Improper integral

Question

$\int_{-1}^{1}\frac{e^{x}}{1+x}dx$

Anyone know how to solve this? I have tried to find a substitute, or a function to compare with, but I don't have any clue whatsoever what to do. Wondered if someone could give me a hint.

Answer 1

Hint:

As $\forall x: e^x\geq x$, we have:

$$\int_{a}^b\frac {e^x}{1+x}dx \geq \int_{a}^b \frac {x}{1+x}dx =\left(x-\ln(x+1)\right)\big|^{x=b}_{x=a} $$

Answer 2

Hint: $$ \int_{a}^1\frac{e^x}{x+1}\,dx\ge e^{-1}\int_{a}^{1}\frac{1}{1+x}\,dx $$ for $-1<a<1$.