1. **State the problem:** We need to evaluate the integral $$\int \frac{10}{x \sqrt{x^2 - 36}} \, dx$$. 2. **Recall the formula and substitution:** For integrals involving expressions like $$\sqrt{x^2 - a^2}$$, a common substitution is $$x = a \sec \theta$$, where $$a = 6$$ here. 3. **Apply substitution:** Let $$x = 6 \sec \theta$$, then $$dx = 6 \sec \theta \tan \theta \, d\theta$$. 4. **Rewrite the integral:** $$\sqrt{x^2 - 36} = \sqrt{36 \sec^2 \theta - 36} = \sqrt{36 (\sec^2 \theta - 1)} = 6 \tan \theta$$. 5. Substitute into the integral: $$\int \frac{10}{6 \sec \theta \cdot 6 \tan \theta} \cdot 6 \sec \theta \tan \theta \, d\theta = \int \frac{10}{36 \sec \theta \tan \theta} \cdot 6 \sec \theta \tan \theta \, d\theta$$ 6. Simplify the integrand: $$\int \frac{10 \cancel{6} \sec \theta \tan \theta}{36 \sec \theta \tan \theta} \, d\theta = \int \frac{10}{6} \, d\theta = \int \frac{5}{3} \, d\theta$$ 7. Integrate: $$\frac{5}{3} \theta + C$$ 8. Back-substitute $$\theta$$: Since $$x = 6 \sec \theta$$, then $$\sec \theta = \frac{x}{6}$$, so $$\theta = \sec^{-1} \left( \frac{x}{6} \right)$$. 9. **Final answer:** $$\int \frac{10}{x \sqrt{x^2 - 36}} \, dx = \frac{5}{3} \sec^{-1} \left( \frac{x}{6} \right) + C$$