How do I perform Inverse Laplace on this function? $$
F(S) = \frac{-S+11}{S^2-2S-3}
$$
Howo do I find $f(t)$? What is a good strategy for attacking these types of problems? Thanks a bunch in advance for your help!
 A: Factor the denominator:
$$s^2-2 s-3 = (s-3)(s+1)$$
Then note that
$$\frac{-s+11}{(s-3)(s+1)} = \frac{A}{s-3}+ \frac{B}{s+1}$$
where
$$A+B=-1 \quad \text{and} \quad A-3 B=11$$
so that $A=2$ and $B=-3$.  Thus
$$F(s) = \frac{2}{s-3}-\frac{3}{s+1}$$
Now know that the inverse Laplace transform of $1/(s+a)$ is $e^{-a t}$ when $s \gt -a$.  You should be able to see this from
$$\int_0^{\infty} dt \, e^{-a t} \, e^{-s t} = \frac{1}{s+a}$$
You should be able to take it from here.
Alternatively, you can use the residue theorem if you know some complex analysis.  Here, you see that the poles of $F$ are at $s=-1$ and $s=3$.  The ILT is the sum of the residues of $F(s) e^{s t}$ at the poles:
$$\operatorname*{Res}_{s=-1} \frac{-s+11}{(s-3)(s+1)} e^{s t} = \frac{12}{-4} e^{-t}$$
$$\operatorname*{Res}_{s=3} \frac{-s+11}{(s-3)(s+1)} = \frac{8}{4} e^{3 t}$$
A: Related problem: (I). Use partial fraction to get

$$ -\frac{3}{\left( s+1 \right)}+\frac{2}{\left( s-3 \right)}. $$

Now, I think you can finish it. For alternative approach see here.
Note:

$$ \mathcal{L}\{e^{ax}\}(s) = \frac{1}{s-a}. $$

A: Hints: Let $F:S\mapsto\frac{-S+11}{S^2-2S-3}$
A well known integral: 


*

*$\int\limits_0^\infty\mathrm e^{-St}\mathrm e^{at}\mathrm dt=\frac1{S-a}$ (can you prove this?).


A well known Laplace transform: 


*

*Let $e_a:t\mapsto\mathrm e^{at}$, then $(Le_a)(S)=\frac1{S-a}$ for every $S\gt a$ (can you prove this?).


A decomposition: 


*

*$S^2-2S-3=(S-3)(S+1)$ (do you know why?) hence $F(S)=\frac{u}{S-3}+\frac{v}{S+1}$ for some real numbers $u$ and $v$ (can you compute them?).


A consequence: 


*

*$F=uL(e_3)+vL(e_{-1})$ (can you prove this?), thus $F=L(ue_3+ve_{-1})$ (can you deduce this?).


Note finally that $F$ is a Laplace transform on the domain $(3,+\infty)$ only.
