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 $[arccosh(1/2x)]+C$

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