Closed form of $\int_0^1 \frac{x^n}{1+x}\, dx$

Question

I am trying to evaluate the integral

$$\int_0^1 \frac{x^n}{1+x}\, dx, \;\;\; n \in \mathbb{N}$$

in a closed form.

I tried tackling it using Beta Form $\displaystyle \int_0^1 \frac{t^{n-1}+t^{m-1}}{(1+t)^{m+n}}\, dt$ but unfortunately it does not coincide with formula of Beta. Then I tried the sub $x=1/t-1$ to get the integral into a zero to infinity form but still I cannot get it into a form I wish. The only I could think of next is to apply incomplete Gamma or incomplete Beta. Any other hints? Could we attack it using contour integration when got in the form $0$ to $+\infty$?

Answer 1

Since you already seem to be quite familiar with both the beta and $\Gamma$ functions, allow me to

offer the following result: $~\displaystyle\int_0^1\frac{x^{2a}}{1+x}~dx~=~\frac{H_a-H_{a-\frac12}}2~,~$ where $a>-\dfrac12~$ and $~H_{k-1}~=$

$=~\psi_{_0}(k)~+~\gamma~$ is the generalized harmonic number, expressible in terms of the digamma

function and the Euler-Mascheroni constant.

Answer 2

You can treat this systematically using the following standard identity: $$ \frac{x^{n}+1}{x+1}=x^{n-1}-x^{n-2}+x^{n-3}\cdots +1, n=2k+1 $$ and $$ \frac{x^{n}-1}{x+1}=x^{n-1}-x^{n-2}+x^{n-3}\cdots -1, n=2k $$ So you essentially only need to integrate $\frac{1}{x+1}$ and various polynomial terms over $[0,1]$. But this should not be difficult for you. You can also establish a recurrence relationship if you needed.