$\displaystyle \int \frac{\1}{(x^2*\sqrt{x^2-9}}dx$

$\displaystyle 3sec\theta = x$

$\displaystyle \int \frac{3sec\theta tan\theta}{sec^2\theta\sqrt{9tan^2\theta}}dx$

$\displaystyle \int sec\theta dx = sin\theta+c$

since secant is hypotenuse over adjacent sine theta must be

$\displaystyle \frac{\sqrt{x^2-9}}{x}$

the book has a nine in the denominator and I can't figure out where it comes from.