Answer
$$ e^{-t} \cos (5t)-\dfrac{3}{5} e^{-t}\sin (5t) $$
Work Step by Step
Since, $F(s)=\dfrac{s-2}{s^2+2s+26}=\dfrac{s+1}{s^2+2s+26}-\dfrac{3}{s^2+2s+26}$
The inverse Laplace transform of function can be expressed as:
$F(t)=L^{-1} [\dfrac{s+1}{s^2+2s+26}-\dfrac{3}{s^2+2s+26} ]$
Now, apply the first shifting Theorem.
$f(t)= e^{-t} \cos (5t)-\dfrac{3}{5} e^{-t}\sin (5t) $
