Let $f \in C^1(\mathbb R, \mathbb R)$ and suppose $ \tag{H} \vert f(x) \vert\le \frac{1}{2}\vert x \vert + 3, \quad \forall x \in \mathbb R. $
Then, every solution of $ x'(t)+x(t)+f(x(t))=0, \quad t \in \mathbb R $ is bounded on $[0,+\infty]$.
First of all, I would like to say that the text has been copied correctly: I mean, it's really $[0,+\infty]$ so I suppose the author wants me to prove $\displaystyle \lim_{t \to +\infty} x(t) <\infty$. Indeed, boundedness on $[0,+\infty)$ is quite obvious, because we have global existence (its enough to write the equation as $x'=-x-f(x)$ and to observe that the RHS is sublinear thanks to $(H)$).
So, we have to prove $\displaystyle \lim_{t \to +\infty} x(t) <\infty$. How can we do? I've got some ideas but I can't conclude. First, I observe that the problem is autonomous: this implies that solutions are either constant either monotonic.
First idea: I've fixed $x_0 \in \mathbb R$ and I've written the equivalent integral equation: $ x(t) = x_0 - \int_0^t [x(s)+f(x(s))]ds $
Taking the absolute value and making some rough estimates, we get $ \vert x(t) \vert \le \vert x_0 \vert + \left\vert \int_0^t [x(s)+f(x(s))]ds \right\vert \le \vert x_0 \vert + \int_0^t \frac{3}{2}\left\vert x(s) \right\vert +3 ds $ but now I don't know how to conclude. Gronwall's lemma? But how can I use it?
Second idea: if $x_0 \in \mathbb R$ is s.t. $x_0 +f(x_0) \neq 0$, the solution of the Cauchy problem is not constant. I can divide both members of equation and I obtain (integrating on $[0,t]$) $ -t=\int_0^t \frac{dx}{x+f(x)} $ Now I let $t \to +\infty$ but... what can I conclude?