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)#