1. The problem is to find the inverse Laplace transform of the function $$\frac{12s}{(s-5)(s+1)}$$. 2. The formula used is the partial fraction decomposition for rational functions in Laplace transforms: $$\frac{12s}{(s-5)(s+1)} = \frac{A}{s-5} + \frac{B}{s+1}$$ where $A$ and $B$ are constants to be determined. 3. Multiply both sides by the denominator $(s-5)(s+1)$: $$12s = A(s+1) + B(s-5)$$ 4. Expand the right side: $$12s = A s + A + B s - 5 B = (A + B)s + (A - 5B)$$ 5. Equate coefficients of like terms: - Coefficient of $s$: $12 = A + B$ - Constant term: $0 = A - 5B$ 6. Solve the system: From $0 = A - 5B$, we get $A = 5B$. Substitute into $12 = A + B$: $$12 = 5B + B = 6B \implies B = 2$$ Then $A = 5 \times 2 = 10$. 7. So the partial fractions are: $$\frac{12s}{(s-5)(s+1)} = \frac{10}{s-5} + \frac{2}{s+1}$$ 8. The inverse Laplace transform of $\frac{1}{s-a}$ is $e^{at}$. 9. Therefore, $$\mathcal{L}^{-1}\left\{\frac{10}{s-5}\right\} = 10 e^{5t}$$ $$\mathcal{L}^{-1}\left\{\frac{2}{s+1}\right\} = 2 e^{-t}$$ 10. The final answer is: $$x(t) = 10 e^{5t} + 2 e^{-t}$$ This function is the inverse Laplace transform of the given rational function.