How to turn Riemann sum into definite integral?

Question

I'm having trouble with turning this Riemann sum into a definite integral. I'm not sure as to where to start.

$$\lim_{n \to \infty}\dfrac{3}{n}\sum_{k = 1}^{n}\dfrac{3k/n}{(3k/n+3)^2}$$

Original image: https://i.stack.imgur.com/kaydt.png

Answer 1

Hint: $$\int_0^1 f(x)dx=\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^n f(k/n)$$ (for $f:[0,1]\to\mathbb{R}$ continuous).

Answer 2

More explicitly, as Luiz Cordeiro stated, replace $\frac1{n}$ outside the sum by $dx$ and $\frac{k}{n}$ by $x$ inside.

This changes $\lim_{n \to \infty}\dfrac{3}{n}\sum_{k = 1}^{n}\dfrac{3k/n}{(3k/n+3)^2} $ into $\int_0^1 3\dfrac{3x\,dx}{(3x+3)^2} =\int_0^1 \dfrac{x\,dx}{(x+1)^2} $.