Trigonometric Substitution 
I am having trouble with this problem even though everything I did seemed right to me since we went over a similar one in my class. I used the method of setting up a triangle, my hypotenuse is $\sqrt{54+9x^2}$ and my sides are $\sqrt{54}$ and $3x$. I got $\tan(t)=3x/\sqrt{54}$ so 
$$x=\sqrt{54} \tan(t) \frac{1}{3}$$
which left 
$$\sec(t) = \frac{\sqrt{54+9x^2}}{\frac{\sqrt{54}}{3}}$$ 
and then 
$$\frac{\sqrt{54}}{3} \sec(t) = \sqrt{54+9x^2}.$$
This left me with a simplified $6 \int \sec^3(t) \, dt$. After using the reduction formula my answer was 
$$3 \tan(t) \sec(t) +3 \ln |\sec(t) + \tan(t)| +C$$ 
and then I plugged back in with my $x$ values. If anyone can help it would be greatly appreciated!

Sorry, I misunderstood your question the first time round.
First, I'm getting 
$$9 \sec \theta \tan \theta + 9 \ln | \sec \theta + \tan \theta | + C $$
as my intermediate answer.  From there, you need to take out all the thetas and put in $x$.  From what you wrote, I'm assuming you see how we get $\tan \theta = x / \sqrt{6}$ and $\sec \theta = \sqrt{(x^2 + 6) / 6}$. (If not, let me know.)  Therefore, we rewrite the above expression as
$$
\\ 9 \cdot \frac{\sqrt{x^2 + 6}}{\sqrt{6}} \cdot \frac{x}{\sqrt{6}} + 9 \ln \left| \frac{\sqrt{x^2 + 6}}{\sqrt{6}} + \frac{x}{\sqrt{6}} \right| + C = \frac{3}{2}x\sqrt{x^2 + 6} + 9 \ln \left| \frac{\sqrt{x^2 + 6} + x}{\sqrt{6}} \right| + C \
$$
A lot of times with these kinds of integrals the trick is with the $\ln$. Remember that $\ln A / B = \ln A - \ln B$; this lets us simplify the expression like so:
$$\frac{3}{2}x\sqrt{x^2 + 6} + 9 \ln \left| \sqrt{x^2 + 6} + x \right| + C$$
Doing this always seems weird to me, but $-9 \ln \sqrt{6}$ is a constant, after all...
