Inverse laplace transform using equating coeffecients method

Question

Please see attached image. Could you please help me do the inverse of this laplace transfrom. I'm using the method of trying to equate the coeffecients.

Thanks

Answer 1

$$\frac{\color{blue}{3(s^2-3s-2)}}{s(s+1)(3s+1)}=\frac{A}{s}+\frac{B}{s+1}+\frac{C}{3 s+1}=\frac{\color{blue}{A(s+1)(3s+1)+Bs(3s+1)+Cs(s+1)}}{s(s+1)(3s+1)}$$

Thus we must have $$ 3s^2-9s-6=(3A+3B+C) s^2+(4 A+B+C) s+A $$ and equating the same power we have $3A+3B+C=3$, $4 A+B+C=-9$ and $A=-6$, and then $B=3$ and $C=12$ that is $$ \overline{x}(s)=\frac{3(s^2-3s-2)}{s(s+1)(3s+1)}=-\frac{6}{s}+\frac{3}{s+1}+\frac{12}{3 s+1}=-\frac{6}{s}+\frac{3}{s+1}+\frac{4}{s+\frac{1}{3}} $$ and $$ x(t)=-6u(t)+3 u(t)\mathrm e^{-t}+4 u(t) \mathrm e^{-t/3} $$