$$\int \frac{\sqrt{4x^2-1}}{x^3}\ dx$$
So, make the substitution
$ x = \sqrt{a \sec \theta}$, which simplifies to $a \tan \theta$.
$2x = \sqrt{1} \sec \theta$,
$ d\theta = \dfrac{\sqrt{1}\sec\theta\tan\theta}{2}$
$\int \dfrac{\sqrt{1}\tan\theta}{(\sqrt{1}\sec\theta)^3} d\theta$
Am I making the correct substitutions here? Substituting $d\theta$ a quantity of $(\sqrt{1}\sec\theta)$ will cancel from the denominator. Somewhere along the line I need to use the identity $\sin(2\theta)=2\sin(\theta)\cos(\theta).$