91Ó°ÊÓ

Given rational expression is

$\frac{x^2-11x-18}{x(x^2+3x+3)}$

partial fraction decomposition of a rational expression

$$\begin{gathered} \frac{P\left(x\right)}{Q\left(x\right)}=\frac{A_1}{x-a_1}+\frac{A_2}{x-a_2}+⋯+\frac{A_n}{x-a_n} \\ \frac{x^2-11x-18}{x(x^2+3x+3)}=\frac{A}{x}+\frac{Bx+C}{(x^2+3x+3)}⋯\left(i\right) \\ \frac{x^2-11x-18}{x(x^2+3x+3)}=\frac{A(x^2+3x+3)}{x(x^2+3x+3)}+\frac{\left(Bx+C\right)x}{x(x^2+3x+3)} \\ {x}^2-11x-18=A(x^2+3x+3)+\left(Bx+C\right)x \\ {x}^2-11x-18=A{x}^2+3Ax+3A+B{x}^3+Cx \\ {x}^2-11x-18=(A+B)x^2-(-3A-C)x-(-3A)⋯\left(ii\right) \end{gathered}$$

Compare the constants in equation ii

$$\begin{gathered} 18=-3A \\ A=-6 \end{gathered}$$

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

$$\begin{gathered} 11=-3A-C \\ 11=-3(-6)-C \\ C=7 \end{gathered}$$

Compare the coefficient of $x^2$in equation ii and Substitute the expression for A

$$\begin{gathered} 1=A+B \\ 1=-6+B \\ B=7 \end{gathered}$$

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

$$\begin{gathered} \frac{x^2-11x-18}{x(x^2+3x+3)}=\frac{A}{x}+\frac{Bx+C}{(x^2+3x+3)} \\ \frac{x^2-11x-18}{x(x^2+3x+3)}=\frac{-6}{x}+\frac{7x+7}{(x^2+3x+3)} \\ \frac{x^2-11x-18}{x(x^2+3x+3)}=\frac{-6}{x}+\frac{7(x+1)}{(x^2+3x+3)} \end{gathered}$$

So the partial fraction decomposition of a rational expression is $\frac{x^2-11x-18}{x(x^2+3x+3)}=\frac{-6}{x}+\frac{7(x+1)}{(x^2+3x+3)}$