How do you find the integral of #1/sqrt(x^2 -4)#?

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Answer 1 · 2015-07-01T18:10:29

One substitution here

#int 1/sqrt(x^2-4)dx = int 1/(sqrt(4(1/4x^2-1)))dx = 1/2int1/(sqrt(1/4x^2-1))dx#

#u = 1/2x#

#du = 1/2dx#

#u^2 = 1/4x^2#

integral become :

#int 1/sqrt(u^2-1)du#

we can recognize the derivative of #arccosh(u)#

#[arccosh(u)]+C#

don't forget that #u = 1/2x#

So we have #[arccosh(1/2x)]+C#