Chapter 10 - The Laplace Transform and Some Elementary Applications - 10.5 The First Shifting Theorem - Problems - Page 695: 39

$$ 2e^{-5t} \cos (7t)- e^{-5t}\sin (7t) $$

Since, $F(s)=\dfrac{2s+3}{(s+5)^2+49}=\dfrac{(2s+10)-7}{(s+5)^2+7^2}-\dfrac{7}{(s+5)^2+7^2}$ The inverse Laplace transform of function can be expressed as: $F(t)=L^{-1} [\dfrac{(2s+10)-7}{(s+5)^2+7^2}-\dfrac{7}{(s+5)^2+7^2}]$ Now, apply the first shifting Theorem. $f(t)= 2e^{-5t} \cos (7t)- e^{-5t}\sin (7t) $