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

What you should end with is

$\sec\theta = \dfrac x3\quad$ and $\quad\tan \theta = \dfrac{\sqrt{ x^2 - 9}}{3}$.

Then you have $$\begin{align} \log \Big| \frac 13\left(x + \sqrt{x^2 - 9}\right)\Big| + C & = \log|x +\sqrt{x^2 - 9}| -\log 3 + C \\ \\ & = \log|x + \sqrt{x^2 - 9}| + C'\end{align}$$