Answer
The statement is False.
$$
\frac{x^{2}-4}{x \left(x^{2}+4\right) }
$$
It can be put in the form:
$$
\frac{A}{x}+\frac{Bx+C}{x^{2}+4}
$$
In fact, $A = −1 , B = 2 $ and $C = 0$
Work Step by Step
Resolve into partial fractions
$$
\frac{x^{2}-4}{x \left(x^{2}+4\right) }
$$
The partial fraction decomposition takes the form
$$
\begin{aligned}
\frac{x^{2}-4}{x \left(x^{2}+4\right) } & =\frac{A}{x}+\frac{Bx+C}{x^{2}+4}\\
&=\frac{A\left(x^2+4\right)+x\left(Bx+C\right) } {x \left(x^{2}+4\right) }
\end{aligned}
$$
multiplying both sides by the denominator, we get:
$$
x^2-4=A\left(x^2+4\right)+x\left(Bx+C\right) \quad (1)
$$
Substituting $x = 0 $ into this equation gives $A = −1$. Comparing the co-efficient of like powers of $ x $ on both sides in Eq. (1) and using the value $A = −1$, we get $ B = 2 $ and $ C = 0 $
So that:
$$
\frac{x^{2}-4}{x \left(x^{2}+4\right) } =\frac{-1}{x}+\frac{2x}{x^{2}+4}
$$
Thus, the statement is False. It can be put in the form:
$$
\frac{A}{x}+\frac{Bx+C}{x^{2}+4}
$$
In fact, $A = −1 , B = 2 $ and $ C = 0 $
