<s>[INST] 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?
 [/INST] 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 $c_0+c_1x+\cdots+c_dx^d$
</s>