How do I find the partial fraction decomposition of (t^4+t^2+1)/((t^2+1)(t^2+4)^2)t4+t2+1(t2+1)(t2+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}x2+x+1(x+1)(x+4)2=Ax+1+Bx+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}=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)=(A+B)x2+(8A+5B+C)x+(16A+4B+C)(x+1)(x+4)

By comparing the coefficients of the numetaors,

A+B=1A+B=1, 8A+5B+C=18A+5B+C=1, and 16A+4B+C=116A+4B+C=1.

By solving the equations for AA, BB, and CC,

A=1/9A=19, B=8/9B=89, and C=-13/3C=133.

Hence, by putting x=t^2x=t2 back in,

{1/9}/{t^2+1}+{8/9}/{t^2+4}+{-13/3}/{(t^2+4)^2}19t2+1+89t2+4+133(t2+4)2