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! :-)