definite

\int_{0}^{1}\prod_{r=1}^{n}\frac{1}{(x+r)}.\sum_{k=1}^{n}\frac{1}{(x+k)}dx$\\\\\\ Now Let $\prod_{r=1}^{n}\frac{1}{(x+r)}=t$, Then taking $ln$ on both side, we Get\\\\\\ $ln\left \{ \prod_{r=1}^{n}\frac{1}{(x+r)} \right \}=lnt$, Then Diff. both side W.r.to $x$, we get\\\\\\ $-\left \{ \sum_{k=1}^{n}\frac{1}{(x+k)}\right \}dx = \frac{1}{t}dt\Leftrightarrow \left \{ \sum_{k=1}^{n}\frac{1}{(x+k)}\right \}dx = -\frac{1}{t}dt$\\\\\\ Now The Integral convert into $-\int_{0}^{1} t.\frac{1}{t}dt = -t = -\prod_{r=1}^{n}\frac{1}{(x+r)}\right_{0}^{1}$\\\\ $=-\frac{1}{2.3.4....................n}+\frac{1}{1.2.3.4.......................n}=0$