Here,
#I=intsqrt(4x^2-1)/x^2dx#
Let,
#2x=secu=>dx=1/2secutanudu#
#:.4x^2=sec^2u and x^2=1/4sec^2u#
#I=intsqrt(sec^2u-1)/(1/4sec^2u)xx1/2secutanudu#
#=4/2inttanu/sec^2uxxsecutanudu#
#=2inttan^2u/secu du#
#=2int(sec^2u-1)/secu du#
#=2int(secu-1/secu)du#
#=2int(secu-cosu)du#
#=2[ln|secu+tanu|-sinu]+c#
#=2ln|secu+sqrt(sec^2u-1)|-2(sinu/cosuxxcosu)+c#
#=2ln|secu+sqrt(sec^2u-1)|-2tanuxx1/secu+c#
#=2ln|secu+sqrt(sec^2u-1)|-(2sqrt(sec^2u-1))/secu+c#
Subst. back, #secu=2x#, we get
#I=2ln|2x+sqrt(4x^2-1)|-(2sqrt(4x^2-1))/(2x)+c#
#I=2ln|2x+sqrt(4x^2-1)|-sqrt(4x^2-1)/x+c#