How to take the inverse laplace of $F(s)=\frac{s-1}{(s+1)^3}$

Question

I tried to separate $F(s)=\frac{s-1}{(s+1)^3}$ into $\frac{s}{(s+1)^3}-\frac{1}{(s+1)^3}$ and $\mathscr{L}^{-1}\Bigl\{\frac{1}{(s+1)^3}\Bigr\}$ is $\frac{1}{2}e^{-t}t^2$

I can't figure out how to solve $\frac{s}{(s+1)^3}$

Can I use partial fraction expansion and expand it to $\frac{A_1}{(s+1)}+\frac{A_2}{(s+1)^2}+\frac{A_3}{(s+1)^3}$ and then solve for each $A_i$ and then take the inverse laplace of each individual one?

Answer 1

Yes, you can do that, because the Laplace-transform is a linear transformation. But you can also do this:

$$\frac{s-1}{(s+1)^3}=\frac{s+(1-1)-1}{(s+1)^3}=\frac{s+1-2}{(s+1)^3}=\frac{s+1}{(s+1)^3}-\frac{2}{(s+1)^3}=\frac{1}{(s+1)^2}-\frac{2}{(s+1)^3}$$ So your $A_1=0$, $A_2=1$, $A_3=-2$.