1. **State the problem:** Find the inverse Laplace transform of the function $$F(s) = \frac{s + 5}{s^2 - 2s - 3}$$. 2. **Factor the denominator:** The quadratic in the denominator factors as $$s^2 - 2s - 3 = (s - 3)(s + 1)$$. 3. **Set up partial fraction decomposition:** We express $$\frac{s + 5}{(s - 3)(s + 1)} = \frac{A}{s - 3} + \frac{B}{s + 1}$$. 4. **Multiply both sides by the denominator:** $$s + 5 = A(s + 1) + B(s - 3)$$ 5. **Expand and group terms:** $$s + 5 = A s + A + B s - 3 B = (A + B) s + (A - 3 B)$$ 6. **Equate coefficients:** - Coefficient of $$s$$: $$1 = A + B$$ - Constant term: $$5 = A - 3 B$$ 7. **Solve the system:** From $$1 = A + B$$, we get $$A = 1 - B$$. Substitute into $$5 = A - 3 B$$: $$5 = (1 - B) - 3 B = 1 - 4 B$$ $$4 B = 1 - 5 = -4$$ $$B = -1$$ Then $$A = 1 - (-1) = 2$$. 8. **Rewrite the function:** $$F(s) = \frac{2}{s - 3} - \frac{1}{s + 1}$$ 9. **Recall inverse Laplace transforms:** $$\mathcal{L}^{-1}\left\{ \frac{1}{s - a} \right\} = e^{a t}$$ for $$t \geq 0$$. 10. **Apply inverse Laplace transform:** $$f(t) = 2 e^{3 t} - e^{-t}$$. **Final answer:** $$\boxed{f(t) = 2 e^{3 t} - e^{-t}}$$