Chapter 10 - The Laplace Transform and Some Elementary Applications - 10.7 The Second Shifting Theorem - Problems - Page 704: 20

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We are given that $F(s)=\frac{e^{-s}(s+6)}{s^2+9}$ Now,$f(t)= L^{-1}[\frac{e^{-s}(s+6)}{s^2+9}]\\ =L^{-1}[e^{-s}(\frac{s}{s^2+9}+\frac{6}{s^2+9})]\\ =L^{-1}[e^{-s}(L(\cos 3t+2\sin 3t))]\\ =\cos 3 (t-1)u_1(t)+2\sin 3 (t-1)u_1(t)\\ =u_1(t)[\cos 3(t-1)+2\sin 3(t-1)]$