Interesting integral: $\int_0^1{\frac{nx^{n-1}}{x+1}}dx$

Put $y=1+x$ and the integral becomes $$ \int_{1}^2 \frac{n(y-1)^{n-1}}{y} \, dy = \int_1^2 \sum_{k=0}^n n\binom{n-1}{k} (-1)^{n-k-1} y^{k-1} \, dy = \left[ n(-1)^{n-1}\log{y} + \sum_{k=1}^n \frac{n}{k} \binom{n-1}{k} (-1)^{n-k-1} y^k \right]_1^2 \\ = n(-1)^{n-1}\log{2} + \sum_{k=1}^n \frac{n}{k} \binom{n-1}{k} (-1)^{n-k-1}(2^k-1). $$

HINT:

If $\displaystyle I_n=\int_0^1\dfrac{nx^{n-1}}{x+1}dx,$

$$I_{m+1}+I_m=\int_0^1\dfrac{(m+1)x^m+mx^{m-1}}{x+1}dx$$ $$=m\int_0^1x^{m-1}\ dx+\int_0^1\dfrac{x^m}{1+x}dx=1+\int_0^1\dfrac{x^m}{1+x}dx$$

Again if $\displaystyle J_m=\int_0^1\dfrac{x^m}{1+x}dx,$

$$J_{m+1}+J_m=?$$

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \int_{0}^{1}{n\,x^{n - 1} \over x + 1}\,\dd x & = n\int_{0}^{1}{x^{n - 1} - x^{n} \over 1 - x^{2}}\,\dd x \,\,\,\stackrel{x^{2}\ \mapsto\ x}{=}\,\,\, {1 \over 2}\,n\int_{0}^{1}{x^{n/2 - 1} - x^{n/2 - 1/2} \over 1 - x}\,\dd x \\[5mm] & = {1 \over 2}\,n\bracks{% \int_{0}^{1}{1 - x^{n/2 - 1/2} \over 1 - x}\,\dd x - \int_{0}^{1}{1 - x^{n/2 - 1} \over 1 - x}\,\dd x} \\[5mm] & = \bbx{{1 \over 2}\,n\pars{H_{n/2 - 1/2} - H_{n/2 - 1}}} \end{align}

where $\ds{H_{z}}$ is the Harmonic Number.