Finding $lim_{n to infty} int_1^a frac{n}{1+x^n} , dx$

EDIT: I tried to make my answer more rigorous.

Let $u = x^{n}$.

Then we have

$$lim_{n to infty} int_1^a frac{n}{1+x^n} , mathrm dx =lim_{n to infty} int_{1}^{a^{n}} frac{u^{1/n-1}}{1+u} , mathrm du = lim_{n to infty} int_{1}^{infty} boldsymbol{1}_{[1,a^{n}]}(u) , frac{u^{1/n-1}}{1+u} , mathrm du.$$

For $n ge 2$, the integrand is dominated by the integrable function $frac{u^{-1/2}}{1+u}$.

So by appealing to the the dominated convergence theorem, we can conclude that

$$ lim_{n to infty} int_1^a frac{n}{1+x^n} , mathrm dx = int_{1}^{infty} frac{u^{-1}}{1+u} , mathrm du = int_{1}^{infty} left(frac{1}{u} - frac{1}{1+u} right) mathrm du = ln 2.$$

$$begin{align*}int_1^a frac{n}{1+x^n},dx &=int_{1}^afrac{n}{x^n}left(frac{1 }{1+(1/x)^n}right)dx\&=int_1^afrac{n}{x^n}left(1-frac{1}{x^n}+frac{1}{x^{2n}}-cdotsright)dx\&=nint_1^afrac{1}{x^n}-frac{1}{x^{2n}}+frac{1}{x^{3n}}-cdots,dx\&=nleft[frac{1}{n-1}-frac{1}{2n-1}+cdotsright]-underbrace{nleft[frac{a^{1-n}}{n-1}-frac{a^{1-2n}}{2n-1}+cdotsright]}_{to,0}end{align*}$$

$$text{Since}; lim_{ntoinfty}frac{n}{kn-1}=frac{1}{k}:$$

$$lim_{ntoinfty}int_1^a frac{n}{1+x^n}dx=1-frac{1}{2}+frac{1}{3}-cdots =ln 2$$