Laplace transform problem

Find the inverse laplace of

e^(-10s)/s(s^2+2s+2)

Ignore the exponential for a moment and do a partial fraction decomp on the rest.

1/[s(s² + 2s + 2)] = A/s + (Bs + C)/(s² + 2s + 2) ==> A =1/2 , B = -1/2, C = -1.

Next, use the fact that s² + 2s + 2 = (s + 1)² + 1, and -½ s = -½ (s + 1) + ½. So

1/[s(s² + 2s + 2)] = (1/2)/s - (1/2) (s + 1)/[(s + 1)² + 1] - (1/2)/[(s + 1)² + 1].

Now, put the exponential back and use the two shifting theorems

L{e^(at) f(t)} = F(s - a), and L^(-1){ e^(-as) F(s)} = f(t - a)U(t - a)

whenever L{f(t)} = F(s).

L^(-1){ e^(-10s)/[s(s² + 2s + 2)]} =

= ½ U(t - 10) - ½ e^[-(t - 10)] cos(t - 10) U(t - 10) - ½ e^[-(t - 10)] sin(t - 10) U(t - 10)

= ½ [1 - e^[-(t - 10)] cos(t - 10) - e^[-(t - 10)] sin(t - 10)] U(t - 10).