Let $I\_{n,k} = \frac{(n+k)!}{k!(k-1)!(n-k)!}$. This is a sort of generalization of the [Apéry's numbers](http://mathworld.wolfram.com/AperyNumber.html), 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 $C~$?**



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


