91Ó°ÊÓ

Given rational expression is

$\frac{x^3+1}{{(x^2+16)}^2}$

partial fraction decomposition of a rational expression

$$\begin{gathered} \frac{P\left(x\right)}{Q\left(x\right)}=\frac{A_1}{x-a_1}+\frac{A_2}{x-a_2}+⋯+\frac{A_n}{x-a_n} \\ \frac{x^3+1}{{(x^2+16)}^2}=\frac{Ax+B}{(x^2+16)}+\frac{Cx+D}{{(x^2+16)}^2}⋯\left(i\right) \\ \frac{x^3+1}{{(x^2+16)}^2}=\frac{\left(Ax+B\right)(x^2+16)}{(x^2+16)}+\frac{Cx+D}{{(x^2+16)}^2} \\ {x}^3+1=\left(Ax+B\right)(x^2+16)+Cx+D \\ {x}^3+\left(0\right)x^2+\left(0\right)x+1=A{x}^3+B{x}^2+(16A+C)x+16B+D⋯\left(ii\right) \end{gathered}$$

Compare the coefficient of$x^3$in equationii

$A=1$

Compare the coefficient ofin equationii

$B=0$

Compare the coefficient ofxin equationii and Substitute the expression forA

$$\begin{gathered} 0=16A+C \\ 0=16\left(1\right)+C \\ C=-16 \end{gathered}$$

Compare the constants in equationii and Substitute the expression forB

$$\begin{gathered} 1=16B+D \\ 1=16\left(0\right)+D \\ D=1 \end{gathered}$$

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

$$\begin{gathered} \frac{x^3+1}{{(x^2+16)}^2}=\frac{Ax+B}{(x^2+16)}+\frac{Cx+D}{{(x^2+16)}^2} \\ \frac{x^3+1}{{(x^2+16)}^2}=\frac{\left(1\right)x+0}{(x^2+16)}+\frac{(-16)x+1}{{(x^2+16)}^2} \\ \frac{x^3+1}{{(x^2+16)}^2}=\frac{x}{(x^2+16)}+\frac{1-16x}{{(x^2+16)}^2} \end{gathered}$$

So the partial fraction decomposition of a rational expression is$\frac{x^3+1}{{(x^2+16)}^2}=\frac{x}{(x^2+16)}+\frac{1-16x}{{(x^2+16)}^2}$