The value of $\int_0^1\ (\prod_{r=1}^n x+r).(\sum_{k=1}^n \frac{1}{x+k}).dx $ is $$\int_0^1\ (\prod_{r=1}^n x+r).(\sum_{k=1}^n \frac{1}{x+k}).dx $$
My working:
$$\int_0^1\ [(x+1)(x+2)(x+3)...(x+n)]\times[\frac{1}{x+1}\ + \frac{1}{x+2}\ + \frac{1}{x+3}\ +\ ...\ + \frac{1}{x+n}].dx \\$$
$\int_0^1[(x+1)(x+2)(x+3)...(x+n)]\ \times\ \frac{[(x+2)(x+3)..(x+n)\ +\ (x+1)(x+3)..(x+n)\ +\ ...\ +\ (x+1)(x+2)..(x+n-1)]}{(x+1)(x+2)(x+3)...(x+n)}.dx\\
$
Now the numerator of the left expression and the denominator of the right expression will cancel out $\\$
$\int_0^1\require{enclose}\enclose{downdiagonalstrike}{[(x+1)(x+2)(x+3)...(x+n)]}\ \times\ \frac{[(x+2)(x+3)..(x+n)\ +\ (x+1)(x+3)..(x+n)\ +\ ...\ +\ (x+1)(x+2)..(x+n-1)]}{\enclose{downdiagonalstrike}{(x+1)(x+2)(x+3)...(x+n)}}.dx$
Now we are left with
$$\int_0^1 [(x+2)(x+3)..(x+n)\ +\ (x+1)(x+3)..(x+n)\ +\ ...\ +\ (x+1)(x+2)..(x+(n-1))].dx$$
I am unable to think of anything after this step although, I have an intuition that it may have something to do with factorials.

The integrand function is the derivative of $(x+1)(x+2)\cdots(x+n)=(x+1)_n$, hence the outcome simply is
$$\left[(x+1)_n\right]_{0}^{1}=(n+1)!-n!=n\cdot n!.$$
