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

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 $$C~$$?
• What if the arguments of your logs are negative? Shouldn't you have log|▪︎| everywhere? – Wolfgang Feb 1 at 21:28
• notice that $\log(x-w)$ and $\log(w-x)$ only differ by an (imaginary) constant, so this can be absorbed in the $C$ in $f_u(x)$; the only difference between $u=1,2$ and $u>2$ that matters is that the sign in front of the logarithm is different from $(-1)^w$; if I am allowed to change the sign of the definition of $I_{u,w}$ for $u=1,2$, I'm done. – Carlo Beenakker Feb 1 at 21:29
• are you sure your equation (1) is correct? I think $-\sum_{w=1}^u$ should be $+\sum_{w=1}^u$ – Carlo Beenakker Feb 1 at 21:38
• @CarloBeenakker you are correct, fixed. – Descartes Before the Horse Feb 1 at 21:40

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.