Why is this weighted sum of binomials with alternating signs simplifies?

I have stumbled upon this relation but can't figure out why this is so:

$$ \sum_{k=1}^i (-1)^{k-1} \binom{i}{k} \frac{k}{n-i+k} = \frac{1}{\binom{n}{i}} $$

Can anyone shows how this is the case? Is this a known relation?

Since, for $p\geq 1$, $$\int_0^1 x^{p-1} dx= \frac 1 p$$

$$\begin{split} \sum_{k=1}^i (-1)^{k-1} \binom{i}{k} \frac{k}{n-i+k} &= \int_0^1\sum_{k=1}^i (-1)^{k-1} \binom{i}{k} kx^{n-i+k-1}dx\\ &=\int_0^1 x^{n-i}\sum_{k=1}^i (-1)^{k-1} \binom{i}{k} kx^{k-1}dx\\ &= -\int_0^1 x^{n-i}\frac{d}{dx}\left(\sum_{k=0}^i (-1)^{k} \binom{i}{k} x^{k}\right)dx\\ &=-\int_0^1 x^{n-i}\frac{d}{dx}\left[\left(1-x\right)^i\right]dx\\ &= i\int_0^1 x^{n-i}\left(1-x\right)^{i-1}dx\\ \end{split}$$ Using the properties of the Beta function, which can be proven by integration by parts, $$\int_0^1 x^{n-i}(1-x)^{i-1}dx=B(n-i+1, i)=\frac{n+1}{(n-i+1)i}\frac 1 {n+1 \choose i}$$

$$\sum_{k=1}^i (-1)^{k-1} \binom{i}{k} \frac{k}{n-i+k}=\frac{n+1} {n-i+1}\frac 1 {n+1 \choose i}=\frac 1 {n \choose i}$$