Is this an integration?

$\begin{aligned} S & = \lim_{n \to \infty} \sum_{r=1}^n \frac {6n}{9n^2-r^2} \\ & = \lim_{n \to \infty} \sum_{r=1}^n \frac {6n}{(3n+r)(3n-r)} \\ & = \lim_{n \to \infty} \sum_{r=1}^n \left(\frac 1{3n+r} + \frac 1{3n-r} \right) \\ & = \lim_{n \to \infty} \frac 1n \sum_{r=1}^n \left(\frac 1{3+\frac rn} + \frac 1{3-\frac rn} \right) & \small \color{#3D99F6} \text{Using Riemann sums:} \\ & = \int_0^1 \left(\frac 1{3+x} + \frac 1{3-x} \right) dx & \small \color{#3D99F6} \lim_{n \to \infty} \frac 1n \sum_{k=a}^b f \left(\frac kn\right) = \lim_{n \to \infty} \int_\frac an^\frac bn f(x) \ dx \\ & = \log (3+x) - \log (3-x) \bigg|_0^1 \\ & = \log \left(\frac {3+x}{3-x} \right) \bigg|_0^1 \\ & = \boxed{\log 2} \end{aligned}$

Reference: Riemann sums