Find the value of $\lim\limits_{n \to \infty}n\left(\left(\int_0^1\frac{1}{1+x^n}\,\mathrm{d}x\right)^n-\frac{1}{2}\right)$
Let $I(n)$ be given by the integral
$$\begin{align} I(n)&=\int_0^1 \frac{1}{1+x^n}\,dx \tag 1\\\\ \end{align}$$
Then, expanding the integrand of the integral on the right-hand side of $(1)$ in the Taylor series $ \frac{1}{1+x^n}=\sum_{k=0}^\infty (-1)^kx^{nk}$ reveals
$$\begin{align} I(n)&=\sum_{k=0}^\infty \frac{(-1)^k}{nk+1}\\\\ &=1+\frac1n \sum_{k=1}^\infty\frac{(-1)^k}{k+1/n} \tag 2 \end{align}$$
Next, using the fact that $\log(2)= \sum_{k=1}^\infty\frac{(-1)^{k-1}}{k}$ and that $\frac{\pi^2}{12}=-\sum_{k=1}^\infty \frac{(-1)^k}{k^2}$, we can write the series in $(2)$ as
$$\begin{align} \sum_{k=1}^\infty\frac{(-1)^k}{k+1/n} &=-\log(2)+\sum_{k=1}^\infty (-1)^k\left(\frac{1}{k+1/n}-\frac1k\right)\\\\ &=-\log(2)-\frac1n \sum_{k=1}^\infty \frac{(-1)^k}{k(k+1/n)}\\\\ &=-\log(2)-\frac1n \sum_{k=1}^\infty\frac{(-1)^k}{k^2}-\frac1n\sum_{k=1}^\infty (-1)^k\left(\frac{1}{k(k+1/n)}-\frac{1}{k^2}\right)\\\\ &=-\log(2)+\frac{\pi^2}{12n}+O\left(\frac1{n^2}\right) \tag 3 \end{align}$$
Using $(1)-(3)$ yields
$$I(n)=1-\frac{\log(2)}{n}+\frac{\pi^2}{12n^2}+O\left(\frac1{n^3}\right) \tag 4$$
Next, using $(4)$, we can write
$$\begin{align} \left(I(n)\right)^n&=e^{n\log\left(1-\frac{\log(2)}{n}+\frac{\pi^2}{12n^2}+O\left(\frac1{n^3}\right)\right)}\\\\ &=e^{-\log(2)+\frac{\pi^2}{12n}-\frac{\log^2(2)}{2n}+O\left(\frac{1}{n^2}\right)}\\\\ &=\frac12 \left(1+\frac{\pi^2}{12n}-\frac{\log^2(2)}{2n}+O\left(\frac{1}{n^2}\right)\right) \end{align}$$
Finally, we have
$$\begin{align} \lim_{n\to \infty}\left(n\left(\left(I(n)\right)^n-\frac12\right)\right)&=\lim_{n\to \infty}\left(\frac{\pi^2}{24}-\frac{\log^2(2)}{4}+O\left(\frac1n\right)\right)\\\\ &=\bbox[5px,border:2px solid #C0A000]{\frac{\pi^2}{24}-\frac{\log^2(2)}{4}} \end{align}$$
And we are done!
Hint. If one knows the digamma function $\psi=\Gamma'/\Gamma$, one may write, as $n \to \infty$, $$ \begin{align} \int_0^1\frac1{1+x^n}\:dx&=\int_0^1\frac{1-x^n}{1-x^{2n}}\:dx \\\\&\stackrel{u=x^{2n}}=\frac1{2n}\int_0^1\frac{(1-u^{1/2})u^{1/(2n)-1}}{1-u}\:du \\\\&=\frac1{2n}\left[\psi\left(\frac1{2n}+\frac12 \right)-\psi\left(\frac1{2n} \right) \right] \\\\&=1-\frac{\ln 2}{n}+\frac{\pi ^2-6 \ln^2 2}{12n^2}+O\left(\frac{1}{n^3}\right) \end{align} $$ giving, as $n \to \infty$, $$ n\ln \left(1-\frac{\ln 2}{n}+\frac{\pi ^2-6 \ln^2 2}{12n^2}+O\left(\frac{1}{n^3}\right) \right)=-\ln 2+\frac{\pi ^2-6 \ln^2 2}{24n}+O\left(\frac{1}{n^2}\right) $$ that is
$$ n\left[\left(\int_0^1\frac1{1+x^n}\:dx \right)^n-\frac12\right]\to \frac{\pi ^2-6 \ln^2 2}{24}. $$
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
$$\lim_{n \to \infty} n \left [I(n)^n-\frac12 \right ] = \frac{\pi^2}{24} - \frac14 \log^2{2} $$