Limit of integral of arctan $$L = \displaystyle \lim_{{n\to\infty}}\sqrt[n]{\int_0^1{(\arctan(1 + x^n))^n dx}}$$
I started with $0 \le x \le 1$ and I got to $\frac{\pi}{4} \le L \le \arctan 2$
I don't really know if that helps ... The answer is $\arctan 2$.
Also, $$\displaystyle \lim_{{n\to\infty}}\sqrt[n]{\int_0^1{(1+x^n)^n dx}}$$
I saw this as 
$$e^{\lim _{n\to \infty }\frac{1}{n}\ln \left(\int _0^1\left(1+x^n\right)^n\:dx\right)}$$
Can I apply L'hospital somehow here?
 A: Here's a generalization that is no harder than the specific problem.
Thm: Suppose $f$ is continuous, increasing, and nonnegative on $[0,1].$ Define $I_n = \int_0^1 f(x^n)^n\,dx.$ Then
$$\lim_{n\to \infty} I_n^{1/n} = f(1).$$
With $f(x) = \arctan (1+x),$ the theorem shows the limit in the given problem is $f(1)=\arctan 2.$
Proof of the theorem: Since $I_n \le f(1)^n,$ we have $I_n^{1/n} \le f(1)$ for all $n.$ To get a lower bound, make the change of variables $x=y^{1/n}$ to see 
$$I_n =\int_0^1 f(y)^n \, (1/n)y^{1/n-1}\,dy.$$
Let $a\in (0,1).$ Then
$$\frac{f(a)^n}{n}\int_a^1 y^{1/n-1}\,dy < I_n.$$
Because $1\le y^{1/n-1}$ on $(0,1]$ for all $n,$ we then have
$$\frac{f(a)^n}{n}(1-a) < I_n.$$
Thus
$$f(a)\left (\frac{1-a}{n}\right )^{1/n} < I_n^{1/n}.$$
Now as $n\to \infty,$ both $(1-a)^{1/n}, n^{1/n} \to 1.$ Therefore
$$f(a) \le \liminf_{n\to \infty} I_n^{1/n}.$$
This is true for each $a\in (0,1);$ therefore it holds in the limit as $a\to 1^-.$ By the continuity of $f,$ we get
$$ f(1) \le \liminf_{n\to \infty} I_n^{1/n}.$$
Recalling our easy upper bound, we thus have
$$\limsup_{n\to\infty} I_n^{1/n} \le f(1) \le \liminf_{n\to \infty} I_n^{1/n} \le \limsup_{n\to\infty} I_n^{1/n}.$$
All expressions here are thus equal, proving the theorem.
A: Since $\arctan(2)>1$, most of the mass of the integral $\int_{0}^{1}\arctan(1+x^n)^n\,dx $ is concentrated in a left neighbourhood of $x=1$. By considering the Taylor series of $\arctan(1+x^n)^n$ around $x=1$ we get
$$ \arctan(1+x^n)^n = \arctan(2)^n\left[1+\frac{n^2}{5\arctan 2}(x-1)+o(x-1)\right] $$
hence $\lim_{n\to +\infty}\sqrt[n]{\int_{0}^{1}\arctan(1+x^n)^n\,dx }=\arctan(2)$, since we are dealing with convex functions, ensuring that
$$\lim_{n\to +\infty}\sqrt[n]{\int_{0}^{1}\arctan(1+x^n)^n\,dx }=\lim_{n\to +\infty}\sqrt[n]{\int_{0}^{1}\arctan(2)^n \exp\left(\frac{n^2}{5\arctan 2}(x-1)\right)\,dx }.$$
By a similar argument (a simplified version of Laplace's method)
$$ \lim_{n\to +\infty}\sqrt[n]{\int_{0}^{1}(1+x^n)^n\,dx}=2.$$
Here $(1+x^n)^n$ is even more than convex, it is log-convex.
