calculus - How to prove the closed form of the integral $int frac {dx}{prod_{r=0}^n (x+r)}$

I want to derive a closed formula for the integral $$I_n= \int \frac {dx}{\prod_{r=0}^n (x+r)}$$

On writing out first few terms we get

For $n=0$, $$I_0=\ln \vert x\vert+C$$
For $n=1$, $$I_1=\ln \vert x\vert-\ln \vert x+1\vert+C$$
For $n=2$ $$I_2=\frac {1}{2!}\left(\sum_{r=0}^2 (-1)^r\binom {2}{r} \ln \vert x+r\vert\right)+C$$
For $n=3$ $$I_3=\frac {1}{3!}\left(\sum_{r=0}^3 (-1)^r\binom {3}{r} \ln \vert x+r\vert\right)+C$$

Hence for generalized $n$ we have $$I_n=\frac {1}{n!}\left(\sum_{r=0}^n (-1)^r\binom {n}{r} (\ln \vert x+r\vert)\right)+C$$

Now this is just an observation but I want to prove that it is correct. I have tried lot of methods but not useful. Partial fractions would have been most useful but would go out on tedious task which is nearly impossible. Also integration by parts won't help nor any trig substitution. So any ideas are welcome. And ya, this is not a homework question, it's a question which I just saw in a integral challenge paper.

Answer

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}

\newcommand{\ic}{\mathrm{i}}
\newcommand{\mc}[1]{\mathcal{#1}}
\newcommand{\mrm}[1]{\mathrm{#1}}
\newcommand{\pars}[1]{\left(\,{#1}\,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
\newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\int{\dd x \over \prod_{r = 0}^{n}\pars{x + r}} & =

\int{\dd x \over \Gamma\pars{x + n + 1}/\Gamma\pars{x}} =
{1 \over n!}
\int{\Gamma\pars{x}\Gamma\pars{n + 1}\over \Gamma\pars{x + n + 1}}\,\dd x
\\[5mm] & =
{1 \over n!}\int\int_{0}^{1}t^{x - 1}\pars{1 - t}^{n}\,\dd t\,\dd x =
{1 \over n!}\int_{0}^{1}\pars{1 - t}^{n}\int t^{x - 1}\,\dd x
\\[5mm] & =
{1 \over n!}\int_{0}^{1}\pars{1 - t}^{n}
\bracks{{t^{x - 1} \over \ln\pars{t}} + \,\mrm{A}\pars{t}}\,\dd x
\\[5mm] & =

{1 \over n!}\int_{0}^{1}{t^{x - 1}\pars{1 - t}^{n} \over \ln\pars{t}}\,\dd t +
{1 \over n!}\int_{0}^{1}{t^{x - 1}\,\mrm{A}\pars{t} \over \ln\pars{t}}\,\dd t
\end{align}

$\ds{\mrm{A}\pars{t}}$ is an integration "constant" ( it doesn't depend on $\ds{x}$ but, in general, it depends on $\ds{t}$ ).