How it can be shown that:
$$\sum_{k=0}^{\large\lfloor{\frac{n}{2}}\rfloor}\binom{n}{2k}\frac{x^{2k}}{2k+1}=\frac{\left(x+1\right)^{\left(n+1\right)}-\left(1-x\right)^{\left(n+1\right)}}{2x\left(n+1\right)}$$
Where $\lfloor{x}\rfloor$ denotes the floor function.
My try:
$$\begin{align}&\sum_{k=0}^{\large\lfloor{\frac{n}{2}}\rfloor}\binom{n}{2k}\frac{x^{2k}}{2k+1}\\ &=\frac{1}{n+1}\sum_{k=0}^{\large\lfloor{\frac{n}{2}}\rfloor}\binom{n+1}{2k+1}x^{2k}\\&=\frac{1}{\left(n+1\right)x}\sum_{k=0}^{\large\lfloor{\frac{n}{2}}\rfloor}\binom{n+1}{2k+1}\color{red}{x^{2k+1}} \end{align}$$
Which follows from the identity:
$$\sum_{k=0}^{\large\lfloor{\frac{n-1}{2}}\rfloor}\binom{n}{2k+1}=2^{n-1} \;\;\;\;\;\;\; (n \in \mathbb N_{\ge1})$$
But I don't know how to continue with that $\color{red}{x^{2k+1}}$.