Evaluation of $\int \prod_{j=1}^u \frac{x+j}{j-x}~dx$

Question

Let $I_{n,k} = \frac{(n+k)!}{k!(k-1)!(n-k)!}$. This is a sort of generalization of the Apéry's numbers, with $I_{n,n} =$ the $n$-th Apéry number. I am studying integrals of the form: $$f_u(x)=\int \prod_{j=1}^u \frac{x+j}{j-x}~dx.$$ Where $u$ is a natural number. For $u>2$, I have shown that $$\tag{1}f_u(x) = C+ (-1)^ux+\sum_{w=1}^{u}(-1)^w \log(x-w)I_{u,w}.$$ For example, $$f_3(x)=-x-60\log(x-3)+60\log(x-2)-12\log(x-1).$$ I am looking for clarification on two things:

1. My eq. $(1)$ does not work for $u=1,2$. It generates a function that is almost what the integral evaluates to; whereas $f_1(x) = -x-2\log(1-x)$, eq. $(1)$ gives me $-x+2\log(x-1)$. Is there any amendment I can make to $(1)$ to ensure that it holds for all natural $u$?
2. My eq. $(1)$ surely appears like it should be written as one sum, but I have not been able to manipulate the summand to include the $(-1)^u x$ term. Is there a way to rewrite $(1)$ as a single sum, barring $C~$?

Answer 1

I may be mistaken, but I get $$f_u(x)=\int \prod_{j=1}^u \frac{x+j}{j-x}~dx= C+ (-1)^ux+\sum_{w=1}^{u}(-1)^w \log(x-w)I_{u,w}$$ and this is correct for all $u=1,2,3,...$, so it seems issue 1 is resolved.