$\begin{align}  & \frac{{{x}^{3}}-{{x}^{2}}-4}{{{x}^{2}}-1} \\ & \text{This is improper fraction} \\ & {{x}^{2}}-1\overset{x-1}{\overline{\left){\begin{align}  & {{x}^{3}}-{{x}^{2}}+0x-4 \\ & \underline{{{x}^{3}}\text{        }-x\text{       }} \\ & \text{      }-{{x}^{2}}+x-4 \\ & \text{ }\underline{\text{     }-{{x}^{2}}\text{      }+1\text{  }} \\ & \text{                   }x-5 \\\end{align}}\right.}} \\ & \therefore \frac{{{x}^{3}}-{{x}^{2}}-4}{{{x}^{2}}-1}=x-1+\frac{x-5}{{{x}^{2}}-1} \\ & \text{Decomposing }\frac{x-5}{{{x}^{2}}-1}\text{ into partial fraction} \\ & \frac{x-5}{{{x}^{2}}-1}=\frac{x-5}{(x-1)(x+1)}=\frac{A}{(x-1)}+\frac{B}{(x+1)} \\ & \therefore x-5=A(x+1)+B(x-1) \\ & \text{Set }x=1 \\ & 1-5=2A \\ & A=-2 \\ & \text{Set }x=-1 \\ & -1-5=B(-1-1) \\ & B=3 \\ & \frac{x-5}{{{x}^{2}}-1}=\frac{x-5}{(x-1)(x+1)}=\frac{-2}{(x-1)}+\frac{3}{(x+1)} \\ & \therefore \frac{{{x}^{3}}-{{x}^{2}}-4}{{{x}^{2}}-1}=x-1-\frac{2}{(x-1)}+\frac{3}{(x+1)} \\\end{align}$