Question:

**Step 1: Analyze the upper limit of the integral** The upper limit is $u(x) = x \tan^{-1} x$. Let's determine its range and sign.\nFor $x > 0$, $\tan^{-1} x > 0$, so $u(x) > 0$.\nFor $x < 0$, $\tan^{-1} x < 0$, so $u(x) = (-)(-) > 0$.\nFor $x = 0$, $u(0) = 0$.\nThus, $u(x) \ge 0$ for all $x \in \mathbb{R}$, and its minimum value is $0$ at $x=0$. **Step 2: Analyze the integrand** The integrand is $g(t) = \frac{e^{t - \cos t}}{1 + t^{2023}}$. Since $u(x) \ge 0$, the integral is always over a domain $[0, u(x)]$. For $t \ge 0$, $e^{t - \cos t} > 0$ and $1 + t^{2023} > 0$. Thus, $g(t) > 0$ for all $t \ge 0$. **Step 3: Determine the minimum value of the integral** The function $f(x) = \int_0^{u(x)} g(t) dt$. Since $g(t) > 0$ for $t \ge 0$, the integral represents the area under a positive curve from $0$ to $u(x)$. This area is strictly increasing with respect to $u(x)$. Therefore, $f(x)$ is minimized when $u(x)$ is minimized. The minimum of $u(x)$ is $0$, so the minimum of $f(x)$ is $\int_0^0 g(t) dt = 0$.