$\int\limits_{0}^{1}(\prod\limits_{r=1}^{n}(x+r))(\sum\limits_{k=1}^{n}\frac{1}{x+k})dx$ 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...

First let us make a common denominator for the rationals under the sum sign
$$
\left(\prod\limits_{r=1}^n (x+r)\right)
\left(\sum\limits_{k=1}^n \frac{1}{x+k} \right)\\
=
\left(\prod\limits_{r=1}^n (x+r)\right)
\left(
\sum\limits_{k=1}^n
\frac{\prod\limits_{r=1,r\neq k}^n (x+r)}
{\prod\limits_{r=1}^n (x+r)}
\right)
=
\left(
\sum\limits_{k=1}^n
\prod\limits_{r=1,r\neq k}^n (x+r)
\right)
=
\frac{d}{dx}
\left(
\prod\limits_{r=1}^n (x+r)
\right),
$$
where we used the product rule for the last equal sign.
Now can you complete your exercise on your own?
