Solution:
$
\mathrm{X}(S)=\frac{S^2+9 S+1}{S\left[S^2+6 S+8\right]}\\
$
$
=\frac{S^2+9 S+1}{S(S+4)(S+2)}\\
$
$
=\frac{A}{S}+\frac{B}{(S+4)}+\frac{C}{(S+2)}\\
$
$
S^2+9 S+1=A(S+4)(S+2)+\mathrm{B} S(S+2)+\mathrm{C} S(S+4)\\
$
$
\therefore X(S)=\frac{\frac{1}{8}}{S}+\frac{-\frac{19}{8}}{(S+4)}+\frac{\frac{13}{4}}{(S+2)}\\
$
Applying inverse Laplace transform,
$
x(t)=\frac{1}{8} u(t)-\frac{19}{8} e^{-4 t} u(t)+\frac{13}{4} e^{-2 t} u(t)\\
$
i) $\boldsymbol{R e}(\boldsymbol{s})\gt0$
ROC lies a right side of all poles,
$
\therefore x(t)=\frac{1}{8} u(t)-\frac{19}{8} e^{-4 t} u(t)+\frac{13}{4} e^{-2 t} u(t)\\
$
ii) $\operatorname{Re}(s)\lt-4$
ROC lies on the left side of all poles,
$
\therefore x(t)=-\frac{1}{8} u(-t)+\frac{19}{8} e^{-4 t} u(-t)-\frac{13}{4} e^{-2 t} u(-t)\\
$
iii) $-\mathbf{2}\gt\operatorname{Re}(\boldsymbol{s})\gt-\mathbf{4}$
ROC lies left side of poles $s=-2, s=0$ and right side of pole $s=-4$
$
\therefore x(t)=-\frac{1}{8} u(-t)-\frac{19}{8} e^{-4 t} u(t)-\frac{13}{4} e^{-2 t} u(-t)\\
$