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?
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: $$\boxed {h(t)=5e^{2t}+6 \sin (2t)}$$