Step-by-Step Solution
Key Concept: To find the range of a rational function, rearrange as a quadratic in $x$ and apply the discriminant condition for real solutions.
Let $y = \frac{2x}{1+x^2}$. Rearranging: $y(1+x^2) = 2x$, so $yx^2 - 2x + y = 0$. For real solutions in $x$, the discriminant must be non-negative: $\Delta = 4 - 4y^2 \geq 0$, giving $y^2 \leq 1$, thus $-1 \leq y \leq 1$. When $y = \pm 1$, we get $x = \pm 1$, which are real values. Therefore, the range is $[-1, 1]$.
Correct Answer: 1