How does one evaluate $\lim _{n\to \infty }\left(\sqrt[n]{\int _0^1\:\left(1+x^n\right)^ndx}\right)$?

Showing that $2$ is an upper bound for the limit is straightforward, since for $0 \leqslant x \leqslant 1$, we have $(1 +x^n)^n \leqslant 2^n$, and

$$ \left(\int_0^1 (1 + x^n)^n \, dx \right)^{1/n} \leqslant 2, \\ \implies \limsup_{n \to \infty}\left(\int_0^1 (1 + x^n)^n \, dx \right)^{1/n} \leqslant 2.$$

For a useful lower bound, note that the integrand becomes more and more concentrated in a left neighborhood of $x = 1$ as $n$ increases. If $1 - \delta/n < x \leqslant 1$, then

$$1 + x^n > 1 +(1 - \delta/n)^n > 1 + 1 - n(\delta/n) = 2 - \delta,$$

and

$$\int_0^1 (1 + x^n)^n \, dx > \int_{1 - \delta/n}^1 (1 + x^n)^n \, dx \\ > \frac{\delta}{n}(2 - \delta)^n.$$

Hence,

$$\liminf_{n \to \infty} \left(\int_0^1 (1 + x^n)^n \, dx \right)^{1/n} > \liminf_{n \to \infty}\,(2-\delta)\frac{\delta^{1/n}}{n^{1/n}} = 2 - \delta.$$

Since $\delta > 0$ can be arbitrarily small, we have

$$2 \leqslant \liminf_{n \to \infty} \left(\int_0^1 (1 + x^n)^n \, dx\right)^{1/n} \leqslant \limsup_{n \to \infty} \left(\int_0^1 (1 + x^n)^n \, dx\right)^{1/n} \leqslant 2.$$

This shows both that the limit exists and the value is $2$.

Let $$ A_n = \Big( \int_{0}^1 (x^n + 1)^n dx \Big) ^{1/n} $$ $$ A_n^{n} = \int_{0}^1 (x^n + 1)^n dx $$ $$ A_n^{n} = \int_{0}^1 (x^n + 1)^n dx = \int_{0}^1 \Sigma_{i=0}^{n} \binom{n}{i} x^{ni} dx = \Sigma_{i=0}^{n} \binom{n}{i} \int_{0}^1 x^{ni} dx $$

$$ A_n^{n} = \Sigma_{i=0}^{n} \binom{n}{i} \frac{1}{n \cdot i + 1} $$

Hence for the upper bound we have:

$$A_n^{n} \le \Sigma_{i=0}^{n} \binom{n}{i} =2^n $$ $$A_n \le 2 $$

And for the lower bound we have: $$A_n^{n} \ge \Sigma_{i=0}^{n} \binom{n}{i} \frac{1}{n \cdot n + 1} = \frac{2^n}{n^2+1} \ge \frac{2^n}{(n+1)^2} $$

$$A_n \ge \frac{2}{(n+1)^{2/n}} $$

Since $$\lim_{n\to\infty} \frac{2}{(n+1)^{2/n}} = 2 \cdot \lim_{n\to\infty} (n+1)^{-2/n} = 2 \cdot \lim_{n\to\infty} e^{-ln(n+1)/n } = 2 \cdot e^{\lim_{n\to\infty} -ln(n+1)/n } =2 \cdot e^{0} = 2$$

Therefore $$\lim_{n\to\infty} A_n \ge 2$$

Hence $$\lim_{n\to\infty} A_n = 2$$

From $x^n \le 1$ for all $x \in [0,1]$, we get $$\int_0^1(2x^n)^n\,dx\le\int_0^1(1+x^n)^n\,dx \le\int_0^12^n\,dx$$and therefore$$\frac{2^n}{n^2+1}\le\int_0^1(1+x^n)^n\,dx \le2^n$$ Since it is obvious that $\lim _{n\to \infty }\left(\sqrt[n]{\frac{2^n}{n^2+1}}\right)=\lim _{n\to \infty }\left(\sqrt[n]{2^n}\right)=2$, we can deduce that the limit exists and its value is $2$.