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

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Answer 1 · 2014-08-30T13:26:41

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,

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

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