## Precalculus (6th Edition) Blitzer

The partial fraction is $\frac{3}{x}-\frac{3}{\left( x-1 \right)}+\frac{4}{\left( x+1 \right)}$.
\begin{align} & \frac{4{{x}^{2}}-7x-3}{{{x}^{3}}-x}=\frac{4{{x}^{2}}-7x-3}{x\left( {{x}^{2}}-1 \right)} \\ & =\frac{4{{x}^{2}}-7x-3}{x\left( x-1 \right)\left( x+1 \right)} \\ & =\frac{A}{x}+\frac{B}{\left( x-1 \right)}+\frac{C}{\left( x+1 \right)} \end{align} Now, multiply $x\left( x-1 \right)\left( x+1 \right)$ on both sides: \begin{align} & x\left( x-1 \right)\left( x+1 \right)\frac{4{{x}^{2}}-7x-3}{x\left( x-1 \right)\left( x+1 \right)}=x\left( x-1 \right)\left( x+1 \right)\frac{A}{x}+x\left( x-1 \right)\left( x+1 \right)\frac{B}{\left( x-1 \right)}+x\left( x-1 \right)\left( x+1 \right)\frac{C}{\left( x+1 \right)} \\ & 4{{x}^{2}}-7x-3=\left( x-1 \right)\left( x+1 \right)A+x\left( x+1 \right)B+x\left( x-1 \right)C \\ & =\left( {{x}^{2}}+x-x-1 \right)A+B{{x}^{2}}+Bx+C{{x}^{2}}-Cx \\ & =A{{x}^{2}}-A+B{{x}^{2}}+Bx+C{{x}^{2}}-Cx \end{align} $4{{x}^{2}}-7x-3={{x}^{2}}\left( A+B+C \right)+x\left( B-C \right)-A$ …… (1) Then compare the coefficient of equation (1): $A+B+C=4$ …… (2) $B-C=-7$ …… (3) $-A=-3$ …… (4) $A=3$ …… (5) Substitute the value of A in equation (2): $3+B+C=4$ $B+C=1$ …… (6) And subtract equation (6) from equation (3), then $B=-3$ Then, put the value of B in equation (6): \begin{align} & -3+C=1 \\ & C=1+3 \\ & C=4 \end{align} Therefore, $\frac{4{{x}^{2}}-7x-3}{{{x}^{3}}-x}=\frac{3}{x}-\frac{3}{\left( x-1 \right)}+\frac{4}{\left( x+1 \right)}$ Thus, the partial fraction of the provided expression is $\frac{3}{x}-\frac{3}{\left( x-1 \right)}+\frac{4}{\left( x+1 \right)}$.