How to integrate $\int_{3\sqrt{2}}^6 1/\big(t^3\sqrt{t^2-9}\big)\;dt$ Here's the full problem

$$\int_{3\sqrt{2}}^6\frac{1}{t^3\sqrt{t^2-9}}\;dt.$$

The problem for me is what to do with the $-9$. I know that $1 - \sec^2(t) = \tan^2(t)$, so I'm trying to substitute $9\sec(x)$ for $t$ to get rid of the square root. So that works, but when I recalculate the limits, I end up with numbers that don't make any sense. I'm certain that if I were doing this correctly I would end up with nice numbers for the new limits, so something must be wrong.
$$
\begin{align}
t &= 9\sec(x)\\
a &= 9\sec^{-1}(3\sqrt{2})\\
b &= 9\sec^{-1}(6)
\end{align}
$$
These answers should be $\displaystyle\frac{x}{\pi}$ or something to that effect. Sorry if this makes no sense. Is there a way for me to evaluate the arcsec function without using a calculator so I get values I can understand? Would I use the unit circle? How can these limits be in radians but not have $\pi$?

$$
\int_{3\sqrt{2}}^6\frac{1}{t^3\sqrt{t^2-9}}\;dt = \frac12 \int_{3\sqrt{2}}^6\frac{2t\,dt}{t^4\sqrt{t^2-9}} = \frac12 \int_{18}^{36} \frac{du}{u^2\sqrt{u-9}}.
$$
Now let $w=\sqrt{u-9}$, so that $w^2 = u-9$ and $2w\,dw=du$, and $u = w^2+9$.  Then the integral is
$$
\frac12 \int_3^{3\sqrt{3}} \frac{2w\,dw}{(w^2+9)^2 w} = \int_3^{3\sqrt{3}} \frac{dw}{(w^2+9)^2} = \int_3^{3\sqrt{3}} \frac{Aw+B}{w^2+9} + \frac{Cw+D}{(w^2+9)^2} \,dw
$$
You need to do some algebra to find $A$, $B$, $C$, and $D$.  Then you have
$$
\int \frac{w}{w^2+9}\,dw = \frac12\log(w^2+9)+\text{constant}
$$
and
$$
\int \frac{1}{w^2+9}\,dw = \frac13 \arctan\frac w3 + \text{constant}.
$$
The term involving $C$ can be done by the same method as the one that yields a logarithm above, but you don't get a logarithm or anything transcendental.
Finally, we have
$$
\int \frac{dw}{(w^2+9)^2}.
$$
At this point I'd start with $w=3\tan\theta$.
