$\int\frac{dx}{x(x+1)(x+2)\cdot\space...\space\cdot(x+n)}$ I've been trying to solve explicitly the following indefinite integral:
$$\int\frac{dx}{x(x+1)(x+2)\cdot\space...\space\cdot(x+n)}$$
I tried to perform partial fraction decomposition, and after substituting some natural n's, I figured the Binomial Theorem might help here, but I couldn't figure out how to use it. 
Thank you and have a good day/night!
 A: Let
$\begin{array}\\
I_n
&=\int\dfrac{dx}{x(x+1)(x+2)\cdot\space...\space\cdot(x+n)}\\
&=\int\dfrac{dx}{\prod_{k=0}^n (x+k)}\\
\end{array}
$
Let's try partial fractions.
If
$\dfrac1{\prod_{k=0}^n (x+k)}
=\sum_{k=0}^n \dfrac{a_k}{x+k}
$,
then
$1
=\sum_{k=0}^n \dfrac{a_k\prod_{j=0}^n (x+j)}{x+k}
=\sum_{k=0}^n a_k\prod_{j=0,j \ne k}^n (x+j)
$.
Setting
$x = -m, 0 \le m \le n$,
$\begin{array}\\
1
&=\sum_{k=0}^n a_k\prod_{j=0,j \ne k}^n (-m+j)\\
&= a_m\prod_{j=0,j \ne m}^n (-m+j)\\
&= a_m\prod_{j=0}^{m-1} (-m+j)\prod_{j=m+1}^n (-m+j)\\
&= a_m(-1)^m\prod_{j=0}^{m-1} (m-j)\prod_{j=m+1}^n (j-m)\\
&= a_m(-1)^m\prod_{j=1}^{m} j\prod_{j=1}^{n-m} j\\
&= a_m(-1)^mm!(n-m)!\\
\end{array}
$
so
$a_m
=\dfrac{(-1)^m}{m!(n-m)!}
$.
Therefore
$\begin{array}\\
I_n
&=\int\dfrac{dx}{\prod_{k=0}^n (x+k)}\\
&=\int \sum_{k=0}^n \dfrac{a_k}{x+k}dx\\
&=\sum_{k=0}^n a_k\int \dfrac1{x+k}dx\\
&=\sum_{k=0}^n a_k(\ln(x+k)+c_k)\\
&=\sum_{k=0}^n a_k\ln(x+k)+C\\
\end{array}
$
This is undoubtedly well-known,
but I did work it out
independently.
