∫(∏r=1n(x+r)) (∑k=1n1x+k) dx=

Let y=nπr=1(x+r)⇒∫logy=∑r=1nlog(x+r)⇒∫1ydy=∑1x+rdx⇒∫dy=nπr=1(x+r)n∑k=11x+r dx∴I=∫dy=y+c=nπr=1(x+r)+c

∫(∏r=1n(x+r)) (∑k=1n1x+k) dx=

Let y=nπr=1(x+r)⇒∫logy=∑r=1nlog(x+r)⇒∫1ydy=∑1x+rdx⇒∫dy=nπr=1(x+r)n∑k=11x+r dx∴I=∫dy=y+c=nπr=1(x+r)+c

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$\int (\prod_{r = 1}^{n} (x + r)) (\sum_{k = 1}^{n} \frac{1}{x + k}) d x =$

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$\prod_{r = 1}^{n} (x + r) + c$

$\sum_{k = 1}^{n} \frac{1}{x + k} + c$

$\sum_{r = 1}^{n} \log_{e} (x + r) + c$

$x + c$

answer is A.

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Detailed Solution

Let $y = \begin{matrix} n \\ π \\ r = 1 \end{matrix} (x + r)$

$\begin{array}{l} \Rightarrow \int \log y = \sum_{r = 1}^{n} \log (x + r) \\ \Rightarrow \int \frac{1}{y} d y = \sum \frac{1}{x + r} d x \\ \Rightarrow \int d y = \begin{matrix} n \\ π \\ r = 1 \end{matrix} (x + r) \begin{matrix} n \\ \sum \\ k = 1 \end{matrix} \frac{1}{x + r} d x \\ ∴ I = \int d y = y + c \\ = \begin{matrix} n \\ π \\ r = 1 \end{matrix} (x + r) + c \end{array}$