Answer
$\frac{7x-3}{x(x+3)(x-1)}=\frac{1}{x}+\frac{-2}{x+3}+\frac{1}{x-1}$
Work Step by Step
$\frac{7x-3}{(x^3+2x^2-3x)}$, factorising $x^3+2x^2-3x=x(x+3)(x-1)$:
$\frac{7x-3}{x(x+3)(x-1)}=\frac{A}{x}+\frac{B}{x+3}+\frac{C}{x-1}$,
$7x-3=A(x-1)(x+3)+Bx(x-1)+Cx(x+3)=(Ax^2+2Ax-3A)+(Bx^2-Bx)+(Cx^2+3Cx)$,
$7x-3=A+B+C)x^2+(2A-B+3C)x-3A=7x-3$,
$(A+B+C)x^2=0, A+B+C=0$,
$(2A-B+3C)x=7x, 2A-B+3C=7$ and $-3A=-3, A=1$:
substituting $A=1$ back into the equation, the first equation becomes $B+C=-1$ and the second equation becomes $-B+3C=5$:
$\begin{cases}
& B & +C & = -1&\\
& -B & +3C & = 5&\\
& -- & -- & -- \\
& 0 & +4C & = 4
\end{cases}$
$C=1$, substituting back into the equation, $B+1=-1, B=-2$:
thus,
$\frac{7x-3}{x(x+3)(x-1)}=\frac{1}{x}+\frac{-2}{x+3}+\frac{1}{x-1}$
