Interesting integral: $\int_0^1{\frac{nx^{n-1}}{x+1}}dx$ Find the value of $$\int_0^1{\frac{nx^{n-1}}{x+1}}dx.$$
I had no luck while integrating it. I also tried differentiating w.r.t n but still couldn't reach anywhere. Need help.
 A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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\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.

A: Put $y=1+x$ and the integral becomes
$$ \int_{1}^2 \frac{n(y-1)^{n-1}}{y} \, dy = \int_1^2 \sum_{k=0}^n n\binom{n-1}{k} (-1)^{n-k-1} y^{k-1} \, dy = \left[ n(-1)^{n-1}\log{y} + \sum_{k=1}^n \frac{n}{k} \binom{n-1}{k} (-1)^{n-k-1} y^k \right]_1^2 \\
= n(-1)^{n-1}\log{2} + \sum_{k=1}^n \frac{n}{k} \binom{n-1}{k} (-1)^{n-k-1}(2^k-1). $$
A: 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 $\displaystyle J_m=\int_0^1\dfrac{x^m}{1+x}dx,$
$$J_{m+1}+J_m=?$$
A: We can write the integral as
$$
F(n) = \int_0^{\;1} {\frac{{nx^{n - 1} }}{{1 + x}}dx}  = \int_0^{\;1} {\frac{1}{{1 + x}}dx^n }  = \int_0^{\;1} {\left( {1 - \frac{x}{{1 + x}}} \right)dx^n }  = 1 - \int_0^{\;1} {\frac{x}{{1 + x}}dx^n } 
$$
$F(n+1)$ will be
$$
F(n + 1) = \int_0^{\;1} {\frac{{\left( {n + 1} \right)x^n }}{{1 + x}}dx}  = \int_0^{\;1} {\frac{x}{{1 + x}}dx^n }  + \int_0^1 {\frac{{x^n }}{{1 + x}}dx}  = 1 - F(n) + \frac{1}{{n + 1}}F(n + 1)
$$
i.e.:
$$
\begin{array}{l}
 \left( {1 - \frac{1}{{n + 1}}} \right)F(n + 1) = 1 - F(n) \\ 
 \frac{{F(n + 1)}}{{n + 1}} = \frac{1}{n} - \frac{{F(n)}}{n} \\ 
 \left( { - 1} \right)^{\,n} \frac{{F(n + 1)}}{{n + 1}} = \frac{{\left( { - 1} \right)^{\,n} }}{n} + \left( { - 1} \right)^{\,n - 1} \frac{{F(n)}}{n} \\ 
 \end{array}
$$
So we have
$$
\left\{ \begin{array}{l}
 G(n) = \left( { - 1} \right)^{\,n} \frac{{F(n + 1)}}{{n + 1}} \\ 
 G(0) = F(1) = \ln 2 \\ 
 G(n + 1) - G(n) = \frac{{\left( { - 1} \right)^{\,n + 1} }}{{n + 1}} \\ 
 \end{array} \right.
$$
which leads to
$$ \bbox[lightyellow] {  
\begin{array}{l}
 G(n) = \ln (2) + \sum\limits_{k = 0}^{n - 1} {\frac{{\left( { - 1} \right)^{\,k + 1} }}{{k + 1}}}  = \left( { - 1} \right)^{\,n} \Phi \left( { - 1,1,n + 1} \right) \\ 
 F(n) = n\;\Phi \left( { - 1,1,n} \right) \\ 
 \end{array}
}$$
where $\Phi$ denotes the Lerch Transcendent.
In fact, by the integral representation of $\Phi$
$$
\Phi \left( {z,s,a} \right)\mathop  \equiv \limits^{def} \frac{1}
{{\Gamma (s)}}\int_0^\infty  {\frac{{t^{\,s - 1} e^{\, - a\,t} }}
{{1 - z\,e^{\, - \,t} }}dt} 
$$
and
$$
\Phi \left( { - 1,1,n} \right) = \int_{t\, = \,0}^\infty  {\frac{{e^{\, - n\,t} }}
{{1 + \,e^{\, - \,t} }}dt} \quad \xrightarrow{{e^{\, - \,t}  = x}}\quad  - \int_{x = 1}^0 {\frac{{x^{\,n - 1} }}
{{1 + \,x}}dt} 
$$
