To solve the integral 

\[
\int \frac{\sqrt{x^2 - 49}}{x^3} \, dx
\]

we can use a trigonometric substitution, specifically by recognizing that the expression \(x^2 - 49\) suggests a substitution related to the identity \( \sec^2 \theta - 1 = \tan^2 \theta \).

### Step-by-step solution:

1. **Substitution**:  
   Let \( x = 7\sec\theta \), so that \( x^2 - 49 = 49(\sec^2 \theta - 1) = 49\tan^2\theta \), and \( dx = 7\sec\theta\tan\theta d\theta \).

2. **Rewrite the integral**:
   \[
   \int \frac{\sqrt{x^2 - 49}}{x^3} \, dx = \int \frac{\sqrt{49\tan^2\theta}}{(7\sec\theta)^3} \cdot 7\sec\theta\tan\theta \, d\theta
   \]
   Simplify the expression:
   \[
   = \int \frac{7\tan\theta}{343\sec^3\theta} \cdot 7\sec\theta\tan\theta \, d\theta
   \]
   \[
   = \int \frac{49\tan^2\theta}{343\sec^2\theta} \, d\theta
   \]
   \[
   = \frac{49}{343} \int \frac{\tan^2\theta}{\sec^2\theta} \, d\theta
   \]
   Simplify further:
   \[
   = \frac{1}{7} \int \sin^2\theta \, d\theta
   \]
   
3. **Use trigonometric identity**:  
   Recall that \( \sin^2\theta = \frac{1 - \cos(2\theta)}{2} \), so:
   \[
   = \frac{1}{7} \int \frac{1 - \cos(2\theta)}{2} \, d\theta
   \]
   \[
   = \frac{1}{14} \int (1 - \cos(2\theta)) \, d\theta
   \]

4. **Integrate**:
   \[
   = \frac{1}{14} \left( \theta - \frac{1}{2} \sin(2\theta) \right) + C
   \]

5. **Substitute back in terms of \(x\)**:
   Recall that \( x = 7\sec\theta \), so \( \sec\theta = \frac{x}{7} \), and therefore \( \theta = \sec^{-1}\left(\frac{x}{7}\right) \). Also, \( \sin(2\theta) = 2\tan\theta \sec\theta = 2\frac{\sqrt{x^2 - 49}}{x} \).

Thus, the integral becomes:
\[
\frac{1}{14} \left( \sec^{-1}\left(\frac{x}{7}\right) - \frac{\sqrt{x^2 - 49}}{x} \right) + C
\]

This is the final solution.

---

Do you want more details on any of the steps or have any other questions?

Here are 5 related questions for further exploration:
1. How would you solve the integral \( \int \frac{1}{x^3} \, dx \)?
2. Can you derive the trigonometric identity \( \sec^2 \theta - 1 = \tan^2 \theta \)?
3. What happens if we use a different substitution for the integral?
4. How can you calculate \( \sec^{-1}(x) \) in terms of logarithmic functions?
5. What is the geometrical interpretation of this integral?

**Tip**: When faced with square roots of quadratic terms like \(x^2 - a^2\), try using trigonometric substitutions, as these often simplify integrals significantly.

This detailed solution shows how to use trigonometric substitution to solve the integral of sqrt(x^2 - 49) divided by x^3. Key steps include the substitution x = 7sec(θ) and applying the Pythagorean identity sec^2(θ) - 1 = tan^2(θ). Suitable for university-level calculus students.

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