The following differential equation has the maximal interval as $[0,\infty)$:
$$ \dot{x}(t)=-cx^2(t)+\frac{x^2(t)}{1+x^2(t)} $$
where $c>0$.
My try: We need to show the solution lives in a bounded set, then we are done. Since $\frac{x^2(t)}{1+x^2(t)} \leq 1$ we can write
$$ \dot{x}(t)\leq-cx^2(t)+1 $$
I do not know how to find the upper bound for $x(t)$ and how to come up with a lower bound for $x(t)$.
As with any one-dimensional autonomous dynamical system with differentiable right side, it is sufficient to consider the roots and signs between roots of the right side function. The roots are at $x_0=0$ and $c(1+x^2)=1\iff x_\pm=\pm\sqrt{\frac1c-1}$ which are only real for $c\in(0,1]$.
As $x=0$ is a constant solution, any other solution has an invariant sign.
For $c>1$, the right side is negative everywhere for $x\ne 0$, so that all solutions are falling. This means that positive solutions are bounded and thus defined on all of $[0,\infty)$. Negative solutions are unbounded and will reach a point where the quadratic term is dominant over the second bounded term. From that point on there will be a blow-up to $-\infty$.
For $c\le 1$ the global image is the same, only that there will be a strip of bounded solutions between the stationary solutions $x=x_\pm$.