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

Aug 30, 2014

We can now write:

$\frac{{x}^{2} + x + 1}{\left(x + 1\right) {\left(x + 4\right)}^{2}} = \frac{A}{x + 1} + \frac{B}{x + 4} + \frac{C}{{\left(x + 4\right)}^{2}}$

By recombining the fractions,

$= \frac{A {\left(x + 4\right)}^{2} + B \left(x + 1\right) \left(x + 4\right) + C \left(x + 1\right)}{\left(x + 1\right) {\left(x + 4\right)}^{2}}$

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,

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

By solving the equations for $A$, $B$, and $C$,

$A = \frac{1}{9}$, $B = \frac{8}{9}$, and $C = - \frac{13}{3}$.

Hence, by putting $x = {t}^{2}$ back in,

$\frac{\frac{1}{9}}{{t}^{2} + 1} + \frac{\frac{8}{9}}{{t}^{2} + 4} + \frac{- \frac{13}{3}}{{\left({t}^{2} + 4\right)}^{2}}$