The result corresponds to a special case of the Beta-binomial distribution, which can be generalized as follows.
Suppose that $Y \sim {\rm Beta}(\alpha,\beta)$, $\alpha,\beta>0$ real; thus $Y$ has density $f_Y{(p)} = p^{\alpha - 1} (1 - p)^{\beta - 1} /{\rm B}(\alpha ,\beta )$, $p \in [0,1]$, where ${\rm B}(x,y) = \Gamma(x)\Gamma(y)/\Gamma(x+y)$ is the Beta function. Further suppose that $X \sim {\rm binomial}(N,Y)$, $N \in \mathbb{N}$ fixed, meaning that given $Y=p$, $X \sim {\rm binomial}(N,p)$.
Then, $X$ has a Beta-binomial distribution with parameters $N$, $\alpha$, and $\beta$.
By the law of total probability, the probability mass function of $X$ is given, for $k=0,1,\ldots,N$, by
$$
{\rm P}(X=k) = {\rm E}[{\rm P}(X=k|Y)] = \int_0^1 {{N \choose k}p^k (1 - p)^{N - k} f_Y (p)\,{\rm d}p}.
$$
Hence,
$$
\int_0^1 {p^{k + \alpha - 1} (1 - p)^{N - k + \beta - 1} \,{\rm d}p} = \frac{{{\rm B}(\alpha ,\beta )}}{{{N \choose k}}}{\rm P}(X = k).
$$
Explicitly, the left-hand side is given by
$$
\int_0^1 {p^{k + \alpha - 1} (1 - p)^{N - k + \beta - 1} \,{\rm d}p} = {\rm B}(k + \alpha ,N - k + \beta ) = \frac{{\Gamma (k + \alpha )\Gamma (N - k + \beta )}}{{\Gamma (N + \alpha + \beta )}}
$$
(say, by definition of the Beta function), but this is, of course, not the point here: the point is the relation to the
Beta-binomial distribution. In the special case where $\alpha=\beta=1$, we have $Y \sim {\rm uniform}[0,1]$, and by substitution we find that ${\rm P}(X=k)=1/(N+1)$. Hence the "uniform$[0,1]$-binomial distribution" is simply the discrete uniform distribution on $\lbrace 0,1,\ldots,N \rbrace$.
