using the identity cosh^2 t - sinh^2 t =1 i'd got with a sub 4x^2 = sinh^2 t ie 2x = sinh t , so 2 \ dx = cosh t \ dt
that leaves:
int \ sqrt(4x² + 1) \ dx = int \ sqrt{sinh^2 t + 1} \ 1/2 cosh t \ dt
= 1/2 int \ cosh^2 t \ dt
we use the following hyperbolic identities
cosh^2 t = 1/2 (cosh 2t + 1)
so the integral becomes
= 1/4 int \ cosh 2t + 1 \ dt
= 1/4 (1/2 sinh 2t + t) + C
within this t = sinh^{-1} 2x from above
for sinh 2t, look at the above identities again.
sinh 2t = 2 sinh t cosh t
= 2 sinh t sqrt{1 + sinh^2 t)
= 4x sqrt{1+ 4x^2}
so
1/4 (1/2 sinh 2t + t) + C
= 1/4 (1/2 * 4x sqrt{1+ 4x^2} + sinh^{-1} 2x) + C
= 1/4 ( 2x sqrt{1+ 4x^2})+ sinh^{-1} 2x) + C
