integration - Find the value of $limlimits_{n to infty}nleft(left(int_0^1frac{1}{1+x^n},mathrm{d}xright)^n-frac{1}{2}right)$

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.

Answer

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

$$\begin{align} I(n)&=\sum_{k=0}^\infty \frac{(-1)^k}{nk+1}\\\\ &=1+\frac1n \sum_{k=1}^\infty\frac{(-1)^k}{k+1/n} \tag 2 \end{align}$$

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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

$$\begin{align} \sum_{k=1}^\infty\frac{(-1)^k}{k+1/n} &=-\log(2)+\sum_{k=1}^\infty (-1)^k\left(\frac{1}{k+1/n}-\frac1k\right)\\\\ &=-\log(2)-\frac1n \sum_{k=1}^\infty \frac{(-1)^k}{k(k+1/n)}\\\\ &=-\log(2)-\frac1n \sum_{k=1}^\infty\frac{(-1)^k}{k^2}-\frac1n\sum_{k=1}^\infty (-1)^k\left(\frac{1}{k(k+1/n)}-\frac{1}{k^2}\right)\\\\ &=-\log(2)+\frac{\pi^2}{12n}+O\left(\frac1{n^2}\right) \tag 3 \end{align}$$

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Using $(1)-(3)$ yields

$$I(n)=1-\frac{\log(2)}{n}+\frac{\pi^2}{12n^2}+O\left(\frac1{n^3}\right) \tag 4$$

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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!