Evaluation of $\int\frac{1}{\prod_{k=0}^{n}(x+k)}dx$ Evaluation of $\displaystyle \int\frac{1}{\prod_{k=0}^{n}(x+k)}dx$
$\bf{My\; Trial::}$We can write
$\displaystyle \int\frac{1}{\prod_{k=0}^{n}(x+k)}dx = \int \frac{1}{(x+0)\cdot(x+1)\cdot (x+2)..........(x+n)}dx$
Now Using Partial fraction.......
$\Rightarrow \displaystyle \frac{1}{(x+0)\cdot (x+1)\cdot (x+2).........(x+n)} = \frac{A_{1}}{x}+\frac{A_{2}}{(x+1)}+\frac{A_{3}}{(x+2)}+....+\frac{A_{n}}{(x+n)}.$
Now How can I solve after that
Help Required
 A: HINT:
Multiply either sides by $\prod_{r=0}^n(x+r)$
Set $\displaystyle x+s=0\iff x=-s,0\le s\le n$ to find $$1=A_{s+1}\left(\prod_{r=0,r\ne s}^n{r+n}\right)$$
A: $\newcommand{\+}{^{\dagger}}
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Note that
\begin{align}
&\Gamma\pars{x}={\Gamma\pars{x + 1} \over x}
={\Gamma\pars{x + 2} \over x\pars{x + 1}}
=\cdots={\Gamma\pars{x + n + 1} \over x\pars{x + 1}\ldots\pars{x + n}}
\\[3mm]\mbox{such that}&\quad
{1 \over x\pars{x + 1}\ldots\pars{x + n}}
={\Gamma\pars{x} \over \Gamma\pars{x + n + 1}}
\end{align}

\begin{align}&\color{#c00000}{\int{\dd x \over x\pars{x + 1}\ldots\pars{x + n}}}
=\int{\Gamma\pars{x} \over \Gamma\pars{x + n + 1}}\,\dd x
={1 \over n!}\int{\Gamma\pars{x}\Gamma\pars{n + 1}\over
\Gamma\pars{x + n + 1}}\,\dd x
\\[3mm]&={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\,\dd t
\\[3mm]&={1 \over n!}\ \overbrace{%
\int_{0}^{1}\pars{1 - t}^{n}{t^{x- 1} \over \ln\pars{t}}\,\dd t}
^{\ds{\mbox{Set}\ t \equiv \expo{-\xi}\ \imp\ \xi = -\ln\pars{t}}}\ =\
{1 \over n!}\int_{\infty}^{0}\pars{1 - \expo{-\xi}}^{n}{\expo{-\pars{x- 1}\xi}\over -\xi}\,\pars{-\expo{-\xi}}\,\dd\xi
\\[3mm]&={1 \over n!}\int_{0}^{\infty}\pars{1 - \expo{-\xi}}^{n}\expo{-x\xi}\,{\dd\xi \over \xi}
\\[3mm]&=-\,{1 \over n!}\int_{0}^{\infty}\ln\pars{\xi}\bracks{%
n\pars{1 - \expo{-\xi}}^{n - 1}\expo{-x\xi}
-x\pars{1 - \expo{-\xi}}^{n}\expo{-x\xi}}\,\dd\xi
\\[3mm]&={1 \over n!}\bracks{%
x\int_{0}^{\infty}\ln\pars{\xi}\pars{1 - \expo{-\xi}}^{n}\expo{-x\xi}\,\dd\xi
-n\int_{0}^{\infty}\ln\pars{\xi}\pars{1 - \expo{-\xi}}^{n - 1}\expo{-x\xi}\,\dd\xi}
\end{align}

\begin{align}
&\int_{0}^{\infty}\ln\pars{\xi}\pars{1 - \expo{-\xi}}^{m}\expo{-x\xi}\,\dd\xi
=\sum_{k = 0}^{m}{m \choose k}\pars{-1}^{k}
\int_{0}^{\infty}\ln\pars{\xi}\expo{-\pars{k + x}\xi}\,\dd\xi
\\[3mm]&=\sum_{k = 0}^{m}{m \choose k}\pars{-1}^{k}
\lim_{\mu \to 0}\partiald{}{\mu}\int_{0}^{\infty}
\xi^{\mu}\expo{-\pars{k + x}\xi}\,\dd\xi
=\sum_{k = 0}^{m}{m \choose k}\pars{-1}^{k}
\lim_{\mu \to 0}\partiald{}{\mu}\bracks{%
\Gamma\pars{\mu + 1} \over \pars{k + x}^{\mu + 1}}
\\[3mm]&=\sum_{k = 0}^{m}{m \choose k}\pars{-1}^{k + 1}\,
{\gamma + \ln\pars{k + x} \over k + x}
\end{align}

\begin{align}&\color{#c00000}{\int{\dd x \over x\pars{x + 1}\ldots\pars{x + n}}}
\\[3mm]&={1 \over n!}\bracks{x\sum_{k = 0}^{n}{n \choose k}\pars{-1}^{k + 1}\,
{\gamma + \ln\pars{k + x} \over k + x}
-n\sum_{k = 0}^{n - 1}{n - 1 \choose k}\pars{-1}^{k + 1}\,
{\gamma + \ln\pars{k + x} \over k + x}}
\end{align}

$\ds{\large\tt\mbox{It will continue}\ldots}$
A: Partial Fractions gives
$$
\begin{align}
\int\frac1{\prod_{k=0}^n(x+k)}\,\mathrm{d}x
&=\int\sum_{k=0}^n\frac{(-1)^k}{k!(n-k)!}\frac1{x+k}\,\mathrm{d}x\\
&=\frac1{n!}\sum_{k=0}^n(-1)^k\binom{n}{k}\log(x+k)+C
\end{align}
$$
