L^-1((s-5)/((s-1)(s^2)+4))

To find the inverse Laplace transform of the given expression, we can use partial fraction decomposition.The given expression is:\( \frac{s-5}{(s-1)(s^2+4)} \)We can decompose this fraction into simpler fractions:\( \frac{s-5}{(s-1)(s^2+4)} = \frac{A}{s-1} + \frac{Bs+C}{s^2+4} \)Multiplying both sides by the denominator \((s-1)(s^2+4)\), we get:\( s-5 = A(s^2+4) + (Bs+C)(s-1) \)Expanding and equating the coefficients of , we have: Grouping the terms, we get:\( s-5 = (A+B)s^2 + (4A-C)s + (4A-Bs-C) \)Equating the coefficients of , , and the constant term, we have: Solving these equations, we find: Substituting these values back into the partial fractions, we get:\( \frac{s-5}{(s-1)(s^2+4)} = \frac{1/5}{s-1} + \frac{-s/5 - 4/5}{s^2+4} \)Now, we can take the inverse Laplace transform of each term separately:\( L^{-1}\left(\frac{1/5}{s-1}\right) = \frac{1}{5}e^t \)\( L^{-1}\left(\frac{-s/5 - 4/5}{s^2+4}\right) = -\frac{1}{5}\cos(2t) - \frac{2}{5}\sin(2t) \)Therefore, the inverse Laplace transform of the given expression is:\( L^{-1}\left(\frac{s-5}{(s-1)(s^2+4)}\right) = \frac{1}{5}e^t - \frac{1}{5}\cos(2t) - \frac{2}{5}\sin(2t) \)