$\lim_{n\rightarrow\infty} x(1/(n+x)+..+1/(n+nx))=\log(1+x)$ I've been wondering about the following problem for some time:

Show that for any positive value of $n$,
$$
\lim_{n\rightarrow\infty} x\left({1\over n+x}+{1\over n+2x}+...+{1\over n+nx}\right)=\log(1+x)
$$

[This is Example 15 of chapter one of Methods of Mathematical Physics by Jeffreys & Jeffreys, on page 55.]
One idea I had was that I could expand
$$
{1\over n+x}={1\over n}(1-x/n+(x/n)^2+...)
$$
but the sum of the various terms didn't lead to anything useful.
Adding successive terms in the form
$$
{1\over n+x}+{1\over n+2x} = {2n+3x\over (n+x)(n+2x)}
$$
etc. doesn't look promising. Any ideas?
 A: $$
x\,\Bigl(\frac{1}{n+x}+\frac{1}{n+x}+\dots+\frac{1}{n+n\,x}\Bigr)=\frac{x}{n}\,\sum_{k=1}^n\frac{1}{1+x\,\frac{k}{n}}\to x\int_0^1\frac{dt}{1+x\,t}.
$$
A: $$x\sum_{r=1}^n\frac1{n+rx}=\frac1n\sum_{r=1}^n\frac1{\dfrac1x+\dfrac rn}$$
Now see Find $\lim\limits_{n \to \infty} \frac{1}{n}\sum\limits^{2n}_{r =1} \frac{r}{\sqrt{n^2+r^2}}$
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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${\ds{\tt\mbox{Besides the simple 'Riemann-Sum Method'}}}$:

\begin{align}&\color{#66f}{\large%
\lim_{n\ \to\ \infty}\bracks{x\sum_{r\ =\ 1}^{n}{1 \over n + rx}}}
=\lim_{n\ \to\ \infty}\bracks{x\sum_{r\ =\ 1}^{n}\int_{0}^{1}t^{n + rx - 1}\,\dd t}
\\[5mm]&=\lim_{n\ \to\ \infty}\bracks{%
x\int_{0}^{1}t^{n - 1}\sum_{r\ =\ 1}^{n}\pars{t^{x}}^{r}\,\dd t}
=\lim_{n\ \to\ \infty}\bracks{%
x\int_{0}^{1}t^{n - 1}t^{x}\,{t^{nx} - 1 \over t^{x} - 1}\,\dd t}
\\[5mm]&=\lim_{n\ \to\ \infty}\bracks{%
x\int_{0}^{1}{t^{1 + \pars{n - 1}/x} - t^{n + 1 + \pars{n - 1}/x}\over 1 - t}\,
{1 \over x}\,t^{1/x - 1}\,\dd t}
\\[5mm]&=\lim_{n\ \to\ \infty}\bracks{%
\int_{0}^{1}{1 - t^{n\pars{1 + 1/x}}\over 1 - t}\,\dd t
-\int_{0}^{1}{1 - t^{n/x}\over 1 - t}\,\dd t}
\\[5mm]&=\lim_{n\ \to\ \infty}\braces{%
\Psi\pars{n\bracks{1 + {1 \over x}} + 1}
-\Psi\pars{{n \over x} + 1}}
\\[5mm]&=\lim_{n\ \to\ \infty}\braces{%
\ln\pars{n\bracks{x + 1} + x \over x} - \ln\pars{n + x \over x}}
=\lim_{n\ \to\ \infty}\ln\pars{n\bracks{x + 1} + x \over n + x}
\\[5mm]&=\color{#66f}{\large\ln\pars{x + 1}}
\end{align}
