Inverse Laplace via Decomposition Theorem
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:
$$x(t) = 1 + (2t-1)e^{-2t}$$