Finding the limit of a sequence of integrals

I thought it would be instructive to present an approach that relies on integration by parts and first-year calculus tools. To that end, we now proceed.

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Let $I_n$ be the sequence given by

$$I_n=\int_0^1 \frac{2nx^{n-1}}{1+x}\,dx\tag 1$$

Integrating by parts the integral on the right-hand side of $(1)$ with $u=\frac1{1+x}$ and $v=2x^n$ reveals

$$I_n=1+\int_0^1 \frac{2x^n}{(1+x)^2}\,dx\tag2$$

It is easy to see that the integral on the right-hand side of $(2)$ approaches $0$ as $n\to\infty$ (it is bounded below by $\frac1{2(n+1)}$ and bounded above by $\frac2{n+1}$) from which we conclude that

$$\lim_{n\to\infty}I_n=1$$

Hint. Note that for $x\in [0,1]$, $$g_n(x):=nx^{n-1}\leq f_n(x)$$ and therefore $$1=\int_0^1 g_n(x) dx \leq \int_0^1 f_ n(x) dx.$$ As regards the other side, $f_n(x)$ is increasing in $[0,1]$ for $n\ge 3$, and for $a\in (0,1)$, $$f_n(x)\leq h_n(x):=\frac{2na^{n-1}}{a+1}\chi_{[0,a]}(x)+\frac{2nx^{n-1}}{a+1}\chi_{(a,1]}(x).$$ Hence $$\int_0^1 f_ n(x) dx\leq \int_0^1 h_ n(x) dx=\frac{2na^{n}}{a+1}+\frac{2(1-a^n)}{a+1}.$$ Can you take it from here?

The limit is 1. For the opposite side, use $$ \int_{0}^{1}f_{n}(x)dx-1 = \int_{0}^{1}\frac{1-x}{1+x}nx^{n-1}dx \leq \int_{0}^{1} (1-x)nx^{n-1}dx = \frac{1}{n+1}\to 0 $$