Rather strange Inverse Laplace transform
Note that $$ H(s) = \frac{s}{s^2+4s+5} = \frac{(s+2)-2}{(s+2)^2+1^2} = G_1(s+2)- 2G_2(s+2) $$ where $$ G_1(s) = \frac{s}{s^2+1^2}, G_2(s) = \frac{1}{s^2+1^2} $$ We deduce $h(t)= [\cos(t)-2\sin(t)]e^{-2t}$.
Note that $$ H(s) = \frac{s}{s^2+4s+5} = \frac{(s+2)-2}{(s+2)^2+1^2} = G_1(s+2)- 2G_2(s+2) $$ where $$ G_1(s) = \frac{s}{s^2+1^2}, G_2(s) = \frac{1}{s^2+1^2} $$ We deduce $h(t)= [\cos(t)-2\sin(t)]e^{-2t}$.