<s>[INST] Decomposing an inverse Laplace transform The given inverse Laplace transform is:
$$\mathscr{L}^{-1}\left[\frac{5s^2+12s-4}{s^3-2s^2+4s-8} \right]$$
First split it up into three separate fractions and factorize the denominator 
$$\mathscr{L}^{-1} \left[\frac{5s^2}{(s-2)(s^2+4)} \right]+\mathscr{L}^{-1} \left[\frac{12s}{(s-2)(s^2+4)} \right]-\mathscr{L}^{-1} \left[\frac{4}{(s-2)(s^2+4)} \right]$$
Now to compile them into partial fraction decomp. form:
$$\mathscr{L}^{-1} \left[\frac{5s^2}{(s-2)(s^2+4)} \right]=\frac{A}{(s-2)}+\frac{Bs+c}{(s^2+4)}$$
$$\mathscr{L}^{-1} \left[\frac{12s}{(s-2)(s^2+4)} \right]=\frac{A}{s-2}+\frac{Bs+c}{s^2+4}$$
$$\mathscr{L}^{-1} \left[\frac{4}{(s-2)(s^2+4)} \right]=\frac{A}{s-2}+\frac{Bs+c}{s^2+4}$$
Then following through:
$$5s^2=A(s^2+4)+(Bs+c)(s-2)$$
$$12s=A(s^2+4)+(Bs+c)(s-2)$$
$$4=A(s^2+4)+(Bs+c)(s-2)$$
Now to solve for the variables, first with $s=2$:
$$5*4=8A \to A=\frac{20}{8}$$
$$12(2)=8A \to A=3$$
$$4=8A \to A=\frac{1}{2}$$
Now subbing in for $s=0$:
$$0=\frac{20}{8}(0^2+4)+(B(0)+C)(0-2) \to 0=10-2C \to C=5$$
$$12(0)=3(4)+(B(0)+C)(-2) \to 0=12-2C \to C=6$$
$$4=\frac{1}{2}(4)-2C \to 4=2-2C \to C=-1$$
Now to solve for $B$; $s=1$
$$5=\frac{100}{8}-B-5 \to B=\frac{5}{2}$$
$$12=3(1+4)+(B+6)(-1) \to B=-3$$
$$4=\frac{1}{2}(1+4)+(B-1)(-1) \to B=-\frac{1}{2}$$
So now our Laplace transforms are:
$$\mathscr{L}^{-1} \left[\frac{5s^2}{(s-2)(s^2+4)} \right]=\frac{20}{8(s-2)}+\frac{5s+5}{2(s^2+4)}$$
$$\mathscr{L}^{-1} \left[\frac{12s}{(s-2)(s^2+4)} \right]=\frac{3}{s-2}-\frac{3s+6}{s^2+4}$$
$$\mathscr{L}^{-1} \left[\frac{4}{(s-2)(s^2+4)} \right]=\frac{1}{2(s-2)}-\frac{s-1}{2(s^2+4)}$$
Are my calculations correct or have I over complicated the problem?
 [/INST] Note that the denominator is:
$$(s-2)(s^2+4)$$
The numerator can be rewritten as :
$$f(s)=5s^2+12s-4$$
$$f(s)=5(s^2+4)-24+12s$$
$$f(s)=5(s^2+4)+12(s-2)$$
Back to our function :
$$h(t)=\mathscr{L}^{-1}\left[\frac{5s^2+12s-4}{s^3-2s^2+4s-8} \right]
$$
$$h(t)=\mathscr{L}^{-1}\left[\dfrac 5 {s-2}+\dfrac{12}{s^2+4} \right]$$
Finally:
$$\boxed {h(t)=5e^{2t}+6 \sin (2t)}$$
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