Let $x(t),t\in [1,\infty)$ be a nondecreasing positive function satisfying the following inequality: $$ x'(t) \le \int_t^{+\infty} x(s)\frac{k(s)}{s^2}\,ds, $$ for any $t \ge 1$, where $k(t),t\in [1,\infty)$ is a nonincreasing positive function such that $$ \int_1^{+\infty}\frac{k(s)}{s}\,ds <\infty. $$
Can we prove that $x(t)$ is a bounded function?
The general argument is similar to what I gave in my previous answer, which was essentially off by a log due to certain inefficiencies in the estimates. Here I rewrite the argument to get rid of the log loss.
To start, as explained in the previous answer, we immediately have that $x(t)$ is sublinear. Since the inequality concerning $x(t)$ is linear, we can assume, without loss of generality, that $\sup x(t)/t \leq 1$ by a simple rescaling of $x$. This means we can find a sequence of times $1 = t_0, t_1, t_2, \ldots$ defined by
$$ t_i = \inf \{ t\in [1,\infty) : \forall s \geq t, x(s) \leq 2^{-i} s \}. $$
(Note that implicitly since $x$ is differentiable it is also continuous, so $x(t_i) = 2^{-i} t_i$.)
Our goal is to estimate $t_i$. Specifically, we want to show that $2^{-i}t_i$ is bounded.
We will also denote by
$$ K(t) = \int_t^\infty \frac{k(s)}{s} ~ds, \quad K_i = \int_{t_i}^{t_{i+1}} \frac{k(s)}{s} ~ds.$$
We note that $K_i \searrow 0$ and the numbers are in fact summable by assumption.
Integrating by parts the differential inequality for $x'$ we get (as I argued in the previous answer) for $1 \leq a < b$
$$ x(b) - x(a) \leq (b-a) \int_b^\infty \frac{x(s)}{s} \frac{k(s)}{s} ~ds + \int_a^b x(s) \frac{s - a}{s} \frac{k(s)}{s} ~ds $$
Estimating $(s-a)/s \leq 1$ we have that, by Gronwall's inequality
$$ x(b) \leq \left[ x(a) + (b-a) \int_a^\infty \frac{x(s)}{s} \frac{k(s)}{s} ~ds\right] \cdot e^{K(a) - K(b)} $$
This implies, setting $b = t_{i+1}$ and $a = t_i$, that
$$ 2^{-1-i} t_{i+1} \leq \left[ 2^{-i} t_i + (t_{i+1} - t_i) \sum_{j = i}^\infty 2^{-j} K_j \right] e^{K_{i}} $$
(here we rewrote $\int_{t_{i}}^\infty x(s) k(s) s^{-2} ~ds = \sum_{j = i}^\infty \int_{t_j}^{t_{j+1}} x(s)s^{-1} \cdot k(s) s^{-1} ~ds$, and used the decaying bound on $x(s)s^{-1}$ above $t_j$, and the fact that all functions involved are positive.)
Simplify (by the summability of $K_j$ we can assume from here on the indices $i$ are always larger than some sufficiently large $i_0$ such that the two terms in the brackets below are guaranteed to be positive)
$$ \left[ e^{-K_{i}} - \sum_{j = 0}^\infty 2^{-j} K_{i+j} \right] t_{i+1} \leq \left[2 - \sum_{j = 0}^\infty 2^{-j} K_{i+j} \right] t_i $$
So for all sufficiently large $i$ we have the bound (using the convexity of the exponential function)
$$ t_{i+1} \leq 2 t_i \cdot \frac{1}{e^{-K_{i}} - \sum_{j = 0}^\infty 2^{-j} K_{i+j}} \leq \frac{2 t_i}{1 - K_i - \sum_{j = 0}^\infty 2^{-j} K_{i + j}} $$
To show our desired conclusion it suffices to show that the infinite product
$$ \prod_{i = i_0}^\infty (1 - K_i - \sum_{j = 0}^\infty 2^{-j} K_{i+j}) $$
is bounded below away from zero. Now, the summability of $K_i$ implies that we can choose $i_0$ sufficiently large that $\sum_{i = i_0}^\infty K_i < \frac12$. This implies
$$ \sum_{i = i_0}^\infty \ln (1 - K_i - \sum_{j = 0}^\infty 2^{-j} K_{i+j}) \geq - 2\ln 2 \sum_{i = i_0}^\infty \left( K_i + \sum_{j = 0}^\infty 2^{-j} K_{i + j} \right) $$
the first term in the sum is obviously bounded by the summability of $K_i$. For the second term we interchange the order of summation
$$ \sum_{i = i_0}^\infty \sum_{j = 0}^\infty 2^{-j} K_{i + j} = \sum_{j = 0}^\infty \sum_{i = i_0}^\infty 2^{-j} K_{i + j} \leq 2 \sum_{i = i_0}^\infty K_{i+1} < \infty$$
and is also bounded. This concludes the proof.
The argument below doesn't give boundedness.
However, it gives that under the assumption $\int_1^\infty x(s) k(s) s^{-2} ds$ converges, $x(t) = o(t^\lambda)$ for any $\lambda > 0$.
With stronger assumptions on $k$ the argument can also imply boundedness.
First one observes that if the integral is finite, the growth rate is sublinear. This follows from the fact that if the integral $\int_1^\infty x(s) k(s) s^{-2} ds$ converges, then $x'(t) \to 0$ as $t\to\infty$, and hence $x(t) = o(t)$ by, e.g., L'Hopital.
Integrating the bound for $x'(t)$ gives, for arbitrary $1 \leq t_1 < t_2$
$$ x(t_2) - x(t_1) \leq \int_{t_1}^{t_2} \int_t^\infty x(s) k(s) s^{-2} ds~dt $$
Integrating in parts in $t$, using that $dt = d(t - t_1)$
$$ x(t_2) - x(t_1) \leq (t_2 - t_1) \int_{t_2}^\infty x(s) k(s) s^{-2} ds + \int_{t_1}^{t_2} (t - t_1) x(t) k(t) t^{-2} dt $$
Now, for any $\delta, \beta> 0$, there exists $\tau_{\delta,\beta}$ such that for every $s \geq \tau_{\delta,\beta}$, it holds that
$$ x(s) \leq \delta s, \qquad \text{and} \qquad \int_{s}^\infty k(s)s^{-1} ds < \beta $$
since we know that $x$ grows sublinearly and the relevant integral converges. This implies that, for $t_2 > \tau_{\delta,\beta}$,
$$ x(t_2) \leq x(\tau_{\delta,\beta}) + (t_2 - \tau_{\delta,\beta}) \delta \beta + \int_{\tau_{\delta,\beta}}^{t_2} x(s) \cdot k(s) s^{-1} ds $$
By Gronwall's inequality this means that
$$ x(t_2) \leq [ \delta \tau_{\delta,\beta} + (t_2 - \tau_{\delta}) \delta \beta ] e^{\beta} $$
This final inequality can be used to estimate $\tau_{\delta/2,\beta}$. Consider the inequality
$$ [\delta \tau_{\delta,\beta} + (s - \tau_{\delta,\beta}) \delta\beta] e^{\beta} \leq \frac{\delta}{2} s \tag{A}$$
Solving this inequality we see that this is satisfied whenever
$$ \frac{e^{\beta} \tau_{\delta,\beta}}{\frac12 - \beta e^{\beta}} \leq s $$
For convenience write
$$ \frac{e^\beta}{\frac12 - \beta e^\beta} = 2^{1+\sigma(\beta)} $$
and note that $\sigma$ is continuous and $\sigma(0) = 0$.
This implies that $\tau_{\delta/2,\beta} \leq 2^{1 + \sigma(\beta)} \tau_{\delta}$.
Now, iterating this argument, starting from some sufficiently small $\delta$ that we fix, we see can construct an increasing sequence of times $\tau_{2^{-k} \delta,\beta}$ such that for $T \geq \tau_{2^{-k} \delta,\beta}$ it holds
$$ x(T) \leq 2^{-k} \delta T $$
We also have by iteration the bounds
$$ \tau_{2^{-k}\delta,\beta} \leq 2^{k + k\sigma(\beta)} \tau_{\delta,\beta} $$
So we have
$$ x(T) \lesssim_{\delta,\beta} T^{\sigma(\beta)} $$
One can upgrade the estimate to get boundedness if one knows the decay rate of $\int_t^\infty k(s) s^{-1} ds$.
For example, if one knows that this integral is bounded by $C/\ln(t)^2$, then Gronwall will imply an estimate along the lines of
$$ (\tau_{\delta} + (s - \tau_\delta) \frac{C}{(\ln \tau_{\delta})^2}) e^{C/(\ln \tau_\delta)^2} < \frac12 s $$
replacing (A).
Now, by our estimates in the previous section we would've found that
$$ \tau_{2^{-k\delta}} \leq 2^{k(1 + \sigma)} \tau_{\delta} $$
for some $\sigma$. Bootstrapping from this we would get that
$$ \tau_{2^{-k} \delta} \leq 2^{1 + O(k^{-2})} \tau_{2^{-(k-1)} \delta} $$
The summability of the series $(k^{-2})$ then gives boundedness of $x(t)$.
[This argument can carry through as long as we know, for example, that $k(s) \lesssim (\ln (1+s))^{-2-\gamma}$ for some $\gamma > 0$; but seemingly fails for $k(s) = (\ln(1+s))^{-2}$.]
Without requiring that the integral \begin{equation} \int_1^\infty x(s)\frac{k(s)}{s^2}\,ds \tag{1} \end{equation} be finite, the answer is no. Indeed, then one can take e.g. $k(s)=1/s$ and $x(s)=e^s$.
I don't know whether the condition that the integral in (1) be finite changes the answer.
