1. **State the problem:** Find the inverse Laplace transform of $$\frac{s+5}{s^2 - 2s - 3}$$. 2. **Factor the denominator:** The quadratic $$s^2 - 2s - 3$$ factors as $$ (s - 3)(s + 1) $$. 3. **Rewrite the expression:** $$\frac{s+5}{(s-3)(s+1)}$$. 4. **Use partial fraction decomposition:** Assume $$\frac{s+5}{(s-3)(s+1)} = \frac{A}{s-3} + \frac{B}{s+1}$$. 5. **Multiply both sides by the denominator:** $$s + 5 = A(s + 1) + B(s - 3)$$. 6. **Expand and group terms:** $$s + 5 = A s + A + B s - 3 B = (A + B)s + (A - 3B)$$. 7. **Equate coefficients:** - Coefficient of $$s$$: $$1 = A + B$$ - Constant term: $$5 = A - 3B$$ 8. **Solve the system:** From $$1 = A + B$$, we get $$A = 1 - B$$. Substitute into $$5 = A - 3B$$: $$5 = (1 - B) - 3B = 1 - 4B$$ $$4B = 1 - 5 = -4$$ $$B = -1$$ Then $$A = 1 - (-1) = 2$$. 9. **Rewrite the partial fractions:** $$\frac{s+5}{(s-3)(s+1)} = \frac{2}{s-3} - \frac{1}{s+1}$$. 10. **Recall inverse Laplace transforms:** - $$\mathcal{L}^{-1}\left\{ \frac{1}{s-a} \right\} = e^{at}$$. 11. **Apply inverse Laplace transform:** $$f(t) = 2 e^{3t} - e^{-t}$$. **Final answer:** $$\boxed{f(t) = 2 e^{3t} - e^{-t}}$$