How do I find the partial fraction decomposition of #(t^4+t^2+1)/((t^2+1)(t^2+4)^2)# ?

1 Answer

We can now write:

#{x^2+x+1}/{(x+1)(x+4)^2}=A/{x+1}+B/{x+4}+C/{(x+4)^2}#

By recombining the fractions,

#={A(x+4)^2+B(x+1)(x+4)+C(x+1)}/{(x+1)(x+4)^2}#

By simplifying the numertor,

#={(A+B)x^2+(8A+5B+C)x+(16A+4B+C)}/{(x+1)(x+4)#

By comparing the coefficients of the numetaors,

#A+B=1#, #8A+5B+C=1#, and #16A+4B+C=1#.

By solving the equations for #A#, #B#, and #C#,

#A=1/9#, #B=8/9#, and #C=-13/3#.

Hence, by putting #x=t^2# back in,

#{1/9}/{t^2+1}+{8/9}/{t^2+4}+{-13/3}/{(t^2+4)^2}#