we try a tan sub relying on the ID:
#tan^2 psi + 1 = sec^2 psi quad star#
so here let #x^2 = 4 tan^2 z, x = 2 tan z, dx = 2 sec^2 z \ dz#
the integration becomes
#int sqrt(x^2+4) \ dx#
#= int sqrt(4 tan^2 z +4) \ 2 sec^2 z \ dz#
#= int 2 sec z \ 2 sec^2 z \ dz#
#I = 4 int sec^3 z \ dz qquad triangle#
and maybe re-use that ID in #star# again....
#= 4 int sec z ( tan^2 z + 1) \ dz#
#I = 4 int color{red}{tan^2 z sec z} + sec z \ dz qquad square#
we can have a crack at the red bit using IBP
#int tan^2 z sec z \ dz = int tan z (sec z tan z)\ dz#
#= int tan z d/dz (sec z) \ dz#
by IBP
#=sec z tan z - int d/dz (tan z) sec z \ dz#
#=sec z tan z - int sec^3 z \ dz#
#=sec z tan z - I/4# from #triangle#
so re-stating #square#
#I = 4 int color{red}{tan^2 z sec z} \ dz + 4 int\ sec z \ dz#
#= 4(sec z tan z - I/4) + 4 int\ sec z \ dz #
#I = 2 sec z tan z+ 2 int\ sec z \ dz #
and using the standard integral for #sec z#
#I= 2 sec z tan z+ 2 ln (tan z + sec z) + C#
recap
#tan z = x/2#
#sec z = sqrt(x^2/4 + 1)#
#I = 2 x/2 sqrt(x^2/4 + 1)+ 2 ln (x/2 + sqrt(x^2/4 + 1)) + C#
#= x/2 sqrt(x^2 + 4)+ 2 ln (1/2(x + sqrt(x^2 + 4))) + C#
