## Precalculus (6th Edition) Blitzer

The partial fraction decomposition is $x+1-\frac{2}{x}-\frac{2}{{{x}^{2}}}+\frac{2}{x-1}$.
Let us consider the expression, $\frac{{{x}^{4}}-{{x}^{2}}+2}{{{x}^{3}}-{{x}^{2}}}$ Firstly, solve by long division method, {{x}^{3}}-{{x}^{2}}\overset{x+1}{\overline{\left){\begin{align} & {{x}^{4}}-{{x}^{2}}+2 \\ & {{x}^{4}}\text{ }-{{x}^{3}} \\ & -\text{ }+ \\ & \overline{\begin{align} & \text{ }{{x}^{3}}-{{x}^{2}}+2 \\ & \text{ }{{x}^{3}}-{{x}^{2}} \\ & \,\,\,-\text{ }+\text{ } \\ & \overline{\text{ }2} \\ \end{align}} \\ \end{align}}\right.}} So, $\frac{{{x}^{4}}-{{x}^{2}}+2}{{{x}^{3}}-{{x}^{2}}}=x+1+\frac{2}{{{x}^{3}}-{{x}^{2}}}$ Now, partial fraction of $\frac{2}{{{x}^{3}}-{{x}^{2}}}=\frac{2}{{{x}^{2}}\left( x-1 \right)}$ And to find partial fraction decomposition, $\frac{2}{{{x}^{2}}\left( x-1 \right)}=\frac{A}{x}+\frac{B}{{{x}^{2}}}+\frac{C}{x-1}$ …… (1) And multiply both sides by ${{x}^{2}}\left( x-1 \right)$, \begin{align} & 2=Ax\left( x-1 \right)+B\left( x-1 \right)+C{{x}^{2}} \\ & 2=A{{x}^{2}}-Ax+Bx-B+C{{x}^{2}} \\ & 2=\left( A+C \right){{x}^{2}}-\left( A-B \right)x-B \\ \end{align} And equate the terms, \begin{align} & A+C=0 \\ & A-B=0 \\ & B=-2 \\ \end{align} Then, solving the above equation, $A=-2,B=-2,C=2$ Put these values in equation (1), $\frac{2}{{{x}^{2}}\left( x-1 \right)}=-\frac{2}{x}-\frac{2}{{{x}^{2}}}+\frac{2}{x-1}$ Thus, the partial fraction decomposition of the remainder term is $x+1-\frac{2}{x}-\frac{2}{{{x}^{2}}}+\frac{2}{x-1}$.