Finding $\lim_{n \to \infty} \int_1^a \frac{n}{1+x^n} \, dx$
$$\begin{align*}\int_1^a \frac{n}{1+x^n}\,dx &=\int_{1}^a\frac{n}{x^n}\left(\frac{1 }{1+(1/x)^n}\right)dx\\&=\int_1^a\frac{n}{x^n}\left(1-\frac{1}{x^n}+\frac{1}{x^{2n}}-\cdots\right)dx\\&=n\int_1^a\frac{1}{x^n}-\frac{1}{x^{2n}}+\frac{1}{x^{3n}}-\cdots\,dx\\&=n\left[\frac{1}{n-1}-\frac{1}{2n-1}+\cdots\right]-\underbrace{n\left[\frac{a^{1-n}}{n-1}-\frac{a^{1-2n}}{2n-1}+\cdots\right]}_{\to\,0}\end{align*}$$
$$\text{Since}\; \lim_{n\to\infty}\frac{n}{kn-1}=\frac{1}{k}:$$
$$\lim_{n\to\infty}\int_1^a \frac{n}{1+x^n}dx=1-\frac{1}{2}+\frac{1}{3}-\cdots =\ln 2$$
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
$$ \lim_{n \to \infty} \int_1^a \frac{n}{1+x^n} \, \mathrm dx = \int_{1}^{\infty} \frac{u^{-1}}{1+u} \, \mathrm du = \int_{1}^{\infty} \left(\frac{1}{u} - \frac{1}{1+u} \right) \mathrm du = \ln 2.$$
Hint: Put $$x=tan^{2/n}\theta$$ $$ dx=\dfrac2ntan^{\frac2n-1}\theta \times sec^2\theta . d\theta$$
Now => $$\boxed{I_n=\int_\phi^\psi 2tan^{\frac2n-1}\theta . d\theta}$$ $\phi,\psi$ are the limits accordingly. $$\phi=tan^{-1}1=\pi/4$$ and $$\psi=tan^{-1}[a^\frac n2]$$ $$Lt_{n\rightarrow\infty}\ \ \ [\psi]=\pi/2$$
$$Lt_{n\rightarrow\infty}[I_n]=\int_{\pi/4}^{\pi/2}\dfrac{2cos\theta}{sin\theta}.d\theta$$.