Consider a function $h : \mathbb{R} \rightarrow \mathbb{R}$ without singularities such that $h(x) \nearrow \infty$ with $x \rightarrow \infty$. Let $g(x)$ defined by $$ g(x) = \exp(-h(x)) \int_0^x \exp(h(s)) ds. $$
Playing around with Desmos, it seems that if $h(x)$ grows at approximately a linear rate, that is to say $\lim_{x \rightarrow \infty} \frac{h(x)}{x} = c \in \mathbb{R}$, then $g(x)$ remains bounded away from both $0$ and $\infty$. Meanwhile, if $h(x)$ grows faster than every line, then $g(x) \rightarrow 0$, and if $h(x)$ grows slower than every line, $g(x) \rightarrow \infty$.
I believe I have a (partial) proof in the case that $h(x)$ is strictly increasing and differentiable, but I suspect this holds in a much more general setting. How can this be shown? Does there exists results that are related to functions of this form?
Edit: This was my approach for the case $h(x)$ being differentiable and strictly increasing at some minimum rate $\alpha > 0$. Write $h(x) = ax + k(x)$, for $a > 0$ and for some $k(x)$ differentiable and $o(x)$. Since $h(x)$ is non-singular, we have that $\sup_{x \in \mathbb{R}} k'(x) \leq m_1$ (this seems questionable, actually). Thus, we have that $$ \frac{d}{dx} e^{h(x)} \leq (a + m_1) e^{h(x)}. $$
Integrating this becomes $$ e^{h(x)} - e^{h(0)} \leq \int_0^x (a + m_1) e^{h(s)} ds \qquad \Rightarrow \qquad g(x) = e^{-h(x)} \int_0^x e^{h(s)} ds \geq \frac{1 - e^{h(0) - h(x)}}{a + m_1}. $$
Since $h(x)$ is everywhere increasing with a minimum rate $\alpha > 0$, we have that $\inf_{x \in \mathbb{R}} k'(x) \geq m_2$ with $a > m_2$. Performing the same as above and taking $x \rightarrow \infty$, we finally obtain that $$ 0 < \frac{1}{a + m_1} \leq \lim_{x \rightarrow \infty} g(x) \leq \frac{1}{a + m_2} < \infty. $$
I have yet to show that this behaves as predicted if $h(x)$ grows at non-linear rates.
Edit 2: As EnEm pointed out in the comments, it is not difficult to find a function $h(x)$ that grows faster than any line but for which $g(x)$ does not go to $0$. It still remains very likely that $h(x)$ with linear growth still forces $g(x)$ to avoid $0$ and $\infty$ under fairly weak conditions.