Chapter 10 - The Laplace Transform and Some Elementary Applications - 10.7 The Second Shifting Theorem - Problems - Page 705: 25

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We are given that $F(s)=\frac{2e^{-2s}}{(s-1)(s^2+1)}$ Now,$f(t)= L^{-1}[\frac{2e^{-2s}}{(s-1)(s^2+1)}]\\ =L^{-1}[\frac{e^{-2s}}{s-1}-\frac{se^{-2s}}{s^2+1}-\frac{e^{-2s}}{s^2+1}]\\ =u_2(t)[e^{t-2}-\cos (t-2)-\sin (t-2)]$