**Alternative solution.**

We have the following bounds (easy)
$$\frac{1}{\sqrt{1 + s^2}} \le 1 - \frac{20}{69}s^2, \quad \forall s\in [0, 1];$$
and
$$\frac{1}{\sqrt{1 + s^2}} < \frac{1}{12}s^2 - \frac12 s + \frac98, \quad \forall s\in [1, \infty).$$

Thus, we have
\begin{align*}
\int_0^\infty \frac{1}{\mathrm{e}^{sx}\sqrt{1 + s^2}} \,\mathrm{d} s
&< \int_0^1 \frac{1 - 20s^2/69}{\mathrm{e}^{sx}}\,\mathrm{d} s
+ \int_1^\infty \frac{\frac{1}{12}s^2 - \frac12 s + \frac98}{\mathrm{e}^{sx}}\,\mathrm{d} s\\
&= \frac{69x^2 - 40}{69x^3}
+ \frac{-x^2 + 136x + 412}{552x^3\mathrm{e}^x}.
\end{align*}

It suffices to prove that
$$\frac{69x^2 - 40}{69x^3}
+ \frac{-x^2 + 136x + 412}{552x^3\mathrm{e}^x} \le \arctan \frac{1}{x}.$$

We split into two cases.

**Case 1.** $x\ge 1$ and $-x^2 + 136x + 412 < 0$

It suffices to prove that
$$\frac{69x^2 - 40}{69x^3} \le \arctan \frac{1}{x}.$$
Let
$$g(x) := \arctan \frac{1}{x} - \frac{69x^2 - 40}{69x^3}.$$
We have
$$g'(x) = - \frac{17x^2 + 40}{23x^4(x^2 + 1)} < 0.$$
Also, $\lim_{x\to\infty} g(x) = 0$.
Thus, $g(x) \ge 0$ for all $x \ge 1$.

**Case 2.** $x\ge 1$ and $-x^2 + 136x + 412\ge 0$

Since $\mathrm{e}^x = \mathrm{e}\cdot \mathrm{e}^{x-1}
\ge \mathrm{e}(1 + x - 1)$ and $\frac{1}{\mathrm{e}^x} \le \frac{1}{\mathrm{e}x} \le \frac{215389350}{584654827x}$, it suffices to prove that
$$f(x) := \arctan \frac{1}{x} - \frac{69x^2 - 40}{69x^3}
- \frac{-x^2 + 136x + 412}{552x^3}\cdot \frac{215389350}{584654827x} \ge 0.$$
We have
$$f'(x) = \frac{(33 - 25x)(1435929x^3 + 504096475x^2 - 516362220x + 896367800)}{26894122042x^5(x^2 + 1)}.$$
Since $1435929x^3 + 504096475x^2 - 516362220x + 896367800 > 0$ on $x\ge 0$, we have
$f'(x) > 0$ on $(1, 33/25)$ and $f'(x) < 0$ on $(33/25, \infty)$. Also, $f(1) > 0$ and $\lim_{x\to \infty} f(x) = 0$. Thus,
$f(x) \ge 0$ for all $x \ge 1$.

We are done.