Scaled integral of 1/(1+x^n) just above 1

Let $\varepsilon>0$ be fixed. The limit $\lim_{n\to\infty} n\int_1^{1+\varepsilon}\frac{dx}{1+x^n}$ equals $\ln k$ for some constant $k$. Find $k$.

$k=2$. Substitute $x=e^{t/n}$ so $dx\approx \tfrac{1}{n}e^{t/n}dt$; the integral becomes $\int_0^{n\ln(1+\varepsilon)}\frac{e^{t/n}}{1+e^{t}}\,dt\to\int_0^\infty\frac{dt}{1+e^t}=\ln 2$, independent of $\varepsilon$.