I assume you want the inverse of the Laplace Transform in terms of t.
If F(s)=ℒ{f(t)}=s/(s²+2s+5)=s/(s²+2s+1+4)=s/((s+1)²+4), then f(t) is the inverse of this. From a Laplace Transform Table, we see G(s)=ℒ{eᵃᵗcos(bt)}=(s-a)/((s-a)²+b²). Also ℒ{eᵃᵗsin(bt)}=b/((s-a)²+b²). If we put a=-1 and b=2, we have G(s)=(s+1)/((s+1)²+4)=s/((s+1)²+4)+1/((s+1)²+4). So G(s)=F(s)+½ℒ{e⁻ᵗsin(2t)}.
Therefore F(s)=ℒ{e⁻ᵗcos(2t)}-½ℒ{e⁻ᵗsin(2t)} and f(t)=e⁻ᵗ(cos(2t)-½sin(2t)).