Calculate $ f [f (x)] $ Where $ f (x) = \frac {1} {1 + x ^ 2} $, $ f [f (x)] $ is equal to:
a
I was factoring and found $f(f(x))= \frac{1}{1+(\frac{1}{1+x^2})^2} = \frac{1}{\frac{x^4+2x^2+2}{x^4+2x^2+1}}= \frac{x^4+2x^2+1}{x^4+2x^2+2}$, but the answer is x. I am wrong?

Your calculation of 
$$f(f(x))=\frac{x^4+2x^2+1}{x^4+2x^2+2}$$
is correct. Both the numerator and denominator are fourth degree polynomials with a coefficient of $1$ before the $x^4$ term. Therefore, there is a horizontal asymptote at $y=1$.
So, the answer cannot be $f(f(x))=x$. As the graph of $f(f(x))$ is bounded above by $y=1$, it cannot be equal to the value of $x$ when $x>1$.
