How do you integrate $2/sqrt(x^2-4)$ ?

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Answer 1 · 2018-05-26T15:21:05

The answer is $=2ln(|(x/2)+sqrt(x^2-4)/2|)+C$

Perform the substitution

$x=2secu$ , $=>$ , $dx=2secutanudu$

$sqrt(x^2-4)=sqrt(4sec^2u-4)=2tanu$

The integral is

$I=int(2dx)/sqrt(x^2-4)=2int(2secutanudu)/(2tanu)$

$=2intsecudu$

$=2int(secu(secu+tanu)du)/(secu+tanu)$

Let $v=secu+tanu$

$=>$ , $dv=(secutanu+sec^2u)du$

Therefore,

The integral is

$I=2int(dv)/(v)$

$=2lnv$

$=2ln(secu+tanu)$

$=2ln(|(x/2)+sqrt(x^2-4)/2|)+C$

How do you integrate 2/sqrt(x^2-4)2/sqrt(x^2-4)?