Inverse Laplace transformation that is slightly different from known transformation. I have the following inverse laplace transformation:
$L^{-1} =\frac{s}{(s-3)(s-4)(s-12)}$
After looking at the laplace transformations the closest I've found is:
$\frac{ae^{at}-be^{bt}}{a-b} = \frac{s}{(s-a)(s-b)}$
I've been working out a solution for having three variables, but I cant seem to get the correct solution. Is there an identity for this type of transformation?
 A: $$\dfrac{s}{(s-3)(s-4)(s-12)} = -\dfrac{1}{2 (s-4)}+\dfrac{1}{3 (s-3)}+\dfrac{1}{6 (s-12)}$$
$$L^{-1} \bigg (\dfrac{s}{(s-3)(s-4)(s-12)}\bigg) =(-1/2) e^{4t} +(1/3) e^{3t} + (1/6) e^{12t}$$
A: Using partial-fraction expansion, we have 
$$
\frac{s}{(s-3)(s-4)(s-12)} = \frac{A}{(s-3)} + \frac{B}{(s-4)} + \frac{C}{(s-12)}
$$
where $A, B$ and $C$ are obtained as follows:
$$
\begin{align}
A &= \frac{s}{(s-4)(s-12)} \Big|_{s=3} = \frac{1}{3}, \\
B &= \frac{s}{(s-3)(s-12)} \Big|_{s=4} = -\frac{1}{2}, \\
C &= \frac{s}{(s-3)(s-4)} \Big|_{s=12} = \frac{1}{6}. \\
\end{align}
$$
As a result, we get
$$
\frac{s}{(s-3)(s-4)(s-12)} = \frac{1}{3(s-3)} - \frac{1}{2(s-4)} + \frac{1}{6(s-12)}
$$
Now, it is straightforward to use the Laplace inverse table, specifically, $\mathscr{L}^{-1} \left[ \frac{1}{s+a} \right] = e^{-at}u(t)$, so we get 
$$
\begin{align}
\mathscr{L}^{-1}\left[ \frac{s}{(s-3)(s-4)(s-12)} \right] &= 
\mathscr{L}^{-1}\left[ \frac{1}{3(s-3)} \right] - \mathscr{L}^{-1}\left[ \frac{1}{2(s-4)} \right] + \mathscr{L}^{1}\left[ \frac{1}{6(s-12)} \right] \\
&=
\frac{1}{3} \mathscr{L}^{1}\left[ \frac{1}{(s-3)} \right] - \frac{1}{2}\mathscr{L}^{1}\left[ \frac{1}{(s-4)} \right] + \frac{1}{6}\mathscr{L}^{1}\left[ \frac{1}{(s-12)} \right] \\
&=
\frac{1}{3} e^{3t} - \frac{1}{2} e^{4t} + \frac{1}{6} e^{12t}
\end{align}
$$
