$$\int_{2021}^{2022} \frac{\arctan(\frac{x}{2021})+\arctan(\frac{x}{2022})}{x} dx$$ I am sorry if the question lacks context. But I tried to solve it by using the KINGS rule in definite integration. It didnt help with it.
I also tried using following property: $\arctan(x)+\arctan(y)=\arctan\left(\frac{x+y}{1-xy}\right)$ but because of the denominator being $2021$ and $2022$ it increased calculations.
Let $u=\frac{ab}{x}$(this is King's rule for product) with $0<a<b$, then $$I=\int_{a}^{b} \frac{\arctan(\frac{x}{a})+\arctan(\frac{x}{b})}{x}{\rm{d}}x\\ =\int_{a}^{b} \frac{\arctan(\frac{a}{u})+\arctan(\frac{b}{u})}{u} {\rm{d}}u\\ =\int_a^b \frac{\arctan(\frac{a}{x})+\arctan(\frac{b}{x})}{x}{\rm{d}}x.\tag{1} $$ Also we have $$\arctan x+\arctan\frac1x=\frac{\pi}{2},\quad \forall x>0.\tag{2}$$ Hence, with $(1)$ and $(2)$, we have $$I=\frac{1}{2}\left(\int_{a}^{b} \frac{\arctan(\frac{x}{a})+\arctan(\frac{x}{b})}{x}{\rm{d}}x+\int_{a}^{b} \frac{\arctan(\frac{a}{x})+\arctan(\frac{b}{x})}{x}{\rm{d}}x\right)\\=\frac{\pi}{2}\int_{a}^{b}\frac{{\rm{d}}x}{x}= \frac{\pi}{2}\ln\frac{b}{a}.$$
Let $a=2021,b=2022$, we can get $$\int_{2021}^{2022} \frac{\arctan(\frac{x}{2021})+\arctan(\frac{x}{2022})}{x}{\rm{d}}x=\frac{\pi}{2}\ln\frac{2022}{2021}.$$
