[SOLVED] trig substitution integral

Evaluate the integral:

$\displaystyle \int\limits^{2}_{\sqrt2} \frac{dx}{x^3\sqrt{x^2-1}}$

Here is what I am getting in my first step:

$\displaystyle \int\limits^{2}_{\sqrt2} \frac{sec\theta tan\theta}{(sec\theta)^3 \cdot (sec\theta - 1)} d\theta $

Is that the right thing to do with the $\displaystyle x^3$ in the denominator?

After that, assuming it is correct, I simplified the denominator:

$\displaystyle \int\limits^{2}_{\sqrt2} \frac{sec\theta tan\theta}{(sec\theta)^4 - (sec\theta)^3} d\theta$

Is this correct to this point? If yes, where do I go from here?

THanks!!