I would go with a trig sub too. But, if you feel brave, you could try another approach.
Let \(\displaystyle u=\sqrt{x^{2}+9}, \;\ du=\frac{x}{\sqrt{x^{2}+9}}dx\)
That way you can play around with the u and x and note that \(\displaystyle x^{2}=u^{2}-9\)
So, rewriting as \(\displaystyle \frac{3x}{x^{2}\sqrt{x^{2}+9}}dx\)
it becomes:
\(\displaystyle 3\int\frac{1}{u^{2}-9}du\)
Note the difference of two squares. It can be written as \(\displaystyle \frac{3}{(u+3)(u-3)}=\frac{1}{2(u-3)}-\frac{1}{2(u+3)}\), and it is a matter of an LN.
