How do I decompose the rational expression $(x-3)/(x^3+3x)$ into partial fractions?

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Answer 1 · 2018-06-17T12:39:56

The answer is $=-1/x+(x+1)/(x^2+3)$

Perform the decomposition into partial fractions after factorising the denominator

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

$=A/x+(Bx+C)/(x^2+3)$

$=(A(x^2+3)+x(Bx+C))/(x(x^2+3))$

The denominators are the same, compare the numerators

$x-3=A(x^2+3)+x(Bx+C)$

Let $x=0$ , $=>$ , $-3=3A$ , $=>$ , $A=-1$

Coefficients of $x^2$

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

Coefficients of $x$

$1=C$

Therefore,

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

How do I decompose the rational expression (x-3)/(x^3+3x)(x-3)/(x^3+3x) into partial fractions?