$U_n= \int_{0}^{1}\frac{1}{1+x^{n}}dx$ $U_n= \int_{0}^{1}\frac{1}{1+x^{n}}dx$ where
Find $\lim_{n\to \infty} U_n$  can i enter the limit inside ?
$W_n= \int_{0}^{1}\frac{x^n}{1+x^{n}}dx$ and i established this relation by parts:
$W_n= \frac{\ln 2 }{n}- \frac{1}{n} \int_{0}^{1}\ln(1+x^{n})dx$
Now my 2nd part of my question to find
Find $\lim_{n\to \infty} \int_{0}^{1}\ln(1+x^{n})dx$
I multiplied the relation with n but how to find $\lim_{n\to \infty} nw_n$  
 A: a hint for the second question:
Note that $\frac{x^n}{1+x^n}=\sum_{j=1}^{\infty}(-1)^{j+1}x^{jn}$ therefore $$W_n=\sum_{j=1}^{\infty}(-1)^{j+1}\frac{1}{jn+1}$$
A: Hint: $0\le \ln (1+x^n)\le x^n.$
A: $$U_n=\int_{0}^{1}{\dfrac{1}{1+x^{n}}dx}$$
Let $x^{n}=u\to x=u^{\frac{1}{n}}$, hence $dx=\dfrac{1}{n}{u^{\frac{1}{n}-1}}$. Then
$$\int_{0}^{1}{\dfrac{1}{1+x^{n}}dx}=\dfrac{1}{n}\int_{0}^{1}{\dfrac{u^{\frac{1}{n}-1}}{(1+u)^{\frac{1}{n}+(1-\frac{1}{n})}}du}=\dfrac{1}{n}\beta\left(\dfrac{1}{n},1-\dfrac{1}{n}\right)=\dfrac{1}{n}\frac{\Gamma\left(\dfrac{1}{n}\right)\Gamma\left(1-\dfrac{1}{n}\right)}{\Gamma\left(\dfrac{1}{n}+1-\dfrac{1}{n}\right)}$$
$$\dfrac{1}{n}\Gamma\left(\dfrac{1}{n}\right)\Gamma\left(1-\dfrac{1}{n}\right)=\dfrac{\pi}{n\sin\left(\dfrac{\pi}{n}\right)}$$
Then $\lim_{n\to \infty}{U_n}=\lim_{n\to\infty}{\dfrac{\pi}{n\sin\left(\dfrac{\pi}{n}\right)}}=1$
A: the invertion limit / limit (an integral being a limit) is always correct if :  regardless of the parameters, the expression is absolutely convergent. this is a consequence of the theorem that an absolutely convergent series can only converge to one limit, wathever the order of summation. 
in your case : $$\int_0^1 \left|\frac{1}{1+x^n }\right| dx < 1$$
this theorem is the one to use for 99.9% of the limits invertions ( or  integrals or series or derivatives). 
