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?

After Partial Fraction Decomposition, the degree of the denominator of each fraction(not the integral part) must be greater than that of the numerator 
and for terms of the form $\displaystyle\frac a{(x+b)^n}$ can be represented as  
$$\frac{A_1}{x+b}+\frac{A_2}{(x+b)^2}+\cdots+\frac{A_n}{(x+b)^n}$$
if the degrees of the numerator of the original expression,  is greater than that of the denominator by $d(\ge0)$ degree
we shall have integral terms $c_0+c_1x+\cdots+c_dx^d$
