For x ∈ ℝ , let tan - 1 x ∈ - π 2 , π 2 . Then the minimum value of the function…

Given,

$f\left(x\right)={\int }_0^{x{\tan }^{-1}x}\frac{e^{\left(t-\cos t\right)}}{1+t^{2023}}dt$

Now differentiating both side we get,

$f^{\text{'}}\left(x\right)=\frac{e^{\left[x{\tan }^{-1}x-\cos \left(x{\tan }^{-1}x\right)\right]}}{1+{\left(x{\tan }^{-1}x\right)}^{2023}}\times \left(\frac{x}{1+x^2}+{\tan }^{-1}x\right)$

$\Rightarrow f^{\text{'}}\left(x\right)=g\left(x\right)·h\left(x\right)$

$\text{ where }g\left(x\right)=\frac{e^{\left[x{\tan }^{-1}x-\cos \left(x{\tan }^{-1}x\right)\right]}}{1+{\left(x{\tan }^{-1}x\right)}^{2023}}>0 \mathrm{\forall } x$

And $h\left(x\right)=\frac{x}{1+x^2}+{\tan }^{-1}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}x \text{which is} \left\{\begin{array}{c}<0\text{ for }x<0\\ =0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}x=0\\ >0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}x>0\end{array}\right.$

$\therefore f\left(x\right)$ has minimum at $x=0$

And $f{\left(x\right)}_{min}=f\left(0\right)=0$