How would we find the following limit? $$\lim _{n\to \infty }\sum _{i=1}^n\left(\frac{2\left(1+\frac{2i}{n}\right)}{\left(1+\frac{2i}{n}\right)^2+1}\right)\cdot \frac{2}{n}.$$
So I thought the equation of the function was $$y=\frac{2(1+2x)}{(1+2x)^2 + 1} × 2.$$
After simplifying it down, the entire function seem to get cancelled, leaving me with a limit of $0$, but this is incorrect. I think this is because the function I got was incorrect but I am not sure where I have gone wrong.
Any help?

The point of this problem is to recognize it as a Riemann sum. Since we have $1 + 2i/n$ appearing in multiple places, it is natural to break the interval $[1, 3]$ into $n$ slices of uniform length $2/n$, and we have
$$\int_1^3 \frac{2x}{x^2 + 1} \, dx = \ln 5.$$
