration.
You use hyperbolic trig if the quadratic under the square root has a positive coefficient of #x^2# and circular trig if it is negative.
