$\int\limits_{0}^{1}(\prod\limits_{r=1}^{n}(x+r))(\sum\limits_{k=1}^{n}\frac{1}{x+k})dx$ The value of $\int\limits_{0}^{1}(\prod\limits_{r=1}^{n}(x+r))(\sum\limits_{k=1}^{n}\frac{1}{x+k})dx$ is equal to
$(A)n\hspace{1cm}(B)n!\hspace{1cm}(C)(n+1)!\hspace{1cm}(D)n.n!$
I tried:$\int\limits_{0}^{1}(\prod\limits_{r=1}^{n}(x+r))(\sum\limits_{k=1}^{n}\frac{1}{x+k})dx$=$\int\limits_{0}^{1}(x+1)(x+2)...(x+n)(\frac{1}{x+1}+\frac{1}{x+2}+......+\frac{1}{x+n})dx$=$\int\limits_{0}^{1}(x+2)(x+3)...(x+n)+(x+1)(x+3)...(x+n)dx$
I cannot solve it further.Is my approach wrong,I am stuck.What is the right way to solve,Please help...

Differentiate $\prod_{k=1}^n(x+k)$ with respect to $x$, and you get the integrand. So the answer is $(n+1)!-n!=n\cdot n!$.
