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

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Answer 1 · 2017-11-15T08:20:39

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

Perform the decomposition into partial fractions

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

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

The denominators are the same, compare the numerators

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

Let $x=1$

$-2=2A$ , $A=-1$

Coefficients of $x^2$

$1=A-B-2C+D$

Coefficients of $x$

$-2=B+C-2D$

and

$-1=A-B+D$ , $=>$ , $-B+D=0$ , $B=D$

$1=-1-B-2C+B$ , $=>$ , $2C=-2$ , $=>$ , $C=-1$

$-2=B-1-2B$ , $=>$ , $B=1$

$=>$ , $D=1$

Finally,

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

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