Binomial and Uniform joint distribution statistics question. I was just wondering if someone could explain the solution to this example in my textbook for me. I am not really understanding it. The question is, 
X follows Binomial(n,p) and p follows Uniform(0,1). Show, P(X =0) = P(X = 1) = · · · = P(X = n). You are given that the integral of the binomial distribution over (0,1) is equal to  [i!(n-1)!]/(n+1)!.
$P(X=k)=\binom{n}{k}\int_0^1 p^k(1-p)^{n-k}dp$.
Integrate by parts $du=(1-p)^{n-k}dp,\ v=p^k$.
You will get$P(X=k)=\binom{n}{k} \frac{k}{n-k+1}\int_0^1 p^{k-1}(1-p)^{n-k+1}dp=P(X=k-1)$

Therefore all P(X=k) are equal.
$P(X=k)=\binom{n}{k}\int_0^1 p^k(1-p)^{n-k}dp$.
Integrate by parts $du=(1-p)^{n-k}dp,\ v=p^k$.
You will get$P(X=k)=\binom{n}{k} \frac{k}{n-k+1}\int_0^1 p^{k-1}(1-p)^{n-k+1}dp=P(X=k-1)$

Therefore all P(X=k) are equal.