Answer
$\int_{0}^{1} \frac{e^x}{1+x} dx$
Work Step by Step
We know the following
$\int_{a}^{b} f(x)dx=\lim\limits_{n \to \infty}\sum\limits_{i=1}^{n} f(x_{i})\Delta x$
$\Delta x=\frac{a-b}{n}$
$x_{i}=a+i\Delta x$
In the problem we see that $f(x)=\frac{e^x}{1+x}$
and we are given the interval of [0,1] therefore the integral will be from 0 to 1.
Therefore
$\lim\limits_{n \to \infty}\sum\limits_{n=1}^{n} f(\frac{e^{x_i}}{1+x_i})\Delta x=\int_{0}^{1} \frac{e^x}{1+x} dx$
