## Precalculus (6th Edition) Blitzer

The partial fraction decomposition is ${{x}^{3}}+x+\frac{3}{2\left( x-1 \right)}-\frac{1}{2\left( x+1 \right)}$.
Let us consider the expression, $\frac{{{x}^{5}}+2}{{{x}^{2}}-1}$ Firstly, solve by long division method, {{x}^{2}}-1\overset{{{x}^{3}}+x}{\overline{\left){\begin{align} & {{x}^{5}}+2 \\ & {{x}^{5}}\text{ }-{{x}^{3}} \\ & -\text{ }+ \\ & \overline{\begin{align} & \text{ }{{x}^{3}}+2 \\ & \text{ }{{x}^{3}}\text{ }-x \\ & \text{ }-\text{ }+ \\ & \overline{\text{ }x+2} \\ \end{align}} \\ \end{align}}\right.}} So, $\frac{{{x}^{5}}+2}{{{x}^{2}}-1}={{x}^{3}}+x+\frac{x+2}{{{x}^{2}}-1}$ Then, partial fraction of $\frac{x+2}{{{x}^{2}}-1}=\frac{x+2}{\left( x-1 \right)\left( x+1 \right)}$ And to find partial fraction decomposition, $\frac{x+2}{\left( x-1 \right)\left( x+1 \right)}=\frac{A}{x-1}+\frac{B}{x+1}$ …… (1) Then, multiply both sides by $\left( x-1 \right)\left( x+1 \right)$, \begin{align} & x+2=A\left( x+1 \right)+B\left( x-1 \right) \\ & x+2=Ax+A+Bx-B \\ & x+2=\left( A+B \right)x+A-B \\ \end{align} And equate the terms, \begin{align} & A+B=1 \\ & A-B=2 \\ \end{align} Then, solving above equation, $A=\frac{3}{2},B=-\frac{1}{2}$ Put these values in equation (1), $\frac{x+2}{\left( x-1 \right)\left( x+1 \right)}=\frac{3}{2\left( x-1 \right)}-\frac{1}{2\left( x+1 \right)}$ Thus, the partial fraction decomposition of the remainder term is ${{x}^{3}}+x+\frac{3}{2\left( x-1 \right)}-\frac{1}{2\left( x+1 \right)}$.