91Ó°ÊÓ

Given rational expression is

$\frac{x}{x^2+2x-3}$

Take denominator of rational expression

$$\begin{gathered} x^2+2x-3=x^2+3x-x-3 \\ =x(x+3)-1(x+3) \\ =(x-1)(x+3) \end{gathered}$$

So rational expression is

$\frac{x}{x^2+2x-3}=\frac{x}{(x-1)(x+3)}$

partial fraction decomposition of a rational expression

$$\begin{gathered} \frac{P\left(x\right)}{Q\left(x\right)}=\frac{A_1}{x-a_1}+\frac{A_2}{x-a_2}+⋯+\frac{A_n}{x-a_n} \\ \frac{x}{(x-1)(x+3)}=\frac{A}{(x-1)}+\frac{B}{(x+3)}⋯\left(i\right) \\ \frac{x}{(x-1)(x+3)}=\frac{A(x+3)}{(x-1)(x+3)}+\frac{B(x-1)}{(x-1)(x+3)} \\ x=A(x+3)+B(x-1) \\ x+0=(A+B)x+(3A-B)⋯\left(ii\right) \end{gathered}$$

Compare the constants in equation ii

$$\begin{gathered} 0=3A-B \\ B=3A \end{gathered}$$

Compare the coefficient of xin equation ii and Substitute the expression for B

$$\begin{gathered} 1=A+B \\ 1=A+3A \\ A=\frac{1}{4} \end{gathered}$$

so$B=3\left(\frac{1}{4}\right)=\frac{3}{4}$

Substitute the value of A, B,and C in the equation i

$$\begin{gathered} \frac{x}{(x-1)(x+3)}=\frac{A}{(x-1)}+\frac{B}{(x+3)} \\ \frac{x}{(x-1)(x+3)}=\frac{\frac{1}{4}}{(x-1)}+\frac{\frac{3}{4}}{(x+3)} \\ \frac{x}{x^2+2x-3}=\frac{\frac{1}{4}}{(x-1)}+\frac{\frac{3}{4}}{(x+3)} \end{gathered}$$

So the partial fraction decomposition of a rational expression is$\frac{x}{x^2+2x-3}=\frac{\frac{1}{4}}{(x-1)}+\frac{\frac{3}{4}}{(x+3)}$