## Precalculus (6th Edition) Blitzer

The partial fraction decomposition of the rational expression is $\frac{x}{\left( x+1 \right)\left( {{x}^{2}}+9 \right)}=\frac{-1}{10\left( x+1 \right)}+\frac{x+9}{10\left( {{x}^{2}}+9 \right)}$.
It is required to compute the partial fraction decomposition. Factor the denominator by the method of grouping as shown below: \begin{align} & \frac{x}{\left( x+1 \right)\left( {{x}^{2}}+9 \right)}=\frac{A}{x+1}+\frac{Bx+C}{{{x}^{2}}+9} \\ & =\frac{A\left( {{x}^{2}}+9 \right)+\left( Bx+C \right)\left( x+1 \right)}{\left( x+1 \right)\left( {{x}^{2}}+9 \right)} \end{align} $x=A\left( {{x}^{2}}+9 \right)+\left( Bx+C \right)\left( x+1 \right)$ (I) Substitute the value of $x=-1$ in equation (I), to obtain the value of A: \begin{align} & -1=A\left( 10 \right) \\ & A=-\frac{1}{10} \end{align} Again, putting the value of $x=0$ in equation (I), we get, \begin{align} & 0=A\left( 9 \right)+C\left( 1 \right) \\ & C=-9A \\ & C=-9.\left( -\frac{1}{10} \right) \\ & C=\frac{9}{10} \\ \end{align} Again, substitute the value of $x=1$ in equation (I) as follows: \begin{align} & 1=A\left( 10 \right)+\left( B+C \right)\left( 2 \right) \\ & 1=-\frac{1}{10}\left( 10 \right)+\left( B+\frac{9}{10} \right)2 \\ & 2=\left( B+\frac{9}{10} \right)2 \\ & 1=B+\frac{9}{10} \end{align} \begin{align} & B=1-\frac{9}{10} \\ & B=\frac{1}{10} \end{align} Therefore, the partial fraction decomposition of the rational expression is $\frac{x}{\left( x+1 \right)\left( {{x}^{2}}+9 \right)}=\frac{-1}{10\left( x+1 \right)}+\frac{x+9}{10\left( {{x}^{2}}+9 \right)}$.