Answer
False:
Since the numerator has a higher degree than the denominator,
$$
\frac{x\left(x^{2}+4\right)}{x^{2}-4}=x+\frac{8 x}{x^{2}-4}=x+\frac{A}{x+2}+\frac{B}{x-2}
$$
Work Step by Step
We wish to express $$
\frac{x\left(x^{2}+4\right)}{x^{2}-4}
$$ in partial fractions.
The numerator is of degree 3. The denominator is of degree 2. So this fraction is improper. It is better to use "Long Division".
This means that if we are going to divide the numerator by the denominator, we are going to divide a term in $ x^{3}$ by one in $ x^{2}$
which gives rise to a term in x. Consequently, we express the
partial fractions in the form:
$$
\frac{x\left(x^{2}+4\right)}{x^{2}-4}=x+\frac{8 x}{x^{2}-4}=x+\frac{A}{x+2}+\frac{B}{x-2}
$$
Thus, the statement is False.
