## Precalculus (6th Edition) Blitzer

The partial fraction decomposition of the rational expression is $\frac{x}{\left( {{x}^{2}}+4 \right)}-\frac{4x}{{{\left( {{x}^{2}}+4 \right)}^{2}}}$
We know that the expression ${{x}^{2}}+4$ occurs multiple times in the denominator of the rational expression; therefore, for each power of ${{x}^{2}}+4$ assign an undefined constant factor in the denominator by the method of grouping as shown below: $\frac{{{x}^{3}}}{{{\left( {{x}^{2}}+4 \right)}^{2}}}=\frac{Ax+B}{\left( {{x}^{2}}+4 \right)}+\frac{Cx+D}{{{\left( {{x}^{2}}+4 \right)}^{2}}}$ Multiply both sides of the equation by ${{\left( {{x}^{2}}+4 \right)}^{2}}$: \begin{align} & {{\left( {{x}^{2}}+4 \right)}^{2}}\cdot \frac{{{x}^{3}}}{{{\left( {{x}^{2}}+4 \right)}^{2}}}={{\left( {{x}^{2}}+4 \right)}^{2}}\left( \frac{Ax+B}{\left( {{x}^{2}}+4 \right)}+\frac{Cx+D}{{{\left( {{x}^{2}}+4 \right)}^{2}}} \right) \\ & {{\left( {{x}^{2}}+4 \right)}^{2}}\cdot \frac{{{x}^{3}}}{{{\left( {{x}^{2}}+4 \right)}^{2}}}={{\left( {{x}^{2}}+4 \right)}^{2}}\left( \frac{Ax+B}{\left( {{x}^{2}}+4 \right)} \right)+{{\left( {{x}^{2}}+4 \right)}^{2}}\left( \frac{Cx+D}{{{\left( {{x}^{2}}+4 \right)}^{2}}} \right) \end{align} And divide out the common factors: \begin{align} & {{\left( {{x}^{2}}+4 \right)}^{2}}\cdot \frac{{{x}^{3}}}{{{\left( {{x}^{2}}+4 \right)}^{2}}}=\left( {{x}^{2}}+4 \right)\left( {{x}^{2}}+4 \right)\left( \frac{Ax+B}{\left( {{x}^{2}}+4 \right)} \right)+{{\left( {{x}^{2}}+4 \right)}^{2}}\left( \frac{Cx+D}{{{\left( {{x}^{2}}+4 \right)}^{2}}} \right) \\ & {{x}^{3}}=\left( {{x}^{2}}+4 \right)\left( Ax+B \right)+\left( Cx+D \right) \\ & {{x}^{3}}=A{{x}^{3}}+4Ax+B{{x}^{2}}+4B+Cx+D \\ & {{x}^{3}}=A{{x}^{3}}+B{{x}^{2}}+\left( 4A+C \right)x+\left( 4B+D \right) \end{align} Then, equate the coefficients of like terms of the equation to write a system of equations as shown below: $A=1$ (I) $B=0$ (II) $4A+C=0$ (III) $4B+D=0$ (IV) Put $A=1$ in equation (III) to obtain the value of C: \begin{align} & 4A+C=0 \\ & 4\left( 1 \right)+C=0 \\ & C=-4 \end{align} Put $B=0$ in equation (IV) to obtain the value of D: \begin{align} & 4B+D=0 \\ & 4\left( 0 \right)+D=0 \\ & D=0 \end{align} Put the values of $A,\,\,B,\,\,C,\,\,\text{and }D$ in the given equation, and define the partial fraction decomposition: \begin{align} & \frac{{{x}^{3}}}{{{\left( {{x}^{2}}+4 \right)}^{2}}}=\frac{Ax+B}{\left( {{x}^{2}}+4 \right)}+\frac{Cx+D}{{{\left( {{x}^{2}}+4 \right)}^{2}}} \\ & =\frac{\left( 1 \right)x+0}{\left( {{x}^{2}}+4 \right)}+\frac{\left( -4 \right)x+0}{{{\left( {{x}^{2}}+4 \right)}^{2}}} \\ & =\frac{x}{\left( {{x}^{2}}+4 \right)}-\frac{4x}{{{\left( {{x}^{2}}+4 \right)}^{2}}} \end{align} Hence, the partial fraction decomposition of the rational expression is $\frac{x}{\left( {{x}^{2}}+4 \right)}-\frac{4x}{{{\left( {{x}^{2}}+4 \right)}^{2}}}$.