Find the partial fraction decomposition of $(x^3+x^2+x+2)/(x^4+x^2)$ ?

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Answer 1 · 2017-01-23T05:29:13

$(x^3+x^2+x+2)/(x^4+x^2)=1/x+2/x^2-1/(x^2+1)$

As factors of $x^4+x^2=x^2(x^2+1)$ , partial fraction decomposition of $(x^3+x^2+x+2)/(x^4+x^2)$ will be of of the form

$(x^3+x^2+x+2)/(x^4+x^2)hArrA/x+B/x^2+(Cx+D)/(x^2+1)$

= $(Ax(x^2+1)+B(x^2+1)+x^2(Cx+D))/(x^4+x^2)$

= $(Ax^3+Ax+Bx^2+B+Cx^3+Dx^2)/(x^4+x^2)$

= $((A+C)x^3+(B+D)x^2+Ax+B)/(x^4+x^2)$

Comparing the coefficients, we get $B=2$

and as $B+D=1$ i.e. $D=-1$ . Further, $A=1$ ,

as $A+C=1$ we have $C=0$

$(x^3+x^2+x+2)/(x^4+x^2)=1/x+2/x^2-1/(x^2+1)$

Find the partial fraction decomposition of (x^3+x^2+x+2)/(x^4+x^2)(x^3+x^2+x+2)/(x^4+x^2)?