How do you express $(x+1)/[(x^2+1)^2(x^2)]$ in partial fractions?

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Answer 1 · 2017-01-20T10:35:21

The answer is $=1/x^2+1/x+(-x-1)/(x^2+1)^2+(-x-1)/(x^2+1)$

Let's perform the decomposition into partial fractions

$(x+1)/((x^2+1)^2x^2)=A/x^2+B/x+(Cx+D)/(x^2+1)^2+(Ex+F)/(x^2+1)$

$=(A(x^2+1)^2+B(x)(x^2+1)^2+(Cx+D)x^2+(Ex+F)(x^2+1)x^2)/((x^2+1)^2x^2)$

Equalising the numerators

$x+1=A(x^2+1)^2+B(x)(x^2+1)^2+(Cx+D)x^2+(Ex+F)(x^2+1)x^2$

Let $x=0$ , $=>$ , $1=A$

Coefficients of $x$ , $1=B$

Coefficients of $x^2$ , $0=2A+D+F$

Coefficients of $x^3$ , $0=2B+C+E$

coefficients of $x^4$ , $0=A+F$ , $=>$ , $F=-A=-1$

$D=-2A-F=-2+1=-1$

Coefficients of $x^5$ , $0=B+E$ , $=>$ , $E=-B=-1$

$C=-2B-E=-2+1=-1$

Therefore,

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

How do you express (x+1)/[(x^2+1)^2(x^2)](x+1)/[(x^2+1)^2(x^2)] in partial fractions?