Answer
The statement is True.
$$
\frac{x^{2}+4}{x\left(x^{2}-4\right) }
$$
can be put in the form:
$$
\frac{A}{x}+\frac{B}{x+2}+\frac{C}{x-2}
$$
In fact, $A = −1, B = 1~and ~C = 1$
Work Step by Step
Resolve into partial fractions:
$$
\frac{x^{2}+4}{x\left(x^{2}-4\right) }
$$
We’ll first factor the denominator and then get the form of the partial fraction decomposition.
$$
\begin{aligned} \frac{x^{2}+4}{x\left(x^{2}-4\right) }& =\frac{x^2+4}{x\left(x+2\right)\left(x-2\right)}\\
&=\frac{A}{x}+\frac{B}{x+2}+\frac{C}{x-2} \\
&=\frac{A\left(x+2\right)\left(x-2\right)+Bx\left(x-2\right)+Cx\left(x+2\right) }{x\left(x+2\right)\left(x-2\right)}
\end{aligned}
$$
multiplying both sides by the denominator, we get:
$$
x^2+4=A\left(x+2\right)\left(x-2\right)+Bx\left(x-2\right)+Cx\left(x+2\right)
$$
Substituting $x = 0 $ into this equation gives $A = −1$, substituting $x = -2 $ gives $ B = 1$, and substituting $x =2 $ gives $ C = 1$ so that
$$
\frac{x^{2}+4}{x\left(x^{2}-4\right) }=\frac{-1}{x}+\frac{1}{x+2}+\frac{1}{x-2}
$$
Thus, the statement is True. In fact, $A = −1, B = 1, C = 1$