Evaluating $ \int_{-1}^{1} \arctan \left(e^{x}\right) d x $

Question

Using the substitution x=ln(t), the integral becomes $$ I=\int_{e^{-1}}^{e} \frac{\operatorname{Arctan}(t)}{t} \cdot d t $$ which has no antiderivative expressed with usual functions.

1. First, i want to prove that $$\frac{\operatorname{Arctan}(t)}{t} $$ has no antiderivative. I thought about using Darboux theorem, but this function is unfortunately continuous on R.
2. Can we really evaluate the integral I, even when we don't know an antiderivative of the function? I tried expressing the integral in term of limit of Riemann sums $$ \int_{-1}^{1} \operatorname{arctan}\left(e^{x}\right) \cdot d x=\lim _{n \rightarrow+\infty} \frac{2}{n} \cdot \sum_{k=0}^{n} \operatorname{Arctan}\left(e^{-1+k\left(\frac{2}{n}\right)}\right) $$ but i don't know how to do with k . Usually when we can't find an antiderivative to evaluate an integral, we try to bound the integral by the "mean formula". When applying this to my function i get the following:
   As our function f is continuous, $$ \begin{aligned} \exists c \in[-1 ; 1]: \int_{1}^{1} \operatorname{Arctan}\left(e^{x}\right) \cdot d x=\operatorname{Arctan}\left(e^{c}\right) \cdot(1-(-1)) \\&=2 \text { Arctan }\left(e^{c}\right) \\ (-1 \leqslant c \leqslant 1) \Rightarrow\left(e^{-1} \leq e^{c} \leq e\right) & \\ \Rightarrow\left(2 \cdot \operatorname{Arctan}\left(e^{-1}\right)\right.&\leqslant I \leqslant 2 \text { Arctan}\left(e^{1}\right) \end{aligned} $$ . Is this last suggestion sufficient to answer the question?

Answer 1

The key point is that for $x\in \mathbb{R}$, $\arctan(e^{-x})=\arctan\left(\frac{1}{e^x}\right)=\frac{\pi}{2}-\arctan{e^x}$.

$$\begin{array}{ccc} \int_{-1}^1\arctan(e^x)dx&=&\int_{-1}^{0}\arctan(e^x)\,dx+\int_{0}^{1}\arctan(e^x)\,dx\\ &=&\int_{1}^{0}-\arctan{e^{-t}}dt+ \int_{0}^{1}\arctan(e^x)\,dx\\ &=&\int_{0}^{1}\arctan(e^x)+\arctan(e^{-x})\,dx\\ &=&\int_{0}^{1}\frac{\pi}{2}\,dx\\ \end{array}$$

Therefore $$\int_{-1}^1\arctan(e^x)dx=\frac{\pi}{2}$$