The function $y=\frac{1}{1+x^2}$ is decreasing in the interval:

A function is decreasing where its first derivative is negative.
$$y^{\prime }=\frac{d}{dx}(1+x^2{)}^{-1}=-(1+x^2{)}^{-2}\cdot 2x=-\frac{2x}{(1+x^2{)}^2}$$
For $y$ to be decreasing, $y^{\prime }<0$:
$$-\frac{2x}{(1+x^2{)}^2}<0⟹2x>0⟹x>0$$
So the function is decreasing for $x\in (0,\infty )$.