# Prove that $\int_{2017}^{2018} (\arctan(\ln(x))-\ln(\arctan(x)))\,\mathrm dx>0$

I want to prove that $$\int_{2017}^{2018} (\arctan(\ln(x))-\ln(\arctan(x)))\,\mathrm{d}x > 0.$$

Here is my idea so far:

By the fundamental theorem of calculus, the above mentioned function $$f(x) = \arctan(\ln(x))-\ln(\arctan(x))$$ is a continuous function at $[2017,2018]$ , therefore there exists a function $F(x)$ such that $F'(x)=f(x)$, therefore it is enough to prove that $$F(2018)-F(2017)>0$$

My idea was proving that $F(x)$ is an increasing function at $[2017,2018]$ by showing that $f(x)$ has positive values at $[2017,2018]$, but I always end up with complicated equations.

My aim is to prove that without using a calculator at all…

Help much appreciated!

For $x > \mathrm{e^2}$,$$\arctan(\ln x) > \arctan(\ln \mathrm{e^2}) = \arctan 2\\ >\arctan \sqrt{3} = \frac{π}{3} > 1 > \ln \frac{π}{2} > \ln(\arctan x).$$