Answer
$\frac{2x-3}{x(x^2+3)}=\frac{-1}{x}+\frac{x+2}{x^2+3}$
Work Step by Step
$\frac{2x-3}{x^3+3x}$, factorising $x^3+3x=x(x^2+3)$.
$\frac{2x-3}{x(x^2+3)}=\frac{A}{x}+\frac{Bx+C}{x^2+3}$,
$A(x^2+3)+(Bx+C)x$,
$Ax^2+3A+Bx^2+Cx$,
$(A+B)x^2+Cx+3A=2x-3$,
$\begin{cases}
A+B=0\\
C=2\\
3A=-3
\end{cases}$
thus, $A=-1$, substituting back into the Equation, $-1+B=0, B=1$.
Therefore,
$\frac{2x-3}{x(x^2+3)}=\frac{-1}{x}+\frac{x+2}{x^2+3}$
