fine the limit : $\lim_{ n \to \infty }\frac{1}{n}\int_{0}^{n}{ \frac{x\ln(1+\frac{x}{n})}{1+x}}=?$ fine the limit :
$$\lim_{ n \to \infty }\frac{1}{n}\int_{0}^{n}{ \frac{x\ln(1+\frac{x}{n})}{1+x}}=?$$
My Try:
in the http://www.integral-calculator.com
$$I=\int_{}^{}{ \frac{x\ln(1+\frac{x}{n})}{1+x}}=\left(x+n\right)\ln\left(\left|x+n\right|\right)+\left(\ln\left(n\right)-\ln\left(n-1\right)\right)\ln\left(\left|x+1\right|\right)+\operatorname{Li}_2\left(-\dfrac{x+1}{n-1}\right)+\left(-\ln\left(n\right)-1\right)+c$$
now?
 A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\lim_{n \to \infty}\bracks{%
{1 \over n}\int_{0}^{n}{x\ln\pars{1 + x/n} \over 1 + x}\,\dd x} =
\lim_{n \to \infty}\bracks{%
{1 \over n}\int_{0}^{n}\ln\pars{1 + {x \over n}}\,\dd x -
{1 \over n}\int_{0}^{n}{\ln\pars{1 + x/n} \over 1 + x}\,\dd x}
\\[5mm] = &\
2\ln\pars{2} - 1\ -\
\underbrace{\lim_{n \to \infty}\bracks{%
{1 \over n}\int_{0}^{1}{\ln\pars{1 + x} \over 1/n + x}\,\dd x}}_{\ds{=\ 0}}\ =\
\bbx{\ds{2\ln\pars{2} - 1}}
\end{align}

because

$$
0 < \verts{{1 \over n}\int_{0}^{1}{\ln\pars{1 + x} \over 1/n + x}\,\dd x} <
{1 \over \verts{n}}\verts{\int_{0}^{1}{\ln\pars{1 + x} \over \color{#f00}{0} + x}\,\dd x}
\,\,\,\stackrel{\mrm{as}\ n\ \to\ \infty}{\to}\,\,\, {\large 0}
$$
A: *

*Step 1: make the change of variable $u=\frac{x}{n}$, so that
$$
\frac{1}{n}\int_{0}^{n}{ \frac{x\ln(1+\frac{x}{n})}{1+x}}
= \int_0^1 \frac{nu}{1+nu}\ln(1+u)du
$$

*Step 2: letting $(f_n)_n$ be defined on $[0,1]$ by $f_n(u) =  \frac{nu}{1+nu}\ln(1+u)$, observe that (1) each $f_n$ is integrable on $[0,1]$, (2) $\lvert f_n(u)\rvert \leq g(u)\stackrel{\rm def}{=} \ln(1+u)$ for every $u\in[0,1]$ and $n\geq 1$, and (3) $(f_n)_n$ converges pointwise to $g$ on $(0,1)$.
By the Dominated Convergence Theorem, it follows that
$$
\int_0^1 f_n(u)du \xrightarrow[n\to\infty]{} \int_0^1 g(u)du
$$

*Step 3: compute
$$
\int_0^1 g(u)du = 2\ln 2 -1.
$$
observing that an antiderivative of $g$ is $G(x) = (1+x)\ln(1+x) - x$.
A: Not a new idea, but avoiding dominated convergence: 
$$\frac{1}{n} \int_0^n \frac{ x \ln (1+x/n) }{1+x} dx \underset{y=x/n}{=}\int_0^1   \ln (1+y)\frac{1+ny-1 }{1+ny} dy =\int_0^1 \ln (1+y) dy - \underset{(*)}{\underbrace{\int_0^1 \frac{\ln (1+y)}{1+ny}dy}}. $$ 
Now 
$$0\le (*) \le \ln 2 \int_0^1 \frac{dy }{1+yn}=\ln 2 \frac{1}{n} \ln (1+n)\underset{n\to\infty}{\to} 0.$$ 
Therefore the answer is $\int_0^1\ln (1+y) dy=\ln 4 - 1$. 
