## Precalculus (6th Edition) Blitzer

The partial fraction is $\frac{1}{4\left( x-1 \right)}+\frac{3}{4\left( x+3 \right)}$.
We find the partial fraction of the stated expression as given below, \begin{align} & \frac{x}{{{x}^{2}}+2x-3}=\frac{x}{\left( x-1 \right)\left( x+3 \right)} \\ & =\frac{A}{\left( x-1 \right)}+\frac{B}{\left( x+3 \right)} \end{align} Now, multiply $\left( x-3 \right)\left( 2x+1 \right)$ on both sides: \begin{align} & \left( x-1 \right)\left( x+3 \right)\frac{x}{\left( x-1 \right)\left( x+3 \right)}=\left( x-1 \right)\left( x+3 \right)\frac{A}{\left( x-1 \right)}+\left( x-1 \right)\left( x+3 \right)\frac{B}{\left( x+3 \right)} \\ & x=\left( x+3 \right)A+\left( x-1 \right)B \\ & =Ax+3A+Bx-B \\ & x=x\left( A+B \right)+3A-B \end{align} …… (1) Then, compare the coefficient of equation (1): $A+B=1$ ……(2) $3A-B=0$ …… (3) Then, add equation (2) with equation (3), Then, $A=\frac{1}{4}$ Now, put the value of A in equation (2): \begin{align} & \left( \frac{1}{4} \right)+B=1 \\ & B=1-\frac{1}{4} \\ & =\frac{3}{4} \end{align} Therefore, $\frac{x}{{{x}^{2}}+2x-3}=\frac{1}{4\left( x-1 \right)}+\frac{3}{4\left( x+3 \right)}$ Thus, the partial fraction of the provided expression is $\frac{1}{4\left( x-1 \right)}+\frac{3}{4\left( x+3 \right)}$