How is this derived? In my textbook I find the following derivation:
$$ \displaystyle \lim _{n \to \infty} \dfrac{1}{n} \displaystyle \sum ^n _{k=1} \dfrac{1}{1 + k/n} = \displaystyle \int^1_0 \dfrac{dx}{1+x}$$
I understand that it's $\displaystyle \int^1_0$ but I don't understand the $\dfrac{dx}{1+x}$ part. 

For any real-valued function $f$ continuous on an interval $[a,b]$ with $a<b$, we have $$\int_a^bf(x)\,dx=\lim_{n\to\infty}\left[\frac{b-a}n\sum_{k=1}^nf\left(a+\frac{k}{n}(b-a)\right)\right].$$
In this particular case, what is $f$? $a$? $b$?
P.S.: $\frac{dx}{1+x}$ is another way of writing $\frac{1}{1+x}\,dx$.
