How do I find the partial fraction decomposition of $\frac{t^{4} + t^{2} + 1}{(t^{2} + 1) {(t^{2} + 4)}^{2}}$ $(t^4+t^2+1)/((t^2+1)(t^2+4)^2)$ ?

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How do I find the partial fraction decomposition of $\frac{t^{4} + t^{2} + 1}{(t^{2} + 1) {(t^{2} + 4)}^{2}}$ $(t^4+t^2+1)/((t^2+1)(t^2+4)^2)$ ?

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Answer 1 · 2014-08-30T13:26:41

We can now write:

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

By recombining the fractions,

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

By simplifying the numertor,

$= \frac{(A + B) x^{2} + (8 A + 5 B + C) x + (16 A + 4 B + C)}{(x + 1) (x + 4)}$ $={(A+B)x^2+(8A+5B+C)x+(16A+4B+C)}/{(x+1)(x+4)$

By comparing the coefficients of the numetaors,

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

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

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

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

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

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How do I find the partial fraction decomposition of $\frac{t^{4} + t^{2} + 1}{(t^{2} + 1) {(t^{2} + 4)}^{2}}$ $(t^4+t^2+1)/((t^2+1)(t^2+4)^2)$ ?

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How do I find the partial fraction decomposition of $\frac{t^{4} + t^{2} + 1}{(t^{2} + 1) {(t^{2} + 4)}^{2}}$ $(t^4+t^2+1)/((t^2+1)(t^2+4)^2)$ ?