Identity related to $\sum_{k=0}^{n}\frac{x^k}{\binom{n}{k}}$ How it can be shown that:

$$\sum_{k=0}^{n}\frac{x^k}{\binom{n}{k}}=\left(n+1\right)\left(\frac{x}{x+1}\right)^{n+1}\sum_{k=1}^{n+1}\frac{1+x^k}{\left(1+x\right)k}\left(\frac{1+x}{x}\right)^{k}$$

for $x \ne-1$
I tried Additive Forms of Reciprocal Pascal’s Identity , but could not derive that.

Your result is a kind of generalization of Newton's binomial identity with binomial coefficients replaced by their inverses. 
You will find in page 2 of this reference by Toufik Mansour, University of Haifa) the more general expression :
$$\sum_{k=0}^{n}\frac{a^kb^{n-k}}{\binom{n}{k}}=\frac{n+1}{(a+b)\left(\tfrac{1}{a}+\tfrac{1}{b}\right)^{n+1}}\sum_{k=1}^{n+1}\frac{(a^k+b^k)\left(\tfrac{1}{a}+\tfrac{1}{b}\right)^{k}}{k}\tag{1}$$
with its proof. Nice expression...
It suffices now to take $a=x$ and $b=1$...
I have to admit that the proof is long... and uses generating functions.
Remark : If in the following identity for Beta integrals (see here): 
$$B(x,y)=\int_{t=0}^{t=1}t^{x-1}(1-t)^{y-1}dt=\frac{(x-1)!(y-1)!}{(x+y-1)!}\tag{2}$$
one takes $x-1=k$ and $y-1=n-k$, we deduce that :
$$\frac{1}{\binom{n}{k}}=(n+1)\int_{t=0}^{t=1}t^k(1-t)^{n-k}dt\tag{3}$$
(besides, (3) is mentionned in the referenced article).
It is likely that a (simpler) proof of your identity can be deduced from (3) by multiplying it by $x^k$ and summing from $k=0$ to $k=n$.
