If $I=\sum_{k=1}^{98}\int_{k}^{k+1}\frac{k+1}{x(x+1)}\,dx$, then

If I=∑ₖ₌₁⁹⁸∫ₖᵏ⁺¹(k+1)/(x(x+1)) dx, then $I = \sum_{k=1}^{98}\displaystyle\int_{k}^{k+1} \dfrac{k+1}{x(x+1)}\,dx$ $\text{To find: how } I \text{ compares with } \log_e 99 \text{ and } \tfrac{49}{50}$ $1.\ \text{sandwich } \dfrac{k+1}{x(x+1)} \text{ between } \dfrac{1}{x+1} \text{ and } \dfrac{1}{x}$ $2.\ \text{integrate, then sum (telescoping)}$ $3.\ \text{compare with } \ln 99 \text{ and } \tfrac{49}{50}$ $x \in [k, k+1] \Rightarrow \dfrac{1}{x+1} \le \dfrac{k+1}{x(x+1)} \le \dfrac{1}{x}$ $\dfrac{k+1}{x(x+1)} \le \dfrac{1}{x} \iff \dfrac{k+1}{x+1} \le 1 \iff k+1 \le x+1 \ \checkmark$ $\displaystyle\int_{k}^{k+1}\dfrac{k+1}{x(x+1)}dx \le \int_{k}^{k+1}\dfrac{dx}{x} = \ln\dfrac{k+1}{k}$ $I < \sum_{k=1}^{98}\ln\dfrac{k+1}{k} = \ln\dfrac{99}{1} = \ln 99$ $I < \ln 99 \Rightarrow (2) \text{ is TRUE}, \ (1) \text{ is FALSE}$ $\dfrac{k+1}{x(x+1)} \ge \dfrac{1}{x+1} \iff \dfrac{k+1}{x} \ge 1 \iff x \le k+1 \ \checkmark$ $\displaystyle\int_{k}^{k+1}\dfrac{k+1}{x(x+1)}dx \ge \int_{k}^{k+1}\dfrac{dx}{x+1} = \ln\dfrac{k+2}{k+1}$ $I > \sum_{k=1}^{98}\ln\dfrac{k+2}{k+1} = \ln\dfrac{100}{2} = \ln 50$ $\ln 50 \approx 3.91 > \dfrac{49}{50} = 0.98$ $I > \ln 50 > \dfrac{49}{50} \Rightarrow (4) \text{ is TRUE}, \ (3) \text{ is FALSE}$ $\boxed{\text{Answer: options (2) and (4)}}$