Find the value of $\lim\limits_{n \to \infty}n\left(\left(\int_0^1\frac{1}{1+x^n}\,\mathrm{d}x\right)^n-\frac{1}{2}\right)$ 
Find the value of the limit $$\lim_{n \to \infty}n\left(\left(\int_0^1 \frac{1}{1+x^n}\,\mathrm{d}x\right)^n-\frac{1}{2}\right)$$

I can't solve the integral $\int_0^1 \mathrm{\frac{1}{1+x^n}}\,\mathrm{d}x$. But maybe the questions doesn't require solving the integral. 
Apparently the $\lim_{n \to \infty}(\int_0^1 \mathrm{\frac{1}{1+x^n}}\,\mathrm{d}x)^n$ should be $\frac{1}{2}$ for the question to make sense. That's all I know.

My approach is a little different.  I would sub $x=u^{1/n}$ in the integral and get
$$I(n) = \int_0^1 \frac{dx}{1+x^n} = \frac1n \int_0^1 du \frac{u^{\frac1n-1}}{1+u} = \frac1n \int_0^1 du \, u^{1/n} \left (\frac1u - \frac1{1+u} \right ) \\= 1-\frac1n \int_0^1 du \frac{u^{1/n}}{1+u}$$
We then note that $u^{1/n} = e^{\log{u}/n}$ and that $n$ is large enough for the following series expansion to be valid:
$$I(n) = 1- \sum_{j=0}^{\infty} \frac1{n^{j+1}} \int_0^1 du \frac{\log^j{u}}{1+u} $$
Due to the nature of the limit we are posed, we go out to second order; thus
$$I(n) = 1-\frac{\log{2}}{n} + \frac{\pi^2}{12 n^2} +O \left ( \frac1{n^3} \right ) $$
Then
$$\begin{align}I(n)^n &= e^{n \log{\left [1-\frac{\log{2}}{n} + \frac{\pi^2}{12 n^2} +O \left ( \frac1{n^3} \right )\right ]} } \\ &= e^{n\left [-\frac{\log{2}}{n} + \frac{\pi^2}{12 n^2} - \frac{\log^2{2}}{2 n^2}+O \left ( \frac1{n^3} \right ) \right ] } \\ &= \frac12 \left [1+\left (\frac{\pi^2}{12} - \frac12 \log^2{2} \right ) \frac1n +O \left ( \frac1{n^2} \right ) \right ]\end{align}$$
The sought after limit is then

$$\lim_{n \to \infty} n \left [I(n)^n-\frac12 \right ] = \frac{\pi^2}{24} - \frac14 \log^2{2} $$

