How do I find the partial fraction decomposition of t4+t2+1(t2+1)(t2+4)2 ?
1 Answer
Aug 30, 2014
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 , and16A+4B+C=1 .
By solving the equations for
A=19 ,B=89 , andC=−133 .
Hence, by putting
19t2+1+89t2+4+−133(t2+4)2