Limits of integrals depending on n

1. For $x\in[0,1]$, $0\le\frac{x^n}{1+x}\le x^n$, so $0\le I_n\le\int_0^1x^n\,dx=\frac1{n+1}\to0$.

2. Integrate by parts with $u=\frac1{1+x}$, $v'=x^n$:
$$I_n=\left[\frac{x^{n+1}}{(n+1)(1+x)}\right]_0^1+\frac1{n+1}\int_0^1\frac{x^{n+1}}{(1+x)^2}\,dx=\frac1{2(n+1)}+\frac{J_n}{n+1},$$
where $0\le J_n\le\int_0^1x^{n+1}dx=\frac1{n+2}\to0$. Hence $(n+1)I_n=\frac12+J_n\to\frac12$.

3. $\displaystyle1-\frac1{1+x^n}=\frac{x^n}{1+x^n}\in[0,x^n]$, so
$$0\le1-\int_0^1\frac{dx}{1+x^n}\le\frac1{n+1}\to0.$$

Remark. Part 2 gives the *rate*: $I_n\sim\frac1{2n}$. Most of the integral's mass of $x^n$ sits near $x=1$, where $\frac1{1+x}\approx\frac12$.