How could I prove this?

Question

$\newcommand{\intd}{\,\mathrm{d}}$On an AOPS forum a user solved that $$\int\frac{1}{x(x+1)(x+2)\ldots(x+n)}\intd{x}=\frac1{n!}\sum_{k=0}^n(-1)^k\binom{n}{k}\ln|x+k|+C$$ This to me is absolutely amazing and their solution interesting enough, I am curious as to how I would go about proving this. To me this shouts induction, it is obvious for $n=0$ ($\int\frac1x\intd{x}$) however on the inductive step I could not figure out what to do. Integration by parts did not help and beyond that I am at a loss. I look forward to your responses

Answer 1

BY induction, we prove: $$\frac{1}{x(x+1)(x+2)\ldots(x+n)}=\frac1{n!}\sum_{k=0}^n(-1)^k\binom{n}{k}\frac{1}{x+k}.$$ For $n=1,2$, it is right. Support $n$ is right, now for $n+1$; $$\frac{1}{x(x+1)(x+2)\ldots(x+n)(x+n+1)}=\frac1{n!}\sum_{k=0}^n(-1)^k\binom{n}{k}\frac{1}{x+k}\frac{1}{x+n+1}$$ $$=\frac1{n!}\sum_{k=0}^n(-1)^k\binom{n}{k}\frac{1}{(n+1)-k}\left(\frac{1}{x+k}-\frac{1}{x+n+1}\right)$$ $$=\frac1{(n+1)!}\sum_{k=0}^n(-1)^k\binom{n+1}{k}\left(\frac{1}{x+k}-\frac{1}{x+n+1}\right)$$ $$=\frac1{(n+1)!}\sum_{k=0}^n(-1)^k\binom{n+1}{k}\frac{1}{x+k}-\frac1{(n+1)!}\sum_{k=0}^n(-1)^k\binom{n+1}{k}\frac{1}{x+(n+1)}$$ $$=\frac1{(n+1)!}\sum_{k=0}^n(-1)^k\binom{n+1}{k}\frac{1}{x+k}+\frac1{(n+1)!}(-1)^{n+1}\frac{1}{x+(n+1)}$$ $$=\frac1{(n+1)!}\sum_{k=0}^{n+1}(-1)^k\binom{n+1}{k}\frac{1}{x+k}.$$

Answer 2

We know that $$\dfrac{1}{x(x+1)(x+2)\cdots (x+n)}=\dfrac{a_0}{x}+\dfrac{a_1}{x+1}+\dfrac{a_2}{x+2}+\cdots +\dfrac{a_n}{x+n}$$where $$a_k=\dfrac{1}{-k(-k+1)(-k+2)\cdots (-1)(1)(2)\cdots (n-k-1)(n-k)}=\dfrac{1}{n!}(-1)^k\dfrac{n!}{k!(n-k)!}=\dfrac{1}{n!}(-1)^k\binom{n}{k}$$which by substituting in $\int \dfrac{1}{x+a}dx=\ln|x+a|+C$ yields to $$I=\dfrac{1}{n!}\sum_{i=0}^{n}(-1)^i\binom{n}{i}\ln|x+i|+C$$

Answer 3

If you want to prove this by induction, try write \begin{align*} &\phantom{==}\frac 1{x(x+1)(x+2)\cdots(x+n+1)}\\ &= \frac 1{(x+1)(x+2)\cdots (x+n)} \cdot \frac 1{x(x+n+1)}\\ &=\frac 1{(x+1)(x+2)\cdots (x+n)} \cdot\left (\frac 1x - \frac 1{x+n+1}\right) \cdot \frac 1{n+1}\\ &= \frac 1{n+1} \left( \frac 1{x(x+1)\cdots (x+n)} - \frac 1 {(x+1)(x+2)\cdots (x+n+1)}\right). \end{align*}