Evaluation of $\lim_{n \to \infty} \int_{0}^\infty \frac{1}{1+x^n} dx$
I thought it would be instructive to present a way forward that relies only on elementary calculus tools. To that end, we now proceed.
Enforcing the substitution $x\mapsto x^{1/n}$, we see that for $n>1$
$$\begin{align} \int_0^\infty \frac1{1+x^n}\,dx&=\frac1n \int_0^\infty \frac{x^{1/n}}{x(1+x)}\,dx\tag 1 \end{align}$$
Writing the integral on the right-hand side of $(1)$ as the sum
$$\begin{align} \int_0^\infty \frac{x^{1/n}}{x(1+x)}\,dx&=\int_0^1 \frac{x^{1/n}}{x(1+x)}\,dx+\int_1^\infty \frac{x^{1/n}}{x(1+x)}\,dx\tag2 \end{align}$$
and enforcing the substitution $x\mapsto 1/x$ in the second integral on the right-hand side of $(2)$ reveals
$$\begin{align} \int_0^\infty \frac1{1+x^n}\,dx&=\frac1n \int_0^1 \frac{x^{1/n}+x^{1-1/n}}{x(1+x)}\,dx\tag3 \end{align}$$
Next, using partial fraction expansion, we find that
$$\begin{align} \int_0^\infty \frac1{1+x^n}\,dx&=\color{blue}{\frac1n\int_0^1 \frac{x^{1/n}+x^{1-1/n}}{x}\,dx}-\frac1n\int_0^1 \frac{x^{1/n}+x^{1-1/n}}{1+x}\,dx\\\\ &=\color{blue}{1+\frac1{n-1}}-\frac1n\int_0^1 \frac{x^{1/n}+x^{1-1/n}}{1+x}\,dx\tag4 \end{align}$$
Inasmuch as the integral on the right-hand side of $(4)$ is trivially seen to be bounded in absolute value by $2$, we find that
$$\int_0^\infty \frac1{1+x^n}\,dx=1+O\left(\frac1n\right)$$
Taking the limit as $n\to \infty$, yields the coveted limit
$$\bbox[5px,border:2px solid #C0A000]{\lim_{n\to \infty}\int_0^\infty \frac1{1+x^n}\,dx=1}$$
TOOLS USED: Elementary Integral Theorems, Substitution, Partial Fraction Expansion
You can also show with a beautiful contour integral that for $n \geq 2$
$$ \int_{0}^{+\infty}\frac{\text{d}x}{1+x^n}=\frac{\pi}{n\sin\left(\displaystyle \frac{\pi}{n}\right)} $$
using that $$ \sin\left(\frac{\pi}{n}\right) \underset{(+\infty)}{\sim}\frac{\pi}{n} $$ You find that
$$ \int_{0}^{+\infty}\frac{\text{d}x}{1+x^n} \underset{n \rightarrow +\infty}{\rightarrow}1 $$