How to show the solution for $\dot{x}(t)=-cx^2(t)+\frac{x^2(t)}{1+x^2(t)}$ is bounded? How to show the following differential equation has a bounded solution.:

$$
\dot{x}(t)=-cx^2(t)+\frac{x^2(t)}{1+x^2(t)}
$$

where $c&gt;0$.

My try:
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 $x(t)$.
If you assume at first that x(t) is unbounded as $t \to \infty$, the last term drops out. This leaves us with  $\dot{x}(t) = -cx(t)+1$. Using again that we assume x(t) to be unbounded, the term 1 can be disregarded, so we are in asymptotic approximation left with: $$\dot{x}(t) = -cx(t)^2$$

This has the following solution:

$$x(t) = \frac{1}{c}\left(\frac{1}{t+K_{1}}\right)$$

Clearly, this solution is bounded as $t \to \infty$, so therefore, we are in contradiction with our assumption. 

The only problem we run into is that we require some initial condition on $x$. If we assume this initial condition to be bounded, then I think we are fine.

edit: Corrected $\dot{x}(t) = -cx(t)$ into $\dot{x}(t) = -cx(t)^2$
If you assume at first that x(t) is unbounded as $t \to \infty$, the last term drops out. This leaves us with  $\dot{x}(t) = -cx(t)+1$. Using again that we assume x(t) to be unbounded, the term 1 can be disregarded, so we are in asymptotic approximation left with: $$\dot{x}(t) = -cx(t)^2$$

This has the following solution:

$$x(t) = \frac{1}{c}\left(\frac{1}{t+K_{1}}\right)$$

Clearly, this solution is bounded as $t \to \infty$, so therefore, we are in contradiction with our assumption. 

The only problem we run into is that we require some initial condition on $x$. If we assume this initial condition to be bounded, then I think we are fine.

edit: Corrected $\dot{x}(t) = -cx(t)$ into $\dot{x}(t) = -cx(t)^2$