How do I find the partial fraction decomposition of t4+t2+1(t2+1)(t2+4)2 ?

1 Answer

We can now write:

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

By simplifying the numertor,

=(A+B)x2+(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=19, B=89, and C=133.

Hence, by putting x=t2 back in,

19t2+1+89t2+4+133(t2+4)2