Limits of sum integral in exponential form

Question

I'm stuck with this limit of sum integral. Couldn't find the pattern, what's the most efficient way to solve this?

$\lim \limits_{n \to ∞} \frac{1}{n}\sum_{k=1}^∞\frac{a^{1+\frac{k}{n}}}{a^{1+\frac{k}{n}}+1}$

$\Delta x = \frac{b-a}{n} = \frac{1}{n}$

$b-a=1$

I know that using right riemann sum, $f(a+k\Delta x) = \frac{a^{1+\frac{k}{n}}}{a^{1+\frac{k}{n}}+1}$, please advice on how can I continue.

Answer 1

$$x_k=\dfrac{k}{n}$$ $$\Delta x_k=x_{k+1}-x_{k}=\dfrac{k+1}{n}-\dfrac{k}{n}=\dfrac{1}{n}$$ then with $$f(x)=\dfrac{a^{1+x}}{a^{1+x}+1}$$ we have \begin{align} \lim_{n\to\infty} \frac{1}{n}\sum_{k=1}^\infty\frac{a^{1+\frac{k}{n}}}{a^{1+\frac{k}{n}}+1} &= \lim_{n\to\infty} \sum_{k=1}^\infty f(x_k)\Delta x_k \\ &= \int_0^1 f(x) dx \\ &= \int_0^1\dfrac{a^{1+x}}{a^{1+x}+1}dx \\ &= \left(\dfrac{1}{\ln a}\ln|a^{1+x}+1|\right)_0^1 \\ &= \dfrac{1}{\ln a}\ln\dfrac{a^2+1}{a+1} \end{align}