If Ik=1∑98∫kk+1x(x+1)k+1dx,then:I (log)e99I>5049 (d) I<(log)e99

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If $I k = 1 \sum 98 \int_{k}^{k + 1} \frac{k + 1}{x ( x + 1 )} d x, t h e n :$ $I < \frac{49}{50}$ (b) $I > (lo g)_{e} 99$ $I > \frac{49}{50}$ (d) $I < (lo g)_{e} 99$

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I k = 1 \sum 98 \int_{k}^{k + 1} \frac{k + 1}{x ( x + 1 )} d x, t h e n :

I < \frac{49}{50}

(b)

I > (lo g)_{e} 99

I > \frac{49}{50}

(d)

I < (lo g)_{e} 99

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Question Text	If $I k = 1 \sum 98 \int_{k}^{k + 1} \frac{k + 1}{x ( x + 1 )} d x, t h e n :$ $I < \frac{49}{50}$ (b) $I > (lo g)_{e} 99$ $I > \frac{49}{50}$ (d) $I < (lo g)_{e} 99$
Answer Type	Video solution: 1
Upvotes	151