Boundedness of solutions of $x'+x+f(x)=0$ 
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?
 A: By assumptions
$$
x'(t) = -x(t) -f(x(t)) \leq -x(t) + \frac 1 2 \lvert x(t) \rvert + 3
$$
Let's suppose there is a point $t_0$ such that 
$$
x(t_0) > \max\{x(0), 6\}
$$
Since $x(t_0)$ is positive the following relation is satisfied
$$
x'(t_0) \leq -x(t_0) + \frac 1 2 \lvert x(t_0) \rvert + 3 = -\frac 1 2 x(t_0)  + 3 < 0
$$
So the $\max$ of $x(t)$ on the interval $[0, t_0]$ must occur at $t_1\in (0, t_0)$, but that leads to contradiction because:
$$
0 = x'(t_1) \leq - \frac 1 2 x(t_1) + 3 < 0
$$
where the equal sign is there because $t_1$ is an extremum point of a differentiable function on an open interval.
A similar reasoning shows that $x(t)$ is bounded from below.
A: The goal is to show that the solution is bounded. (I have no idea what the $+\infty$ bit is about, it adds nothing as far as I can tell.)
We first need to show that the solution from any initial condition $x_0$ (at $t_0$) is defined on $[t_0,\infty)$. It is not immediately clear to me that 'we have global existence'. After that, it is fairly straightforward to show boundedness using a 'Lyapunov' function of sorts.
Let $\phi(x) =-x -f(x)$. Since $f$ is $C^1$, we have that $\phi$ is $C^1$, hence Lipschitz on any bounded set. From the usual considerations of ODEs, it is straightforward to establish that, for any $B>0$, if $|x_0| < B$, then there exists a $\delta>0$ (which depends only on $B$ and $f$) such that the solution to $\dot{x} = \phi(x)$ with initial condition $x(t_0) = x_0$ is defined on $[t_0,t_0+\delta]$. If we can establish that $|x(t_0+\delta)| < B$, then it is clear that the solution is defined on $[t_0,\infty)$. Furthermore, the solution is $C^1$.
Using the bound provided for $f$, we have:
If $x\geq 8$, $\phi(x) \leq -x -f(x) \leq -x +\frac{1}{2} |x| +3 = -\frac{1}{2} x+3 \leq -1$.
If $x\leq -8$, $\phi(x) \geq -x -f(x) \geq-x -\frac{1}{2} |x| -3 = -\frac{1}{2} x-3 \geq +1$.
Now let $x$ be as a solution to $\dot{x} = \phi(x)$, and define the 'Lyapunov' function $V(t) = \frac{1}{2} x(t)^2$. Then it is easy to see that $V$ is $C^1$ and $\dot{V}(t) = x(t) \phi(t)$. Consequently, if $|x(t)| \geq 8$, then $\dot{V}(t) \leq -8$.
Thus, if $|x_0| \geq 8$, then the solution will satisfy $|x(t)| < 8$ in finite time.
Now suppose $|x_0| < 8$, and that at some $t \in [t_0,t_0+\delta]$, we have $|x(t)| \geq 8$. Let $t_1$ be the first time at which $|x(t_1)| = 8$ ($x$ is continuous, so this is well defined). Since $V$ is $C^1$, we have, for some suitably small $\epsilon>0$, $\dot{V}(t) \leq -7$, for $t \in [t_1-\epsilon, t_1]$. Since $|x(t_1-\epsilon)| < 8$, this contradicts $|x(t_1)| = 8$, since $V(t_1) = V(t_1-\epsilon)+\int_{t_1-\epsilon}^{t_1} \dot{V}(\tau) d \tau$. 
Hence $|x(t)| < 8$ for all $t \in [t_0,t_0+\delta]$, and from the considerations above, we see that $x$ is defined on $[t_0, \infty)$, and $|x(t)| < 8$, for all $t\geq t_0$.
If $\phi(x(t')) = 0$ for some $t'$, then it is clear that $x(t) = x(t')$, for all $t \geq t'$ (by uniqueness of solution). If $\phi(x(t')) \neq 0$ for all $t$, then it is clear that $\phi(x(t))$ has the same sign as $\phi(x(t'))$ for all $t\geq t'$. Hence either $x(t)$ is increasing and bounded above, or decreasing  and bounded below. Hence $\lim_{t \to \infty} x(t)$ exists (and is in $[-8,8]$, of course).
