Show that $\int_{0}^{1}(1-u)^p u^{p+1} \text{d}u = \frac{\Gamma(p+1)\Gamma(p+2)}{\Gamma(2p+3)}$ Assume $p > 1$. I want to show that 
$$ \int_{0}^{1}(1-u)^p u^{p+1} \text{d}u = \frac{\Gamma(p+1)\Gamma(p+2)}{\Gamma(2p+3)},$$
here the result is from WolframAlpha. 
However, not having any experience with Gamma functions, I'm not really sure where to even start. Wolfram also does not show any step-by-step hints, as it sometimes does. Any hints on how to tackle such integrals?
 A: Per definition
$$
B(x,y) = \int_{0}^1 t^{x-1}(1-t)^{y-1} dt
$$
Applying to your case we get
$$
B(x+1,y+2)
$$
On the other hand 
$$
B(x,y) = \frac{\Gamma(x) \Gamma(y)}{\Gamma (x+y)}
$$
So
$$
B(x+1,y+2) = \frac{\Gamma(x+1) \Gamma(y+2)}{\Gamma (x+y+3)}
$$
Now set $p=y=x$ and read 
https://en.wikipedia.org/wiki/Beta_function#Relationship_between_gamma_function_and_beta_function
for verification.
A: I'll just start from the basics if you aren't aware what the Beta Function is. Essentially, the Beta Function establishes a relationship between the Gamma function$$\Gamma(z)=(z-1)!$$and the integral$$\int\limits_0^1\mathrm dt\, t^{n-1}(1-t)^{m-1}=\operatorname B(m,n)$$More succinctly, we have that$$\operatorname{B}(m,n)=\frac {\Gamma(m)\Gamma(n)}{\Gamma(m+n)}$$Here's a brief elementary proof.

First, start off with $\operatorname{B}(m+1,n+1)$ and use integration by parts such that$$\int\limits_0^1\mathrm dt\, t^n(1-t)^m=\frac m{n+1}\int\limits_0^1\mathrm dt\, t^{n+1}(1-t)^{m-1}$$Another round of integration by parts gives$$\begin{align*}\int\limits_0^1\mathrm dt\, t^n(1-t)^m & =\frac {m(m-1)}{(n+1)(n+2)}\int\limits_0^1\mathrm dt\, t^{n+2}(1-t)^{n-2}\\ & =\frac {m(m-1)(m-2)}{(n+1)(n+2)(n+3)}\int\limits_0^1\mathrm dt\, t^{n+3}(1-t)^{m-3}\end{align*}$$It's pretty clear that a general pattern will emerge. Integrating by parts a total of $m$ parts with respect to $t$ eventually gives$$\begin{align*}\operatorname{B}(n+1,m+1) & =\frac {m(m-1)(m-2)\cdots3\cdot2\cdot1}{(n+1)(n+2)(n+3)\cdots(n+m)}\int\limits_0^1\mathrm dt\, t^{n+m}\\ & =\frac {\Gamma(m+1)}{(n+1)_{m+1}}\end{align*}$$where $(a)_k$ is the rising factorial. It can be shown using simple manipulations that$$(a)_k=\frac {\Gamma(a+k)}{\Gamma(a)}$$So therefore$$\operatorname{B}(n+1,m+1)=\frac {\Gamma(n+1)\Gamma(m+1)}{\Gamma(m+n+2)}$$Now it's just a matter of making the substitution $m\mapsto m-1$ and $n\mapsto n-1$ to get that$$\operatorname{B}(n,m)=\int\limits_0^1\mathrm dt, t^{n-1}(1-t)^{m-1}\color{blue}{=\frac {\Gamma(m)\Gamma(n)}{\Gamma(m+n)}}$$
