How does one evaluate $\lim _{n\to \infty }\left(\sqrt[n]{\int _0^1\:\left(1+x^n\right)^ndx}\right)$? I tried using Lebesgue's dominated convergence theorem and I'm getting $\lim _{n\to \infty }\left(1+\int _0^1x^ndx\:\right)$ which is $1$. But the answer should be 2.

Let $I_n$ be given by 
$$I_n=\int_0^1(1+x^n)^n\,dx \tag1$$
From the binomial theorem,  we can write 
$$(1+x^n)^n=\sum_{k=0}^n\binom{n}{k}x^{nk}\tag 2$$
Using $(2)$ in $(1)$ reveals
$$\begin{align}
I_n&=\int_0^1 \sum_{k=0}^n\binom{n}{k}x^{nk}\,dx\\\\
&=\sum_{k=0}^n\binom{n}{k}\frac{1}{1+nk}\tag3
\end{align}$$
Clearly from $(3)$, we obtain the estimates for $(I_n)^{1/n}$
$$\frac{2}{(1+n^2)^{1/n}}\le (I_n)^{1/n}\le 2 \tag 4$$
whereupon applying the squeeze theorem to $(4)$ yields the coveted limit
$$\lim_{n\to \infty}\left(\int_0^1(1+x^n)^n\,dx\right)^{1/n}=2$$
as was to be shown!

NOTE:
The bounds given in $(4)$ follow by letting $k=0$ and $k=n$ in the term $\frac{1}{1+nk}$ in the binomial expansion of $(1+x^n)^n$.  Since $x\le 1$ for $x\in [0,1]$, these bounds are tantamount to the bounds $2x^n\le 1+x^n\le 2$ in $(1)$ (as used by @didgogns), which lead immediately to $(4)$.  
