Inverse Laplace via Decomposition Theorem

The original query is to find x(t) with a Laplace transform of X(s) = 4(s+1) / s(s+2)^2. The poles are s_1 = 0 and s_2 = -2 (with a multiplicity of 2).

Because the second pole is not simple, its x(t) should be solved via the Decomposition Theorem, correct?

I.e. lim(s->s_{2}) = d/ds [ X(s) (s-s_{2})^2 e^st ]

Continuing to solve, I reach the following and am unsure of how to continue:

$$lim_{(s->s_{2})} \frac{d}{ds} \left(\frac{4(s+1)}{s}\right) * e^st$$

Because x(t) for the first pole is x(t) = 1, and x(t) for the second (non-simple pole) is x(t) = (2t-1)*e^(-2t), the final solution is the sum of the two:

$$x(t) = 1 + (2t-1)*e^(-2t)$$

How is this reached (i.e. how is the solution for the second pole reached, from the point where I left off)?

You should do a partial fraction expansion (also named partial fraction decomposition) instead.

Using partial fraction expansion, you can write the Laplace domain polynomial as:

$$X(s) = \frac{4(s+1)}{s(s+2)^2} = \frac{A}{s}+\frac{B}{s+2}+\frac{C}{(s+2)^2}$$

Solving for the coefficients is simple and you can show that

$$A = 1, B = -1, C = 2$$

And so

$$X(s) = \frac{4(s+1)}{s(s+2)^2} = \frac{1}{s}-\frac{1}{s+2}+\frac{2}{(s+2)^2}$$

Doing the inverse Laplace transform term by term is easy and can be easily found in a table. The inverse Laplace transform of X(s) is now then

$$x(t) = 1 - e^{-2t} + 2te^{-2t}$$

Re-arrange the terms to get your final expression: $$x(t) = 1 + (2t-1)e^{-2t}$$