<s>[INST] Asymptotic expansion at order 2 of $\int_0^1 \frac{x^n}{1+x} \, dx$ I'd like to get an asymptotic expansion of $\int_0^1 \frac{x^n}{1+x} \, dx$ at order two in $\frac{1}{n}$.
I'm able to prove that $$\lim\limits_{n \to \infty} n \int_0^1 \frac{x^n}{1+x} \, dx = \frac{1}{2}$$ which provides an asymptotic expansion at order $1$. How can I go one step further? Even better, is there a way to get an asymptotic expansion at any order $m$?
 [/INST] Method 1.
An elementary approach.
You may just integrate by parts twice,
$$
\begin{align}
I_n&=\int_0^1\frac{x^n}{1+x}\:dx
\\&=\left. \frac{x^{n+1}}{(n+1)}\frac{1}{1+x}\right|_0^1+\frac{1}{(n+1)}\int_0^1\frac{x^{n+1}}{(x+1)^2}\:dx\\
&=\frac1{2(n+1)}+\frac{1}{n+1}\int_0^1\frac{x^{n+1}}{(x+1)^2}\:dx\\
&=\frac1{2(n+1)}+\frac{1}{n+1}\left(\left. \frac{x^{n+2}}{(n+2)}\frac{1}{(1+x)^2}\right|_0^1+\frac{2}{(n+1)}\int_0^1\frac{x^{n+1}}{(x+1)^3}\:dx \right)\\
&=\frac1{2(n+1)}+\frac1{4(n+1)(n+2)}+\frac{2}{(n+1)^2}\int_0^1\frac{x^{n+1}}{(x+1)^3}\:dx
\end{align}
$$ but
$$
0\leq \int_0^1\frac{x^{n+1}}{(x+1)^3}\:dx\leq\int_0^1x^{n+1}dx=\frac{1}{(n+2)} 
$$ thus
$$
\frac{2}{(n+1)^2}\int_0^1\frac{x^{n+1}}{(x+1)^3}\:dx=\mathcal{O}\left(\frac{1}{n^3} \right)
$$
Finally, as $n \to \infty$, 

$$ I_n=\int_0^1\frac{x^n}{1+x}dx=\frac1{2n}-\frac{1}{4n^2}+\mathcal{O}\left(\frac{1}{n^3} \right).$$ 

$$
$$
Method 2.
One may use the standard integral representation of the digamma function and its asymptotics, as $n \to \infty$, 

$$
\begin{align}
I_n=\int_0^1\frac{x^n}{1+x}\:dx&=\int_0^1\frac{x^n-x^{n+1}}{1-x^2}\:dx\\
&=\frac12\int_0^1\frac{(1-t^{n/2})-(1-t^{(n-1)/2})}{1-t}\:dt\\
&=\frac12\psi\left(\frac{n}{2}+1\right)-\frac12\psi\left(\frac{n}{2}+\frac12\right)\\
&=\frac1{2n}-\frac{1}{4n^2}+\mathcal{O}\left(\frac{1}{n^3} \right). 
\end{align}
$$

</s>