To find the integral

$\int \frac{\sqrt{x^2 - 9}}{x^3} \, dx$

we'll proceed with a trigonometric substitution. Since we have a term of the form $x^2 - 9$ under the square root, let's use the substitution $x = 3 \sec \theta$, which will simplify $x^2 - 9$ to a perfect square.

1. Substitute $x = 3 \sec \theta$:

   • Then $dx = 3 \sec \theta \tan \theta \, d\theta$.
   • Substituting $x = 3 \sec \theta$ gives: $x^2 = 9 \sec^2 \theta \Rightarrow x^2 - 9 = 9 \sec^2 \theta - 9 = 9 (\sec^2 \theta - 1) = 9 \tan^2 \theta.$
   • Therefore, $\sqrt{x^2 - 9} = 3 \tan \theta$.

2. Rewrite the integral in terms of $\theta$: Substituting into the integral, we get $\int \frac{\sqrt{x^2 - 9}}{x^3} \, dx = \int \frac{3 \tan \theta}{(3 \sec \theta)^3} \cdot 3 \sec \theta \tan \theta \, d\theta.$

3. Simplify the expression: Simplifying each term: $= \int \frac{3 \tan \theta}{27 \sec^3 \theta} \cdot 3 \sec \theta \tan \theta \, d\theta.$ $= \int \frac{9 \tan^2 \theta}{27 \sec^2 \theta} \, d\theta = \int \frac{\tan^2 \theta}{3 \sec^2 \theta} \, d\theta.$ Now, since $\sec^2 \theta = 1 + \tan^2 \theta$, rewrite $\sec^2 \theta$ as follows: $= \int \frac{\tan^2 \theta}{3 (1 + \tan^2 \theta)} \, d\theta.$

4. Use the substitution $u = \tan \theta$: Let $u = \tan \theta$, then $du = \sec^2 \theta \, d\theta$, and the integral becomes $\int \frac{u^2}{3 (1 + u^2)} \cdot \frac{du}{\sec^2 \theta}.$