The value of int_{0}^{1} left( prod_{r=1}^{n} | vedclass

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(D) Let $f(x) = \prod_{r=1}^{n} (x+r)$.Then $\ln(f(x)) = \sum_{k=1}^{n} \ln(x+k)$.Differentiating both sides with respect to $x$,we get $\frac{f'(x)}{

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$n \cdot n!$

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(D) Let $f(x) = \prod_{r=1}^{n} (x+r)$.Then $\ln(f(x)) = \sum_{k=1}^{n} \ln(x+k)$.Differentiating both sides with respect to $x$,we get $\frac{f'(x)}{f(x)} = \sum_{k=1}^{n} \frac{1}{x+k}$.Thus,$f'(x) = f(x) \sum_{k=1}^{n} \frac{1}{x+k}$.The integral becomes $\int_{0}^{1} f'(x) dx = [f(x)]_{0}^{1}$.$f(1) = \prod_{r=1}^{n} (1+r) = 2 \cdot 3 \cdot \dots \cdot (n+1) = (n+1)!$.$f(0) = \prod_{r=1}^{n} (0+r) = 1 \cdot 2 \cdot \dots \cdot n = n!$.Therefore,the value of the integral is $(n+1)! - n! = n!(n+1-1) = n \cdot n!$.

The value of $\int_{0}^{1} \left( \prod_{r=1}^{n} (x+r) \right) \left( \sum_{k=1}^{n} \frac{1}{x+k} \right) dx$ equals

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