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#