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
