Let $x$ be a positive scalar variable whose time derivative satisfies $$\dot{x}(t)\leq \exp \left\{\left(\int_{0}^{t}\frac{1}{x(\tau)} \mathrm{d} \tau \right)\right\},$$ where $\cdot$ denotes the absolute value. Does the above inequality imply that $x(t)$ is a bounded function? (Note that we do not need to know about the explicit upper bound for $x(t)$ ). Thanks.
Integrate a few times: Using that $x > 0$ we have that $\dot{x}(t) \leq \exp (0) = 1$ which implies that $x(t) \leq M + t$ for some $M > 0$. Plug it back in we have $$ \dot{x}(t) \leq \exp \left(  \int_0^t \frac{1}{M+\tau} \mathrm{d}\tau \right) = \exp \left(  \ln (M+t) + \ln M\right) = \frac{M}{M+t} $$ this implies that $$ x(t) \leq M + M \left[\ln (M+t)  \ln M\right] $$ for which we can very, very roughly estimate by $$ x(t) \leq 2\sqrt{M' + t} $$ for a sufficiently large $M'$. Plugging this back in we have that $$ \dot{x}(t) \leq \exp \left(  \int_0^t \frac{1}{2\sqrt{M'+\tau}} \mathrm{d}\tau\right) = \exp \left( \sqrt{M'}  \sqrt{M' + t}\right) $$ Using that $e^{\sqrt{t}}$ is integrable from $0$ to $\infty$ to some constant, we have that $x$ must be bounded. 

