Integration by trig substitution - why is my answer wrong?

Attempting integral:

$$-\int \frac{dx}{\sqrt{x^2-9}}$$

Let $x = 3\ sec\ \theta$ so that under the square root we have:

$$\sqrt{9\ sec^2\ \theta - 9}$$

$$\sqrt{9(sec^2\ \theta - 1)}$$

$$\frac{dx}{d\theta} =3\ sec\ \theta\ tan\ \theta$$

The $1/3$ and the $3$ cancel eachother out outside the integral, so we have:

$$-\int \frac{sec\ \theta\ tan\ \theta\ d\theta}{tan\ \theta}$$

The $tan\ \theta$ terms cancel, so we're left to integrate $sec\ \theta$ which is equal to:

$$- \ln\ (tan\ \theta + \ sec\ \theta) + C$$

Since $tan\ \theta = \sqrt{x^2-9}\ $ since it replaced it in the integral, and $sec\ \theta = \frac{x}{3}$, the answer is:

$$-\ ln\ (\sqrt{x^2-9}\ + \frac{x}{3})+C$$

However, using an online calculator, the answer turned out to be:

$$-\ ln\ (\sqrt{x^2-9}\ + x)+C$$

It seemed to come about due to their substitution of $u = \frac{x}{3}$ we led them to get the standard integral of $sec^{-1}x$, which would imply that

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

becomes $$\sqrt{u^2 - 1}$$

And I just don't see how. Can someone explain why what they did is valid and/or where my mistake was?