Partial fractions question.... Can someone comment on the following as to whether this partial fractions can be true or not? I'm concerned I've come across a trick question.
1) $\frac{x(x^2+4)}{x^2-4} = \frac{A}{x+2} + \frac{B}{x-2}$
Reason would lead me to believe the answer is yes, but mathematically I'm unsure if I'm correct?
If we take the original function...
$\frac{x(x^2+4)}{x^2-4}$
... multiply out the brackets and use long division, I get...
$x + \frac{8x}{x^2-4}$
Factorising the denominator gives us (x+2)(x-2), so in partial fractions we get...
2) $\frac{x(x^2+4)}{x^2-4} = x + \frac{A}{x+2} + \frac{B}{x-2}$
However, reason tells me that since $x = \frac{x(x+2)}{x+2}$ then equation 1, the value of A can be $x(x+2)$ plus the value of A in equation 2. i.e. Thus converting equation 2 to be...
2.1) $\frac{x(x^2+4)}{x^2-4} = \frac{x(x+2) + A}{x+2} + \frac{B}{x-2}$
So basically what I'm asking is can the partial fractions be altered in these ways? Or is there a trick I'm missing somewhere?

No: we cannot have $$\dfrac{x(x^2 + 4)}{x^2-4} = \dfrac A{x-2} +\dfrac B{x+2}$$
Why not? $$\dfrac{x^3 + 4x}{x^2 - 2} \neq \dfrac{A(x+2) + B(x-2)}{x^2 - 4}$$
The original numerator is a polynomial of degree $3$. The "transformed" numerator is of degree at most $1$.
Recall that $A, B$ represent constants only, not polynomials (of degree greater than $0)$.
Before invoking partial fractions, we want to get the degree of the numerator less than the degree of the denominator. We can do this, as you do, by polynomial long division. $$\frac{x(x^2 + 4)}{x^2 - 4} = \frac{x^3 + 4x}{x^2 - 4}= x + \dfrac{8x}{x^2 - 4}$$
Now the degree in the numerator is less than the degree in the denominator of our remaining rational function. So we can now use partial fractions.
$$x + \dfrac{8x}{x^2 - 4} = x + \dfrac A{x-2} + \dfrac B{x+2}$$ where $$A(x+2) + B(x-2) = 8x$$
