Write the partial fraction decomposition of the rational expression.

$\frac{2x+3}{x^4-9x^2}$

Consider the given question,

$\frac{2x+3}{x^4-9x^2}$

Factor the denominator of the given ration expression,

$$\begin{gathered} x^4-9x^2=\left(x^2-3x\right)\left(x^2+3x\right) \\ {x}^4-9x^2=x^2\left(x-3\right)\left(x+3\right) \end{gathered}$$

Decompose the rational expression,

$\frac{2x+3}{x^4-9x^2}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x+3}+\frac{D}{x-3}$

Where, Aand B are to be determined.

Multiply every side by $x^2\left(x+3\right)\left(x-3\right)$,

role="math" localid="1647594117392" $$\begin{gathered} \frac{2x+3}{x^4-9x^2}·x^2\left(x+3\right)\left(x-3\right)=\left[\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x+3}+\frac{D}{x-3}\right]·x^2\left(x+3\right)\left(x-3\right) \\ 2x+3=Ax\left(x^2-9\right)+B{x}^2\left(x^2-9\right)+C{x}^2\left(x^2-3\right)+D{x}^2\left(x+3\right) \\ 2x+3=\left(A+C+D\right)x^3+\left(B-3C+3D\right)x^2+\left(-9Ax-9B\right) \end{gathered}$$

The equation obtained is an identity in x.

Equate the coefficients of like powers of x,

$\left\{\begin{array}{l}A+C+D=0......\left(i\right)\\ B-3C+3D=0......\left(ii\right)\\ -9A=2......\left(iii\right)\\ -9B=3......\left(iv\right)\end{array}\right.$

These four equations can be solved to get the values of A, B, C, D.

Divide both sides of equation (iii) by $-9$,

$$\begin{gathered} \frac{-9A}{-9}=\frac{2}{-9} \\ A=-\frac{2}{9} \end{gathered}$$

Divide both sides of equation (iv) by $-9$,

$$\begin{gathered} \frac{-9B}{-9}=\frac{3}{-9} \\ B=-\frac{1}{3} \end{gathered}$$

Substitute the value of Bin equation (ii),

$$\begin{gathered} -\frac{1}{3}-3C+3D=0 \\ -3C+3D=\frac{1}{3}.......\left(v\right) \end{gathered}$$

Substitute the value of Ain equation (i),

$$\begin{gathered} -\frac{2}{9}+C+D=0 \\ C+D=\frac{2}{9}......\left(vi\right) \end{gathered}$$

Multiply both sides by $3$,

role="math" localid="1647594679992" $$\begin{gathered} 3\left(C+D\right)=3·\frac{2}{9} \\ 3C+3D=\frac{2}{3}.......\left(vii\right) \end{gathered}$$

On adding equations (v) and (vii), we get,

$$\begin{gathered} \left(-3C+3D\right)+\left(3C+3D\right)=\frac{1}{3}+\frac{2}{3} \\ 6D=1 \\ D=\frac{1}{6} \end{gathered}$$

Substitute the value of Din equation (vi),

role="math" localid="1647594755526" $$\begin{gathered} C+\frac{1}{6}=\frac{2}{9} \\ C=\frac{1}{18} \end{gathered}$$

Considering the previous steps,

$$\begin{gathered} A=-\frac{2}{9}, \\ B=-\frac{1}{3}, \\ C=\frac{1}{18}, \\ D=\frac{1}{6} \end{gathered}$$

Therefore, the partial fraction decomposition is give below,

$\frac{2x+3}{x^4-9x^2}=\frac{-{\displaystyle \frac{2}{9}}}{x}+\frac{-{\displaystyle \frac{1}{3}}}{x^2}+\frac{{\displaystyle \frac{1}{18}}}{x+3}+\frac{{\displaystyle \frac{1}{6}}}{x-3}$