Answer
The partial fraction is $\frac{3}{x}-\frac{3}{\left( x-1 \right)}+\frac{4}{\left( x+1 \right)}$.
Work Step by Step
$\begin{align}
& \frac{4{{x}^{2}}-7x-3}{{{x}^{3}}-x}=\frac{4{{x}^{2}}-7x-3}{x\left( {{x}^{2}}-1 \right)} \\
& =\frac{4{{x}^{2}}-7x-3}{x\left( x-1 \right)\left( x+1 \right)} \\
& =\frac{A}{x}+\frac{B}{\left( x-1 \right)}+\frac{C}{\left( x+1 \right)}
\end{align}$
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 $\frac{3}{x}-\frac{3}{\left( x-1 \right)}+\frac{4}{\left( x+1 \right)}$.
