$\text{ If }I=\sum _{k=1}^{98}{\int }_k^{k+1}\frac{k+1}{x(x+1)}dx\text{ , then }$

$\begin{array}{l}I=\sum _{k=1}^{98}\underset{k}{\overset{k+1}{\int }}\frac{k+1}{x\left(x+1\right)}dx\\ =\sum _{k=1}^{98}\left(k+1\right)log\left(\frac{x}{x+1}\right)\left|{ }_k^{k+1}\right.\\ =2\log \frac{2^2}{1.3}+3.\log \frac{3^2}{2.4}+4.\log \frac{4^2}{3.5}+\mathrm{.....}+99\log \frac{{99}^2}{98.100}\\ =\log \frac{2^4{.3}^6{.4}^8{.5}^{10}{.6}^{12}-----{98}^{196}{.99}^{198}}{2^3{3}^6{.4}^8{.5}^{10}-----{98}^{196}{.99}^{98}{.100}^{99}}\\ =\log \left(\frac{{2.99}^{100}}{{100}^{99}}\right) <\log {}_{e}99 \left( \sin ce \frac{2\times {99}^{100}}{{100}^{99}}=99\times 2\times {\left(\frac{99}{100}\right)}^{99}\approx 99\times 2\times 0.36=99\times 0.72<99\right)\end{array}$