How do you use substitution to integrate $1/sqrt(x^2 -4)$ ?

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Answer 1 · 2015-06-23T17:39:02

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

Let x= 2sec u. so that dx= 2sec u tan u du. The given integral becomes $int 1/sqrt (4sec^2 u -4) 2 sec u tan u du$

= $int secu du$

= ln lsecu +tan u I +C

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

How do you use substitution to integrate 1/sqrt(x^2 -4) 1/sqrt(x^2 -4)?