Summation of integration terms If $I = \sum\limits_{k = 1}^{98} {\int\limits_k^{k + 1} {\frac{{k + 1}}{{x\left( {x + 1} \right)}}} dx} $ then
(A) $I>\ln99$
(B) $I<\ln99$
(C) $I<\frac{49}{50}$
(D) $I>\frac{49}{50}$
The official answer is B and D
My approach is as follow
$$I = \sum\limits_{k = 1}^{98} {\int\limits_k^{k + 1} {\frac{{k + 1}}{{x\left( {x + 1} \right)}}} dx} $$
$$\frac{A}{x} + \frac{B}{{x + 1}} = \frac{{A\left( {x + 1} \right) + Bx}}{{x\left( {x + 1} \right)}} = \frac{{x\left( {A + B} \right) + A}}{{x\left( {x + 1} \right)}}$$
$$A = k + 1\text{ and } B =  - \left( {k + 1} \right)$$
$$I = \sum\limits_{k = 1}^{98} {\int\limits_k^{k + 1} {\left( {\frac{{k + 1}}{x} - \frac{{k + 1}}{{x + 1}}} \right)} dx}  \Rightarrow I = \sum\limits_{k = 1}^{98} {\left( {k + 1} \right)\int\limits_k^{k + 1} {\left( {\frac{1}{x} - \frac{1}{{x + 1}}} \right)} dx} $$
$$I = \sum\limits_{k = 1}^{98} {\left( {k + 1} \right)\left. {\ln\frac{x}{{x + 1}}} \right|_k^{k + 1}}  \Rightarrow I = \sum\limits_{k = 1}^{98} {\left( {k + 1} \right)\left( {\ln\frac{{k + 1}}{{k + 2}} - \ln\frac{k}{{k + 1}}} \right)}$$ $$\Rightarrow I = \sum\limits_{k = 1}^{98} {\left( {k + 1} \right)\left( {\ln\frac{{{{\left( {k + 1} \right)}^2}}}{{k\left( {k + 2} \right)}}} \right)} $$
Not able to approach from here

You have
$$\sum_{k = 1}^{98} \int_k^{k + 1} \frac{{k + 1}}{{x\left( {x + 1} \right)}} dx  \le \sum_{k = 1}^{98} \int_k^{k + 1} \frac{dx}{x} = \int_1^{99} \frac{dx}{x} = \ln 99$$
hence B is true. Also
$$\begin{aligned}\sum_{k = 1}^{98} \int_k^{k + 1} \frac{{k + 1}}{{x\left( {x + 1} \right)}} dx  &\ge \sum_{k = 1}^{98} \int_k^{k + 1} \frac{dx}{x+1} \\
&\ge \sum_{k = 1}^{98} \int_k^{k + 1} \frac{dx}{100}\\
&=\frac{98}{100}= \frac{49}{50}
\end{aligned}$$
hence D is also true.
