$\dfrac1{x^2+3}=x^{-2}\left(1+\dfrac{3}{x^2}\right)^{-1} = x^{-2}\left(1-\dfrac{3}{x^2}+\dfrac{9}{x^4}+\cdots\right)$
But this can also be factored into:
$\dfrac{1}{3}\left(1+\dfrac{x^2}{3}\right)^{-1} = \dfrac{1}{3}\left(1-\dfrac{x^2}{3}+\dfrac{x^4}{9}+\cdots\right)$
What's wrong here? The two answers are the opposite of each other.