Evaluate the limit of $(n+1)\int_0^1x^n\ln(1+x)\,dx$ when $n\to\infty$

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 $$ 0 \leq \int_0^1 \frac{x^{n+1}}{x+1}dx \leq \int_0^1 \frac{x^{n+1}}{1}dx = \frac{1}{n+2} $$

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\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.