The set of critical points of the function $f(x)$ given by $f(x)=x-\log _e x+\int\limits_2^x\left(\frac{1}{t}-2-2 \cos 4 t\right) d t$, is

$\left\{\frac{\pi}{6}+\frac{n \pi}{2}: n=0,1,2, ....\right\}$

We have,

$f(x)=x-\log _e x+\int\limits_2^x\left(\frac{1}{t}-2-2 \cos 4 t\right) d t$

$\Rightarrow f'(x)=1-\frac{1}{x}+\left(\frac{1}{x}-2-2 \cos 4 x\right)$

$\Rightarrow f'(x)=-1-2 \cos 4 x$

Now,

$f'(x)=0$

$\Rightarrow \cos 4 x=-\frac{1}{2}=\cos \frac{2 \pi}{3}$

$\Rightarrow 4 x=2 n \pi \pm \frac{2 \pi}{3}, n \in Z \Rightarrow x=\frac{n \pi}{2} \pm \frac{\pi}{6}, n \in Z$

But, f(x) is defined for x > 0. Therefore,

$f'(x)=0 \Rightarrow x=\frac{n \pi}{2}+\frac{\pi}{6}, n \in N, x=\frac{\pi}{6}$

Hence, the set of critical points of f(x) is

$\left\{\frac{n \pi}{2}+\frac{\pi}{6}: n=0,1,2,...\right\}$