Answer
The partial fraction is $\frac{1}{4\left( x-1 \right)}+\frac{3}{4\left( x+3 \right)}$.
Work Step by Step
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 $\frac{1}{4\left( x-1 \right)}+\frac{3}{4\left( x+3 \right)}$
