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

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Answer 1 · 2018-04-30T17:44:42

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

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

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

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

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