I am trying to find if a closed form formula exists for the following summation:
$$\sum_{i=0}^{n} {n\choose i}\frac{x^{i}}{i+k}$$
where $x$ and $k$ can be any real numbers. I tried to search but the closest I found was the following identity (from vol.2 at https://www.math.wvu.edu/~gould/):
$$\sum_{i=0}^{n} {n\choose i}\frac{x^{i}}{i+1}=\frac{(x+1)^{n+1}-1}{(n+1)x}$$
Any help would be greatly appreciated.
A partial solution. Let $S$ be the desired sum. Then observe that for $k > 0$, \begin{align*} x^k\cdot S&=\sum_{i=0}^{n} {n\choose i}\frac{x^{i+k}}{i+k}= \sum_{i=0}^{n} {n\choose i}\int_{0}^x t^{i+k-1}\,dt\\ &=\int_{0}^x \sum_{i=0}^{n} {n\choose i} t^{i+k-1} =\int_{0}^x t^{k-1}(1+t)^n\,dt. \end{align*} If $k\geq 1$ is a natural number, we can repeatedly integrate by parts to evaluate the last integral.