Evaluate the limit of $(n+1)\int_0^1x^n\ln(1+x)\,dx$ when $n\to\infty$ 
Evaluate the following limit : $$\lim_{n\to \infty}\left[(n+1)\int_0^1x^n\ln(1+x)\,dx\right].$$

We have , $$\lim_{n\to \infty}\left[(n+1)\int_0^1x^n\ln(1+x)\,dx\right]$$
$$=\lim_{n\to \infty}\int_0^1\ln(1+x)\,d(x^{n+1})$$Now put , $x^{n+1}=y$. Then , $$=\lim_{n\to \infty}\int_0^1\ln\left(1+y^{\frac{1}{n+1}}\right)\,dy$$Let , $\displaystyle g_n(y)=\ln\left(1+y^{\frac{1}{n+1}}\right)$.

Edit  :
Then , $\displaystyle g(y)=\lim_ng_n(y)=\ln 2$ in $(0,1]$. 
Now , $\displaystyle \sup_{x\in (0,1]}|g_n(y)-g(y)|=\sup_{x\in (0,1]}\ln\left(\frac{1+y^{\frac{1}{n+1}}}{2}\right)=0$. ( As , $y^{\frac{1}{n+1}}$ is monotone increasing function in $(0,1]$ , so $\ln\left(\frac{1+y^{\frac{1}{n+1}}}{2}\right)$ is also monotone increasing in $(0,1]$ and so it attains its maximum value at $y=1$ . ) So , $\{g_n(y)\}$ converges uniformly to $\ln 2$.

Then ,we can show that $g_n(y)$ converges uniformly to $\ln 2$ in $(0,1]$ and hence the given limit is $\ln 2$.

Is this correct ? Does there any other technique to evaluate the limit ?

 A: It seems that you don't actually have uniform convergence, as you claim.  In particular, $x^{1/n}$ does not converge to $1$ uniformly on $(0,1]$.  However, you have monotone convergence, which is sufficient by the monotone convergence theorem.
Another way to evaluate the limit: with integration by parts, we have
$$
\int_0^1(n+1)x^n \ln(x+1)\,dx = 
(1)^{n+1}\ln(2) - \int_0^1 \frac{x^{n+1}}{x+1}dx
$$
And, we have
$$
0 \leq \int_0^1 \frac{x^{n+1}}{x+1}dx \leq \int_0^1 \frac{x^{n+1}}{1}dx = \frac{1}{n+2}
$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
&\bbox[10px,#ffd]{\lim_{n\to \infty}
\bracks{\pars{n + 1}\int_{0}^{1}x^{n}\ln\pars{1 + x}\,\dd x}} =
\lim_{n\to \infty}
\bracks{\pars{n + 1}\int_{0}^{1}\pars{1 - x}^{n}\ln\pars{2 - x}\,\dd x}
\\[5mm] = &\
\lim_{n\to \infty}
\bracks{\pars{n + 1}\int_{0}^{1}\exp\pars{n\ln\pars{1 - x}}
\ln\pars{2 - x}\,\dd x}
\\[5mm] = &\
\lim_{n\to \infty}
\bracks{\pars{n + 1}\int_{0}^{\infty}\exp\pars{-nx}\ln\pars{2}\,\dd x}
\qquad\pars{~Laplace'\!s\ Method~}
\\[5mm] = &\
\ln\pars{2}\lim_{n \to \infty}\bracks{\pars{n + 1}{1 \over n}} =
\bbx{\ln\pars{2}}
\end{align}

See Laplace's Method.

A: Let $U_1,U_2,\dots$ be i.i.d. uniform (0,1) random variables and set 
$M_n=\max(U_1,\dots, U_{n})$. Then $M_{n+1}$ has density $(n+1) x^n$ for $x\in(0,1)$. As $n\to\infty$ we have  $M_{n+1}\to 1$ in distribution, so 
$$\int_0^1 (n+1) x^n \ln(1+x)\,dx=\mathbb{E}(\ln(1+M_{n+1}))\to \ln(1+1)=\ln(2).$$

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