Converting a complex Riemann sum into a definite integral I'm having some trouble converting this Riemann sum into a definite integral on the interval [0,1]. Any help/explanations would be much appreciated. Thank you. 
$$\sum_{i=1}^n   \frac{1}{n(2+\frac in)ln(2+\frac in)}$$
 A: Hint. One may obtain, as $n \to \infty$,
$$
\sum_{i=1}^n   \frac{1}{n(2+\frac in)\ln(2+\frac in)} \to \int_0^1\frac{dx}{(2+x)\ln(2+x)}=\int_0^1\frac{(\ln(2+x))'}{\ln(2+x)}\:dx
$$ where we have applied the Riemann sum result
$$
\sum_{i=1}^n \frac1n  f\left(\frac in\right) \to \int_0^1f(x)\:dx
$$ to
$$
f(x)=\frac{1}{(2+x)\ln(2+x)}.
$$
A: $\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
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 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

If you like an 'intuitive approach', write:
  \begin{align}
{i \over n} & \equiv x_{i}\quad\mbox{and}\quad
\pars{~\Delta x \equiv x_{i + 1} - x_{i} = {1 \over n} \implies n\Delta x = 1~}
\end{align}

Your sum becomes:
\begin{align}
&\sum_{i = 1}^{n}{1 \over n\pars{2 + i/n}\ln\pars{2 + i/n}} =
\sum_{i = 1}^{n}{1 \over n\pars{2 + x_{i}}\ln\pars{2 + x_{i}}}\,n\Delta x =
\sum_{x_{i}\ \in\ \Omega}{\Delta x \over \pars{2 + x_{i}}\ln\pars{2 + x_{i}}}
\\ & \mbox{where}\quad\Omega \equiv \braces{{1 \over n},{2 \over n},\ldots,1}
\end{align}
As $\ds{n \to \infty}$, $\ds{\Delta x \to 0}$ such that you'll arrive to an integral. The rest of the history is given in $\texttt{@Olivier Oloa}\,\,\,$ answer.
