How to prove that $\frac{e-1}{2e} \le \int_0^1 \frac{e^{-x}}{1+x}dx \le \ln 2$

Question

Prove: $$ \ \frac{e-1}{2e} \le \int_0^1 \frac{e^{-x}}{1+x}dx \le \ln 2 $$

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Attempt:

I can clearly see that $\ 2e \ge 1+x \ge 1 $ for every $\ 0 \le x \le 1 $ but $\ \frac{e^{-x}}{1+x} \ge \ln 2 $ for $\ x = 1 $

Answer 1

First inequality: in the domain, $\frac{1}{1+x} \geq \frac{1}{2}$. Thus $\int_0^1{\frac{e^{-x}}{1+x}} \geq \frac{1}{2}\int_0^1{e^{-x}}=\frac{e-1}{2e}$.

Second inequality: in the domain, $e^{-x} \leq 1$. Thus $\int_0^1{\frac{e^{-x}}{1+x}} \leq \int_0^1{\frac{1}{1+x}} = \log(2)$.