Answer
$\frac{x-3}{x(x^2+3)}=\frac{-1}{x}+\frac{x+1}{x^2+3}$
Work Step by Step
$\frac{x-3}{x^3+3x}$, factorising $x^3+3x=x(x^2+3)$.
$\frac{x-3}{x(x^2+3)}=\frac{A}{x}+\frac{Bx+C}{x^2+3}$,
$A(x^2+3)+Bx+C(x)$,
$Ax^2+3A+Bx^2+Cx$,
$(A+B)x^2+Cx+3A=x-3$,
$A+B=0$, $C=1$, $3A=-3$.
thus, $A=-1, B=1$
Therefore,
$\frac{x-3}{x(x^2+3)}=\frac{-1}{x}+\frac{x+1}{x^2+3}$,
