Chapter 10 - The Laplace Transform and Some Elementary Applications - 10.7 The Second Shifting Theorem - Problems - Page 705: 23

See below

We are given that $F(s)=\frac{e^{-4s}(s+3)}{s^2-6s+13}$ Now,$f(t)= L^{-1}[\frac{e^{-4s}(s+3)}{s^2-6s+13}]\\ =L^{-1}[e^{-4s}(\frac{e^{s-3}}{(s-3)^2+4}+\frac{6}{(s-3)^2+4})]\\ =L^{-1}[e^{-4s}][L(e^{3t}(\cos 2t+3\sin 2t))]\\ =u_4(t)[e^{3(t-4)}\cos 2(t-4)+3e^{t(-4)}\sin 2(t-4)]$