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Question Number 65681 by aliesam last updated on 01/Aug/19

$${\int }_0^1\left(\underset{r=1}{\prod ^n}(x+r)\right)\left(\underset{k=1}{\sum ^n}\frac{1}{x+k}\right)dx$$

Answered by Tanmay chaudhury last updated on 02/Aug/19

$${\int }_0^1(x+1)(x+2)(x+3)\dots (x+n)\times (\frac{1}{x+1}+\frac{1}{x+2}+\frac{1}{x+3}+\dots +\frac{1}{x+n})dx$$$$y=(x+1)(x+2)(x+3)\dots (x+n)$$$$lny=ln(x+1)+ln(x+2)+ln(x+3)+..+ln(x+n)$$$$\frac{1}{y}\times \frac{dy}{dx}=\frac{1}{x+1}+\frac{1}{x+2}+\frac{1}{x+3}+..+\frac{1}{x+n}$$$$\frac{dy}{dx}=(x+1)(x+2)(x+3)\dots (x+n)\times (\frac{1}{x+1}+\frac{1}{x+2}+..+\frac{1}{x+n})$$$$dy=\left[\underset{r=1}{\prod ^n}(x+r)\times \underset{n=1}{\sum ^n}\frac{1}{x+r}\right]dx$$$${\int }_0^1\mathrm{\Pi }(x+r)\times \underset{k=1}{\sum ^n}\frac{1}{x+k}dx$$$$={\int }_0^{1}d(x+1)(x+2)..(x+n)$$$$\mid (x+1)(x+2)(x+3)..(x+n){\mid }_0^1$$$$=\left(2\right)\left(3\right)\left(4\right)\dots (1+n)-(1\times 2\times 3..\times n)$$$$=(n+1)!-n!$$$$=(n+1)!-n!$$$$=(n+1)n!-n!$$$$=(n+1-1)\times n!$$$$=n\times n!$$$$$$

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