91Ó°ÊÓ

Given rational expression is

$\frac{x^2+2x+3}{{(x^2+4)}^2}$

partial fraction decomposition of a rational expression

$$\begin{gathered} \frac{P\left(x\right)}{Q\left(x\right)}=\frac{A_1}{x-a_1}+\frac{A_2}{x-a_2}+⋯+\frac{A_n}{x-a_n} \\ \frac{x^2+2x+3}{{(x^2+4)}^2}=\frac{Ax+B}{(x^2+4)}+\frac{Cx+D}{{(x^2+4)}^2}…\left(i\right) \\ \frac{x^2+2x+3}{{(x^2+4)}^2}=\frac{\left(Ax+B\right)(x^2+4)}{{(x^2+4)}^2}+\frac{Cx+D}{{(x^2+4)}^2} \\ {x}^2+2x+3=\left(Ax+B\right)(x^2+4)+Cx+D \\ \left(0\right)x^3+x^2+2x+3=A{x}^3+B{x}^2+(4A+C)x+(4B+D)…\left(ii\right) \end{gathered}$$

Compare the coefficient of $x^3$in equation ii

$A=0$

Compare the coefficient of $x^2$in equation ii

$B=1$

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

$$\begin{gathered} 2=4A+C \\ 2=4\left(0\right)+C \\ C=2 \end{gathered}$$

Compare the constants in equation ii and Substitute the expression for B

$$\begin{gathered} 3=4B+D \\ 3=4\left(1\right)+D \\ D=-1 \end{gathered}$$

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

$$\begin{gathered} \frac{x^2+2x+3}{{(x^2+4)}^2}=\frac{Ax+B}{(x^2+4)}+\frac{Cx+D}{{(x^2+4)}^2} \\ \frac{x^2+2x+3}{{(x^2+4)}^2}=\frac{\left(0\right)x+\left(1\right)}{(x^2+4)}+\frac{\left(2\right)x+(-1)}{{(x^2+4)}^2} \\ \frac{x^2+2x+3}{{(x^2+4)}^2}=\frac{1}{(x^2+4)}+\frac{2x-1}{{(x^2+4)}^2} \end{gathered}$$

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