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## Homework Statement

What is the integral of sqrt(x^2-1)/x dx?

## Homework Equations

## The Attempt at a Solution

∫ √(x^2 - 1) dx / x

let x = sec u: u = sec^-1(x) and tan u = √(x^2 - 1)

dx = sec u tan u du

now the integral becomes

∫ √sec^2(u) - 1) sec u tan u du / sec u

= ∫ tan u tan u du

= ∫ tan^2(u) du

= ∫ (sec^2(u) - 1) du

= ∫ (sec^2(u) du - ∫ du

= tan u - u + c

back substitute u = sec^-1(x) and tan u = √(x^2 - 1)

= √(x^2 - 1) - sec^-1(x) + c

However, the correct answer should be arccot(sqrt(x^2-1))+sqrt(x^2-1) + c. Can someone help me find what went wrong?