Why Doesn't This Integral $\int \frac{\sqrt{x^2 - 9}}{x^2} \ dx$ Work? I am trying to solve this integral, which is incorrect compared to Wolfram|Alpha. Why doesn't my method work?

Find $\int \frac{\sqrt{x^2 - 9}}{x^2} \ dx$

Side work:

$\sin{\theta} = \frac{3}{x}$
$x = \frac{3}{\sin{\theta}} = 3 \csc{\theta}$
$dx = -3 \csc{\theta}\cot{\theta} \ d\theta$

$-\int \frac{\sqrt{9\csc^2{\theta} - 9}}{27 \csc^3{\theta}} \cdot 3 \csc{\theta}\tan{\theta} \ d\theta$
$-\int \frac{3 \cdot \sqrt{\csc^2{\theta} - 1}}{27 \csc^3{\theta}} \cdot 3 \csc{\theta}\cot{\theta} \ d\theta$
$-\int \frac{\cot{\theta}\csc{\theta}\cot{\theta}}{3 \csc^3{\theta}} \ d\theta$
$-\frac{1}{3} \int \cos^2{\theta} \ d\theta$
$-\frac{1}{3} \int \frac{1}{2} \left(1 + \cos{\left(2\theta\right)}\right) \ d\theta$
$-\frac{1}{6} \int 1 + \cos{\left(2\theta\right)} \ d\theta$
$-\frac{1}{6} \int \ d\theta + \frac{1}{6} \int \cos{\left(2\theta\right)} \ d\theta$

Sidework:
$u = 2\theta$
$du = 2 \ d\theta$

$-\frac{1}{6} \int \ d\theta + \frac{1}{3} \int \cos{u} \ du$
$-\frac{\theta}{6} + \frac{1}{3} \cdot \sin{u} + C$
$-\frac{\theta}{6} + \frac{1}{3} \cdot \sin{2\theta} + C$

Sidework:
$\sin{\left(2\theta\right)} = 2 \sin{\theta} \cos{\theta}$
$\sin{\theta} = \frac{3}{x}$
$\cos{\theta} = \frac{\sqrt{x^2 - 9}}{x}$
$\theta = \sin^{-1}{\left(\frac{3}{x}\right)}$

$-\frac{\sin^{-1}{\left(\frac{3}{x}\right)}}{6} + \frac{2}{3} \cdot \frac{3}{x} \cdot \frac{\sqrt{x^2 - 9}}{x} + C$
$-\frac{\sin^{-1}{\left(\frac{3}{x}\right)}}{6} + \frac{2\sqrt{x^2 - 9}}{x^2} + C$

Thank you for your time.

Added: It has been pointed out that there may be some ambiguity about the integral the OP is trying to find.  The introduction has an $x^2$ at the bottom, but in the solution the OP seems to be using $x^3$. That is much easier.  The substitution $x=3\sec\theta$ transforms the integrand into a constant times $\frac{\tan^2\theta}{\sec^2\theta}$, which is just $\sin^2\theta$.
Original post: If we are going to use a trigonometric substitution, I would let $x=3\sec\theta$. (Secant is a bit more familiar than cosecant. You are therefore less likely to differentiate $\sec\theta$ incorrectly; incorrect differentiation of $\csc\theta$ spoiled our integration.) 
Then $dx=3\sec\theta\tan\theta\,d\theta$. Plug in. From now on, I will forget about constants. So everything is a little wrong, but only by a constant factor.  
On top we get $(\tan\theta)(\sec\theta\tan\theta)$. On the bottom we get $\sec^2\theta$. 
So we are interested in 
$$\frac{\tan^2\theta\sec\theta}{\sec^2\theta}.$$
Put everything in terms of sines and cosines. We get $\dfrac{\sin^2\theta}{\cos\theta}$.  
There is a trick that one can use when there is an odd power of sine or cosine. In this case, multiply top and bottom by $\cos\theta$, and replace the $\cos^2\theta$ now at the bottom by $1-\sin^2\theta$.
We have arrived at 
$$\int \frac{\cos\theta \,\sin^2\theta}{1-\sin^2\theta}\,d\theta.$$
Make the substitution $u=\sin\theta$.
We arrive at
$$\int \frac{u^2}{1-u^2}\,du.$$
The integrand is a rational function. It can be rewritten as 
$$\frac{1}{1-u^2}-1.$$
But (partial fractions) we find that $\dfrac{1}{1-u^2}=\dfrac{1}{2}\left(\dfrac{1}{1+u}+\dfrac{1}{1-u}\right)$. 
Remark: As usual, there are many other approaches. For example, we can change $\frac{\tan^2\theta}{\sec\theta}$ to $\frac{\sec^2\theta-1}{\sec\theta}$, and then to $\sec\theta -\cos\theta$.  If we happen to know the integral of $\sec\theta$, we are essentially finished.  More strikingly, if we guess somehow that $F(x)$ is an antiderivative, then the magic substitution $u=F(x)$ solves the problem.
