Question
The fraction $\dfrac{x^{2}-9}{x^{2}{(x^{2}+9)}^{2}}$ can be expressed as $\dfrac{A}{x}+\dfrac{B}{{(x^{2}+9)}^{2}}$, where A and B are constants?
My attempt to solve
I understand as an identity, that is to say: $$\dfrac{x^{2}-9}{x^{2}{(x^{2}+9)}^{2}}=\dfrac{A}{x}+\dfrac{B}{{(x^{2}+9)}^{2}}\;\forall x\neq0$$ If $x=3$, then $0=\dfrac{A}{3}+\dfrac{B}{{(3^{2}+9)}^{2}}=\dfrac{A}{3}+\dfrac{B}{18^{2}}\implies 0=6\cdot18\cdot A+B\ldots(\alpha)$
If $x=-3$, then $0=-\dfrac{A}{3}+\dfrac{B}{{({-3}^{2}+9)}^{2}}=-\dfrac{A}{3}+\dfrac{B}{{18}^{2}}\implies 0=A\cdot6\cdot18-B\ldots(\beta)$
Then $\alpha$ and $\beta$: $A=B=0$.
Finally $\dfrac{x^{2}-9}{x^{2}{(x^{2}+9)}^{2}}=\dfrac{A}{x}+\dfrac{B}{{(x^{2}+9)}^{2}}=\dfrac{0}{x}+\dfrac{0}{{(x^{2}+9)}^{2}}=0$
This is correct?
Thank you.