91Ó°ÊÓ

Consider the given question,

$\frac{x^2+9}{x^4-2x^2-8}$

The denominator can be factorized as,

$x^2=u$

Then,

$$\begin{gathered} u^2-2u-8=0 \\ u=4,-2 \end{gathered}$$

Therefore,

$$\begin{gathered} x^4-2x^2-8=\left(x^2-4\right)\left(x^2+2\right) \\ {x}^4-2x^2-8=\left(x-2\right)\left(x+2\right)\left(x^2+2\right) \end{gathered}$$

The denominator has a repeated non-reducible quadratic factors. Then we decompose the rational expression as follows,

$$\begin{gathered} \frac{x^2+9}{x^4-2x^2-8}=\frac{x^2+9}{\left(x+2\right)\left(x-2\right)\left(x^2+2\right)} \\ =\frac{A}{x-2}+\frac{B}{x+2}+\frac{Cx+D}{x^2+2} \end{gathered}$$

Take the numerator,

$$\begin{gathered} x^2+9=A\left(x+2\right)\left(x^2+2\right)+B\left(x-2\right)\left(x^2+2\right)+\left(Cx+D\right)\left(x-2\right)\left(x+2\right) \\ {x}^2+9=A\left(x^3+2x+2x^2+4\right)+B\left(x^3+2x-2x^2-4\right)+\left(Cx+D\right)\left(x^2-4\right) \\ {x}^2+9=A{x}^3+2Ax+2A{x}^2+B{x}^3+2Bx-2B{x}^2-4B+C{x}^3-4Cx+D{x}^2-4D \\ {x}^2+9=\left(A+B+C\right)x^3+\left(2A-2B+D\right)x^2+\left(2A+2B-4C\right)x+\left(4A-4B-4D\right) \end{gathered}$$

The equation obtained is an identity in x.

Equate the coefficients of like powers of x to get,

$\left\{\begin{array}{l}A+B+C=0......\left(i\right)\\ 2A-2B+D=1......\left(ii\right)\\ 2A+2B-4C=0......\left(iii\right)\\ 4A-4B-4D=9......\left(iv\right)\end{array}\right.$

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

Subtract equation (iii) and equation (i) and multiply it by $2$,

role="math" localid="1647596401594" $$\begin{gathered} 2A+2B-4C-2\left(A+B+C\right)=0 \\ -6C=0 \\ C=0 \end{gathered}$$

Subtract equation (iv) and equation (ii) and multiply it by $2$,

$$\begin{gathered} 4A-4B-4D-2\left(2A-2B+D\right)=9-2·1 \\ 4A-4B-4D-4A+4B-2D=7 \\ -6D=7 \\ D=-\frac{7}{6} \end{gathered}$$

Substract equation (ii) and equation (i) and multiply it by $2$,

$$\begin{gathered} 2A-2B+D-2\left(A+B+C\right)=1 \\ 2A-2B+D-2A-2B-2C=1 \\ -4B-\frac{7}{6}=1 \\ B=-\frac{13}{24} \end{gathered}$$

Substitute the values of B, Cin equation (i),

$$\begin{gathered} A\left(-\frac{13}{24}\right)+0=0 \\ A-\frac{13}{24}=0 \\ A=\frac{13}{24} \end{gathered}$$

Therefore, the partial fraction decomposition is give below,

$\frac{x^2+9}{x^4-2x^2-8}=\frac{13}{24\left(x-2\right)}-\frac{13}{24\left(x-2\right)}-\frac{7}{6\left(x^2+2\right)}$