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Question Number 165757 by Zaynal last updated on 07/Feb/22

	Answered by Mathspace last updated on 08/Feb/22
	$p(x)=Π_(n=1) ^(100) (x+n) ⇒((p^′ (x))/(p(x))) =Σ_(n=1) ^(100) (1/(x+n)) ⇒ ∫_0 ^1 {Π_(n=1) ^(100) (x+n)(Σ_(n=1) ^(100 ) (1/(x+n)))dx =∫_0 ^1 p(x).((p^′ (x))/(p(x)))dx=∫_0 ^1 p^′ (x)dx =p(1)−p(0) =Π_(n=1) ^(100) (n+1)−Π_(n=1) ^(100) n =Π_(n=2) ^(101) n−100! =101!−100! =101.100!−100! =100×(100)!$
	$p (x) = \prod_{n = 1}^{100} (x + n) \Rightarrow \frac{p^{^{'}} (x)}{p (x)}$ $= \sum_{n = 1}^{100} \frac{1}{x + n} \Rightarrow$ $\int_{0}^{1} {\prod_{n = 1}^{100} (x + n) (\sum_{n = 1}^{100} \frac{1}{x + n}) d x$ $= \int_{0}^{1} p (x) . \frac{p^{^{'}} (x)}{p (x)} d x = \int_{0}^{1} p^{^{'}} (x) d x$ $= p (1) - p (0)$ $= \prod_{n = 1}^{100} (n + 1) - \prod_{n = 1}^{100} n$ $= \prod_{n = 2}^{101} n - 100!$ $= 101! - 100!$ $= 101.100! - 100!$ $= 100 \times (100)!$
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