Does this integral converge? If so, then how to evaluate it? I encountered the following integral when I was practicing some problems. Here it is :
$$\int_0^1 \frac{x^n}{1+x} dx$$ I have absolutely no idea on how to proceed with this integral. I tried taking the absolute value of the integral and then comparing this integral with some related integrals like:
$$\left\vert \int_0^1 \frac{x^n}{1+x} dx\right\vert \leq \int_0^1 \left\vert \frac{x^n}{1+x}\right\vert dx$$ But since we are integrating between $0$ and $1$, the $1+x$ has to be greater than $1$ and thus can be removed and so I came up with this:
$$\int_0^1 \left\vert\frac{x^n}{1+x}\right\vert dx \leq \underbrace{\int_0^1 x^n dx}_{\frac{1}{n+1}}$$ And so our argument thus becomes :
$$\left\vert \int_0^1 \frac{x^n}{1+x} dx\right\vert \leq \int_0^1 \left\vert \frac{x^n}{1+x}\right\vert dx \leq \frac{1}{n+1}$$ This is all I did and don't know how to proceed further
Any help/hint is appreciated. Thanks in advance
 A: HINT
I am going to assume that $n\in\mathbb{N}$ in this answer.
If $n$ is odd, then
$$\begin{align}\int_0^1 \frac{x^n}{1+x} dx&=\int_0^1\frac{x^n+x^{n-1}-x^{n-2}-x^{n-3}+\cdots+x+1-1}{x+1}dx\\
&=\int_0^1 x^{n-1}-x^{n-2}+\cdots-x+1-\frac{1}{x+1}dx\\
&=\left[\frac{1}{n}x^n-\frac{1}{n-1}x^{n-1}+\cdots-\frac{1}{2}x^2+x-\ln(1+x)\right]_0^1\\
&=\sum_{k=1}^n\frac{(-1)^{n+1}}{n}-\ln2.
\end{align}$$
Incidentally, since the integrand is always positive in the interval $[0,1]$ this demonstrates rather nicely that $$\ln 2<\sum_{k=1}^n\frac{(-1)^{n+1}}{n}$$ for any odd positive integer $n$.
Can you cover the case when $n$ is even on your own now?

I hope that helps. If you have any questions please don't hesitate to ask :-)
A: Definitely not an answer. Maybe I'm completely off here, but it does look like
$$ \int _0 ^1 x^{a-1}(1+x)^{b-1}dx =: B(a,b) $$
provided $Re(a),Re(b)>0$. So for small $\alpha>0$ we would have this written as $$\int _0^1 x^{(n+1)-1}(1+x)^{\alpha-1} dx = B(n+1,\alpha) $$
But I have no idea how this behaves as $\alpha\to 0$.
