Integral $\int\frac{\sqrt{4x^2-1}}{x^3}dx$ using trig identity substitution! 
$$\int \frac{\sqrt{4x^2-1}}{x^3}\ dx$$

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 $\sin(2\theta)=2\sin(\theta)\cos(\theta).$

Setting
$x=\frac{\sqrt{\theta^{2}+1}}{2}$
$dx=\frac{\theta}{2\sqrt{\theta^{2}+1}}$,
and the funcion to integrate is:
$\frac{4\theta^{2}}{(\theta^{2}+1)^{2}}$,
whose integral is:
$2. atan(\theta)-\frac{2\theta}{\theta^{2}+1}$.
Substituting the value of
$\theta=\sqrt{4x^{2}-1}$,
we get:
$I=2. atan(\sqrt{4x^{2}-1})-\frac{\sqrt{4x^{2}-1}}{2x^{2}}$.
