1. **State the problem:** We need to evaluate the integral $$\int \frac{5}{|x| \sqrt{x^2 - 16}} \, dx.$$\n\n2. **Recall the formula and rules:** The integral involves an expression with absolute value and a square root of a quadratic. We note that $$|x| = \begin{cases} x, & x > 0 \\ -x, & x < 0 \end{cases}.$$ Also, $$\sqrt{x^2 - a^2}$$ suggests a hyperbolic substitution or a trigonometric substitution for $$x^2 - a^2$$ form.\n\n3. **Consider the domain:** Since $$\sqrt{x^2 - 16}$$ is real only if $$|x| \geq 4$$, and the denominator has $$|x|$$, we consider $$x > 0$$ and $$x < 0$$ separately.\n\n4. **Case 1: $$x > 0$$:** Then $$|x| = x$$ and the integral becomes $$\int \frac{5}{x \sqrt{x^2 - 16}} \, dx.$$\n\n5. **Substitution:** Let $$x = 4 \sec \theta$$, so $$dx = 4 \sec \theta \tan \theta \, d\theta$$ and $$\sqrt{x^2 - 16} = \sqrt{16 \sec^2 \theta - 16} = 4 \tan \theta.$$\n\n6. **Rewrite the integral:**\n$$\int \frac{5}{4 \sec \theta \cdot 4 \tan \theta} \cdot 4 \sec \theta \tan \theta \, d\theta = \int \frac{5}{16 \sec \theta \tan \theta} \cdot 4 \sec \theta \tan \theta \, d\theta = \int \frac{5}{16} \cdot 4 \, d\theta = \int \frac{5}{4} \, d\theta.$$\n\n7. **Integrate:**\n$$\int \frac{5}{4} \, d\theta = \frac{5}{4} \theta + C.$$\n\n8. **Back-substitute:** Since $$x = 4 \sec \theta$$, $$\theta = \sec^{-1} \frac{x}{4}$$, so\n$$\int \frac{5}{x \sqrt{x^2 - 16}} \, dx = \frac{5}{4} \sec^{-1} \frac{x}{4} + C, \quad x > 0.$$\n\n9. **Case 2: $$x < 0$$:** Then $$|x| = -x$$ and the integral becomes $$\int \frac{5}{-x \sqrt{x^2 - 16}} \, dx = - \int \frac{5}{x \sqrt{x^2 - 16}} \, dx.$$\n\n10. Using the same substitution as above, the integral evaluates to\n$$- \frac{5}{4} \sec^{-1} \frac{x}{4} + C, \quad x < 0.$$\n\n**Final answer:**\n$$\int \frac{5}{|x| \sqrt{x^2 - 16}} \, dx = \frac{5}{4} \operatorname{sgn}(x) \sec^{-1} \frac{|x|}{4} + C,$$ where $$\operatorname{sgn}(x)$$ is the sign function.