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Find the inverse Laplace transform?

Find the inverse Laplace transform of F(s)=(8s^2-4s+12)/(s(s^2+4)).

Answer: f(t)=3-2 sin 2t+5 cos 2t

A/s+(Bs+C)/(s^2+4)

8s^2-4s+12=A(s^2+4)+(Bs+C)(s)=As^2+4A+Bs^2+Cs=s^2(A+B)+Cs+4A

8=A+B

C=-4

A=3

B=5

L^-1 (3/s)+L^-1 ((5s-4)/(s^2+4))

=3+ (now I'm stucked)

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