Find Laplace transform for frac{s+8}{s^2+4s+5}

Step 1
Formula Used:
$L\{e^{at}\sin bt\}=\frac{b}{(s-a{)}^2+b^2}$
$L\{e^{at}\cos bt\}=\frac{s-a}{(s-a{)}^2+b^2}$
Step 2
$\text{Consider }F(s)=\frac{s+8}{s^2+5s+5}$
$=\frac{s+8}{(s+2{)}^2+1}$
$=\frac{s+2+6}{(s+2{)}^2+1}$
$=\frac{(s+2)}{(s+2{)}^2+1}+\frac{6}{(s+2{)}^2+1}$
Step 3
$\text{Consider }F(s)=\frac{(s+2)}{(s+2{)}^2+1}+\frac{6}{(s+2{)}^2+1}$
We now apply Inverse Laplace Transform, we get
$L^{-1}\{F(s)\}=L^{-1}\left\{\frac{(s+2)}{(s+2{)}^2+1}\right\}+6L^{-1}\left\{\frac{1}{(s+2{)}^2+1}\right\}$
$\Rightarrow f(t)=e^{-2t}\cos t+6e^{-2t}\sin t$
$=e^{-2t}(\cos t+6\sin t)$
Step 4
ANSWER
: The inverse Laplace Transform of F(s) is
$f(t)=e^{-2t}(\cos t+6\sin t)$