Calculate $\lim_{n\rightarrow\infty} I_n$ knowing that the sequence $(I_n)_{n\ge 2}$ is $I(n) = \int_1^{\sqrt{2}} \sqrt[n] {x^n+x^{n+2}} dx $ 
Calculate
  $\lim_{n\rightarrow\infty}   I_n$ knowing that the sequence $(I_n)_{n\ge 2}$ is $I_n = \int_1^{\sqrt{2}} \sqrt[n] {x^n+x^{n+2}} dx $

Now, my attempt was like this :
$\int_1^{\sqrt{2}} \sqrt[n] {x^n(1+x^2)} dx $ and $\int_1^{\sqrt{2}} x\sqrt[n] {1+x^2} dx $ . From here I was thinking to do a substitution like $x=\tan(u)$ and using the formula $1+\tan^2(u)=\sec(u)$ but I don't think this is the way to go . Also, I was thinking that maybe I don't need to evaluate the integral directly and maybe use Dominated convergence theorem . Any help will be appreciated !

$I_n=\int_{1}^{√2}(x^n+x^{n+2})^{1/n}dx;$
$\int_{1}^{√2}x(1+1^2)^{1/n} \lt I_n \lt \int_{1}^{√2}x(1+(√2)^2)^{1/n}dx;$
$2^{1/n}\int_{1}^{√2}xdx\lt I_n < (1+2)^{1/n}\int_{1}^{√2}xdx.$
Integrate and take the limit.
Used: 
$0 \le f(x):=(1+1^2)^{1/n}x  \le x(1+x^2)^{1/n} \le x(1+√2^2)=:g(x),$ 
then $\int_{1}^{√2}f(x)dx \le I_n \le \int_{1}^{√2}g(x)dx$ (Monotony of Riemann Integral).
