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.
The sum is a Riemann Sum for the given integral. As $n\to\infty$, $$ \sum_{k=1}^n\frac1{1+k/n}\frac1n $$ tends to the sum of rectangles $\frac1{1+k/n}$ high and $\frac1n$ wide. This approximates the integral $$ \int_0^1\frac1{1+x}\mathrm{d}x $$ where $x$ is represented by $k/n$ and $\mathrm{d}x$ by $\frac1n$.
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 $\frac{1}{1+x}\,dx$.
The sum $\dfrac{1}{n} \sum\limits ^n _{k=1} \dfrac{1}{1 + \frac{k}{n}}$ is a Riemann sum of the function $f(x)=\dfrac{1}{1+x}$ with respect to the partition $0<\dfrac{1}{n}<\dfrac{2}{n}<\ldots<\dfrac{k}{n}\ldots<\dfrac{n}{n}=1.$
The integral $\int_0^1{\frac{1}{x+1}}dx$ exists, therefore the lower sum is equal to the upper sum, which in turn is equal to this integral. If we take the partition $\{\frac{k}{n}: k=0,1,..n \}$ of $[0,1]$ then the by taking the lower sums of $f(x)=\frac{1}{1+x}$ we have: $$L_n=\sum_{k=1}^{n}Δχ_κminf(χ_κ)=\sum_{k=1}^{n}\frac{1}{n}\frac{1}{1+\frac{k}{n}}$$ If you take limits, on the left side you are going to have the integral.
