$$X(s)=\frac{4s+1}{s(2s^2+2s+1)}$$
Please somebody show me how to factor this out step by step so I can take the inverse laplace using tables. I already have the answer: $1+(3e^{0.5t})\sin0.5t-(e^{0.5t})\cos0.5t$
The partial fraction is difficult to handle. I've tried repeatedly and still can't get it. I need detailed steps please.
On
Write the expression as a sum: $$ \frac{A}{s} + \frac{Bs + C}{2s^2 + 2s + 1} = \frac{4s+1}{s(2s^2 + 2s + 1}. $$ Combine terms: $$ \frac{A(2s^2 + 2s + 1) + (Bs + C)s}{s(2s^2 + 2s + 1} = \frac{4s+1}{s(2s^2 + 2s + 1}. $$ The two numerators must be equal, so $$ A(2s^2 + 2s + 1) + (Bs + C)s = 4s+1 \\ 2As^2 + 2As + A + Bs^2 + Cs = 4s+1 \\ (2A+B)s^2 + (2A+C)s + A = 4s+1 \\ $$
from which you see that $$ A =1\\ 2A + C = 4\\ 2A + B = 0 $$ That gives $A = 1, C = 2, B = -2$. Thus $$ \frac{4s+1}{s(2s^2 + 2s + 1} = \frac{A}{s} + \frac{Bs + C}{2s^2 + 2s + 1} \\ = \frac{1}{s} + \frac{-2s + 2}{2s^2 + 2s + 1}. $$
Integrating the first term is easy. The second needs to be converted slightly: $$ \frac{-2s + 2}{2s^2 + 2s + 1} = \frac{(-2s - 1) + 3}{2s^2 + 2s + 1} \\ = \frac{(-2s - 1)}{2s^2 + 2s + 1} + \frac{3}{2s^2 + 2s + 1} $$
The first term now yields to thte substitution $u = 2s^2 + 2s + 1, du = 2(2s + 1)$. The last term has to be solved with an arctan-type substitution via completion of the square.
The quadratic factor $2s^2+2s+1$ is irreducible. You should write it as $2[s^2+s+1/2]=2[(s+1/2)^2+1/4]$. From there, you should be able to get it.