Prove that $\lim n\int_1^a\frac{1}{1+x^n}dx=\ln 2$ My problem is that for a given $a>1$, we have that
$$\lim_{n\to\infty}n\int_{1}^{a}\frac{1}{1+x^n}dx=\ln 2$$
The natural idea seems to be to add and substract $x^n$ from the numerator and we obtain easily that
$$n\int_{1}^{a}\frac{1}{1+x^n}dx=n(a-1)-a\ln(1+a^n)+\ln2+\int_1^a\ln(1+x^n)dx$$
which would sort of explain the $\ln 2$ result but I can't continue from here.
 A: Let $$
I_n =n\int_1^a\frac{1}{1+x^n}\mathrm{d}x.
$$By making change of variables $ x = 1+\frac{u}{n}$, we obtain
$$
I_n =\int_0^{n(a-1)}\frac{1}{1+\left(1+\frac{u}{n}\right)^n}\mathrm{d}u=\int_0^{\infty}\frac{1_{\{u\le n(a-1)\}}}{1+\left(1+\frac{u}{n}\right)^n}\mathrm{d}u.
$$ Since $\left(1+\frac{u}{n}\right)^n\le \left(1+\frac{u}{n+1}\right)^{n+1} \longrightarrow e^u$ for all $u\ge 0$, we find
$$
\frac{1_{\{u\le n(a-1)\}}}{1+\left(1+\frac{u}{n}\right)^n}\le \frac{1}{1+\left(1+\frac{u}{2}\right)^2},\quad\forall n\ge 2.
$$ The RHS is an integrable function, hence by dominated convergence theorem it follows
$$\begin{eqnarray}
\lim_{n\to\infty} I_n& =&\int_0^\infty \frac{1}{1+e^u}\mathrm{d}u\\&=&\int_1^\infty \frac{1}{s(s+1)}\mathrm{d}s\tag{$e^u=s$}\\&=&\lim_{n\to\infty}\int_1^n \left(\frac{1}{s}-\frac{1}{s+1}\right)\mathrm{d}s\\
&=&\int_1^2 \frac{1}{s}\mathrm{d}s-\lim_{n\to\infty}\int_{n}^{n+1} \frac{1}{s}\mathrm{d}s\\
&=&\int_1^2 \frac{1}{s}\mathrm{d}s=\ln 2.
\end{eqnarray}$$ 
A: I didn't see Song's post until after I posted, so I deleted mine. Then I realized that this answer was a bit shorter, and possibly a bit easier to follow.
$$
\begin{align}
\lim_{n\to\infty}n\int_1^a\frac1{1+x^n}\,\mathrm{d}x
&=\lim_{n\to\infty}\int_1^\infty[x\le a^n]\frac{x^{\frac1n-1}}{1+x}\,\mathrm{d}x\tag1\\
&=\int_1^\infty\frac1{x(1+x)}\,\mathrm{d}x\tag2\\
&=\left.\log\left(\frac{x}{x+1}\right)\right]_1^\infty\tag3\\[6pt]
&=\log(2)\tag4
\end{align}
$$
Explanation:
$(1)$: substitute $x\mapsto x^{1/n}$
$(2)$: dominated convergence; dominated by $\frac{x^{-1/2}}{1+x}$
$(3)$: integrate
$(4)$: evaluate
