In your case, the denominator is already factorized completely, so you can start immediately.
As your factors are non-linear, the partial fraction decomposition looks like this:
Find A, B, C, D so that
(s^3+s-4) / ((s^2+1)(s^2+4)) = (As+B) / (s^2+1) + (Cs+D) / (s^2+4)
Now, solve the equation!
... multiply both sides with the denominator (s^2+1)(s^2+4) in order to "get rid" of fractions...
<=> s^3 + s - 4 = (As+B)(s^2+4) + (Cs+D)(s^2+1)
... expand the expression ...
<=> s^3 + s - 4 = As^3 + Bs^2 + 4As + 4B + Cs^3 + Ds^2 + Cs + D
... compare the expressions with the same power of s....
<=> color(blue)(s^3) + color(red)(0*s^2) + color(green)(s) color(purple)(- 4) = color(blue)(As^3) + color(red)(Bs^2) + color(green)(4As) + 4B + color(blue)(Cs^3) + color(red)(Ds^2) + color(green)(Cs) + color(purple)(D)
At this point, you can split the equation in four different ones: for color(blue)(s^3) terms, color(red)(s^2) terms, color(green)(s) terms and color(purple)("linear") terms:
{ (color(white)(x) 1 = color(white)(x) A + C color(white)(xxx) "(I)" ),
(color(white)(x) 0 = color(white)(x) B + D color(white)(xxi) "(II)" ),
(color(white)(x) 1 = 4A + C color(white)(xxi) "(III)" ),
(-4 = 4B + D color(white)(xx) "(IV)" ) :}
From "(I)" and "(III)", you can compute A and C, e.g. by solving "(I)" for A and plugging the value into "(III)":
=> A = 0, " "C = 1
Similarly, if you e.g. solve "(II)" for B and plug the value into "(IV)", you will compute B and D:
=> B = -4/3, " "D = 4/3
Thus, your partial fraction decomposition looks like this:
(s^3+s-4) / ((s^2+1)(s^2+4)) = (-4/3)/(s^2+1) + (s+4/3) / (s^2+4)
... expand the second fraction by 3 to gain a nicer fraction...
(s^3+s-4) / ((s^2+1)(s^2+4)) = -4/(3(s^2+1)) + (3s+4)/(3(s^2+4))
Hope that this helped! :-)
