Computing $\lim_{n \to \infty} \sqrt[n]{ \int_{0}^{1} (1+x^n)^n dx}$

(Of course I did some numerical experiments before starting on a proof.)

Let $$J_n:=\left(\int_0^1 (1+x^n)^n\>dx\right)^{1/n}\ .$$ From $1\leq 1+x^n\leq 2$ $\>(0\leq x\leq 1)$ it immediately follows that $1\leq J_n\leq 2$ for all $n\geq1$, in particular $\lim\sup_{n\to\infty} J_n\leq2$.

Let a small $\alpha>0$ be given. When $1-{\alpha\over n}\leq x\leq 1$ then Bernoulli's inequality gives $$x^n\geq\left(1-{\alpha\over n}\right)^n\geq1-\alpha\ ,$$ so that we obtain $$1+x^n\geq 2-\alpha\qquad\left(1-{\alpha\over n}\leq x\leq 1\right)\ .$$ From this we conclude $$\int_0^1(1+x^n)^ndx\geq \int_{1-\frac{\alpha}{n}}^1 (2-\alpha)^ndx= {\alpha\over n}(2-\alpha)^n$$ which then implies $$J_n\geq{\root n\of{\alpha\over n}}(2-\alpha)\ .$$ Taking the $\lim\inf$ on both sides proves $\lim\inf_{n\to\infty} J_n\geq 2-\alpha$, and since this is true for all $\alpha>0$ we conclude that $\lim\inf_{n\to\infty} J_n\geq2$.

It follows that in fact $\lim_{n\to\infty} J_n=2$.

Quite trivially $ I_n=\sqrt[n]{\int_{0}^{1}(1+x^n)^n\,dx} $ is less than two.

However, $1+x^n$ is a convex function on $[0,1]$, hence by considering the tangent in $x=1$ we have: $$ I_n \geq \sqrt[n]{\int_{0}^{\frac{2}{n}}(2-nx)^n\,dx}=\sqrt[n]{\frac{1}{n}\int_{0}^{2}(2-x)^n\,dx}=\sqrt[n]{\frac{2^{n+1}}{n}\int_{0}^{1}(1-x)^n\,dx} $$ hence: $$ I_n \geq 2\,\left(\frac{2}{n(n+1)}\right)^{1/n} $$ proves that $\lim_{n\to +\infty}I_n = \color{red}{2}$.