Invert the following system to the time domain using partial fraction. f(s)=frac(6)(s^2(s+2)(s+4))

Step 1
By using partial fraction method and the table of Laplace transform we solve the given problem as follows:
Step 2
${\displaystyle f\left(s\right)=\frac{6}{s^2(s+2)(s+4)}}$ By using partial fraction
${\displaystyle \frac{6}{s^2(s+2)(s+4)}=\frac{A}{s}+\frac{B}{s^2}+\frac{C}{s+2}+\frac{D}{s+4}}$
${\displaystyle \Rightarrow 6=A(s^3+6s^2+8s)+B(s^2+6s+8)+C(s^3+4s^2)+D(s^2+2s^2)}$
${\displaystyle \Rightarrow 6=(A+C+D)s^3+(6A+B+4C+2D)s^2+(8A+6B)s+8B}$
${\displaystyle \Rightarrow A+C+D=0,6A+B+4C+2D=0}$
${\displaystyle 8A+6B=0,8B=6\Rightarrow B=\frac{3}{4}}$
on solving thus we get
${\displaystyle A=\frac{-9}{16},B=\frac{3}{4},C=\frac{3}{4},D=\frac{-3}{16}}$
${\displaystyle \Rightarrow \frac{6}{s^2(s+2)(s+4)}=\frac{\frac{-9}{16}}{s}+\frac{\frac{3}{4}}{s^2}+\frac{\frac{3}{4}}{s+2}+\frac{\frac{-3}{16}}{s+4}}$
Now apply inverse laplace we get
$L^{-1}\left\{\frac{6}{s^2(s+2)(s+4)}\right\}=\frac{-9}{16}{L}^{-1}\left\{\frac{1}{s}\right\}+\frac{3}{4}{L}^{-1}\left\{\frac{1}{s^2}\right\}+\frac{3}{4}{L}^{-1}\left\{\frac{1}{s+2}\right\}-\frac{3}{16}{L}^{-1}\left\{\frac{1}{s+4}\right\}$
${\displaystyle \Rightarrow f\left(t\right)=\frac{-9}{16}+\frac{3}{4}t+\frac{3}{4}{e}^{-2t}-\frac{3}{16}{e}^{-4t}}$