Chapter 10 - The Laplace Transform and Some Elementary Applications - 10.5 The First Shifting Theorem - Problems - Page 695: 35

$6 e^{-t} \sin (t)$

Since, $F(s)=\dfrac{6}{s^2+2s+2}$ The inverse Laplace transform of function can be expressed as: $F(t)=L^{-1} [\dfrac{6}{(s+1)^2+1} ]$ Now, apply the first shifting Theorem. $f(t)= 6 e^{-t} \sin (t)$