91Ó°ÊÓ

The rational expression is,

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

Decomposing the rational expression,

$\frac{x+4}{x^2(x^2+4)}=\frac{A}{x}+\frac{B}{x^2}+\frac{Cx+D}{(x^2+4)}............\left(1\right)$

Multiplying both sides by $x^2(x^2+4)$,

role="math" localid="1647499865407" $$\begin{gathered} x+4=Ax(x^2+4)+B(x^2+4)+(Cx+D)x^2 \\ x+4=A{x}^3+4Ax+B{x}^2+4B+C{x}^3+D{x}^2 \\ x+4=(A+C)x^3+(B+D)x^2+4Ax+4B............\left(2\right) \end{gathered}$$

Equating the coefficients of the like powers of $x$, we get,

$$\begin{gathered} A+C=0..........\left(3\right) \\ B+D=0.........\left(4\right) \\ 4A=1............\left(5\right) \\ 4B=4...........\left(6\right) \end{gathered}$$

Solving equation (6), we get, $B=1$.

Solving equation (5), we get, $A=\frac{1}{4}$.

Solving equation (3) by inputting the value of A, we get,

$C=-\frac{1}{4}$.

Solving equation (4), by inputting the value of B, we get,

$D=-1$.

The partial fraction decomposition is,

$\frac{x+4}{x^2(x^2+4)}=\frac{{\displaystyle \frac{1}{4}}}{x}+\frac{1}{x^2}+\frac{-{\displaystyle \frac{1}{4}}x-1}{x^2+4}$

$=\frac{1}{4x}+\frac{1}{x^2}+\frac{-(x+4)}{4(x^2+4)}$

Adding the rational expressions,

$\frac{1}{4x}+\frac{1}{x^2}+\frac{-(x+4)}{4(x^2+4)}=\frac{x(x^2+4)+4(x^2+4)+x^2(-(x+4\left)\right)}{4x^2(x^2+4)}$

$$\begin{gathered} =\frac{x^3+4x+4x^2+16-x^3-4x^2}{4x^2(x^2+4)} \\ =\frac{4x+16}{4x^2(x^2+4)} \\ =\frac{4(x+4)}{4x^2(x^2+4)} \\ =\frac{x+4}{x^2(x^2+4)} \end{gathered}$$

The solution is true.