How do you evaluate the integral #int sqrt(4x^2-1)/x^2#?

1 Answer
Apr 30, 2018

#I=2ln|2x+sqrt(4x^2-1)|-sqrt(4x^2-1)/x+c#

Explanation:

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#