Question

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

Answer 1

`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`