1. **State the problem:** We need to evaluate the integral $$\int \frac{-10}{x \sqrt{x^2 - 1}} \, dx$$. 2. **Recall the formula and substitution:** For integrals involving expressions like $$\sqrt{x^2 - 1}$$, a common substitution is $$x = \sec \theta$$, which implies $$dx = \sec \theta \tan \theta \, d\theta$$ and $$\sqrt{x^2 - 1} = \tan \theta$$. 3. **Apply the substitution:** $$ \int \frac{-10}{x \sqrt{x^2 - 1}} \, dx = \int \frac{-10}{\sec \theta \cdot \tan \theta} \cdot \sec \theta \tan \theta \, d\theta $$ 4. **Simplify the integrand:** $$ = \int -10 \, d\theta $$ 5. **Integrate:** $$ = -10 \theta + C $$ 6. **Back-substitute to x:** Since $$x = \sec \theta$$, then $$\theta = \sec^{-1} x$$. 7. **Final answer:** $$ \int \frac{-10}{x \sqrt{x^2 - 1}} \, dx = -10 \sec^{-1} x + C $$ This completes the integration.