Improper Integral

1. Sep 24, 2012

Bashyboy

1. The problem statement, all variables and given/known data
$\int \frac{dx}{\sqrt{x^2-4}}$

2. Relevant equations

3. The attempt at a solution

I tried trig-substitution, by realizing that $cot\theta = \frac{4}{\sqrt{x^2-4}}$

and that $-4sin\theta = dx$

My answer, though, found after the substitution and integration, is very different from the books: mine is $- \frac{\sqrt{x^2-4}}{x}$, theirs is $ln|x+\sqrt{x^2-4}|$

How do you account for this variation?

2. Sep 24, 2012

Staff: Mentor

You have mistakes in your substitution. One of the legs in your triangle should be 2, not 4. Also, your equation for dx is incorrect.