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
Aug 30, 2014
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 , and16A+4B+C=116A+4B+C=1 .
By solving the equations for
A=1/9A=19 ,B=8/9B=89 , andC=-13/3C=−133 .
Hence, by putting
{1/9}/{t^2+1}+{8/9}/{t^2+4}+{-13/3}/{(t^2+4)^2}19t2+1+89t2+4+−133(t2+4)2