Find $\lim_{n\to\infty}\left(2n\int_{0}^{1}\frac{x^n}{1+x^2}dx\right)^n$ Find this limits
$$\lim_{n\to\infty}\left(2n\int_{0}^{1}\dfrac{x^n}{1+x^2}dx\right)^n\tag{1}$$
since following links post this question have solve  it.$$\lim_{n\to\infty}2n\int_{0}^{\frac{\pi}{4}}\tan^n{x}dx=\dfrac{1}{2}$$
Solving $\lim_{n\to\infty}(n\int_0^{\pi/4}(\tan x)^ndx)$?
Could you suggest a helpful idea with $(1)$ ?
 A: Integration by parts gives:
$$\int_{0}^{1}nx^{n-1}\frac{2x}{1+x^2}\,dx = \left.x^n\frac{2x}{1+x^2}\right|_{0}^{1}-2\int_{0}^{1}x^{n}\frac{(1-x^2)}{(1+x^2)^2}\,dx$$
so performing it twice:
$$\int_{0}^{1}nx^{n-1}\frac{2x}{1+x^2}\,dx = 1-2\left.\frac{x^{n+1}}{n+1}\frac{(1-x)^2}{(1+x^2)^2}\right|_{0}^{1}+2\int_{0}^{1}\frac{x^{n+1}}{n+1}\cdot\frac{2x(3-x^2)}{(1+x^2)^3}\,dx$$
but since $\frac{2x(3-x^2)}{(1+x^2)^3}$ is non-negative and bounded on $(0,1)$, it follows that:
$$\int_{0}^{1}nx^{n-1}\frac{2x}{1+x^2}\,dx = 1+O\left(\frac{1}{n^2}\right),$$
so the wanted limit is just $\color{red}{1}$.

To be more accurate, we may evaluate $I_n = 2n\int_{0}^{1}\frac{x^n}{1+x^2}\,dx$ in terms of the digamma function:
$$ I_n = \frac{n}{2}\left(\psi\left(\frac{n+3}{4}\right)-\psi\left(\frac{n+1}{4}\right)\right)=\sum_{m\geq 1}\frac{4n}{(4m+n-1)(4m+n-3)}$$
or through the simple series representation:
$$ I_n = 2\sum_{m\geq 1}\left(\frac{4m-1}{n+4m-1}-\frac{4m-3}{n+4m-3}\right)$$
to get:
$$ I_n \approx 1-\frac{1}{n^2}$$
for large $n$s.
