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

Hint: $$\sum_1^{98} (k+1)\ln\frac{(k+1)^2}{k(k+2)} \\ = \sum_1^{98} (k+1) \left( \ln\frac{k+1}{k} - \ln \frac{k+2}{k+1} \right) \\ = 2(\ln 2-\ln\frac 32) + 3(\ln \frac 32 -\ln \frac 43) +4(\ln \frac 43 -\ln \frac 54) +\dots +99(\ln \frac{99}{98} -\ln \frac{100}{99}) \\ =2\ln 2 + \big( \ln\frac 32 +\ln \frac 43 +\dots +\ln \frac{99}{98} \big) -99\ln\frac{99}{100} \\ =2\ln 2+\ln\frac{99}{2} -99\ln\frac{99}{100}$$
