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#
