Integrals of the form $\int^1_{0} x^\alpha(1-x)^\beta dx$ where $\alpha,\beta \in \mathbb{R}$. Let $\alpha,\beta \in \mathbb{R}$. I was wondering if there is a systematic way to solve integrals of the following form:
\begin{equation}
\int^1_{0} x^\alpha(1-x)^\beta dx
\end{equation}
I have seen similar kind of integrals many times. Any help/hints would be really appreciated. Thanks!
 A: Begin with the definition of the Gamma function:
$$\Gamma(\alpha+1) = \int_0^{\infty} du\; u^{\alpha} e^{-u} $$
Similarly,
$$\Gamma(\beta+1) = \int_0^{\infty} dv\; v^{\beta} e^{-v} $$
Multiply these together...
$$\begin{align}\Gamma(\alpha+1) \Gamma(\beta+1) &= \int_0^{\infty} du\; \int_0^{\infty} dv\; u^{\alpha} v^{\beta} e^{-u-v}\\ &= \frac1{2^{\alpha+\beta+1}} \int_0^{\infty} dp \; \int_{-p}^p dq \, (p+q)^{\alpha} (p-q)^{\beta} e^{-p} \\ &= \frac1{2^{\alpha+\beta+1}} \int_0^{\infty} dp \; p^{\alpha+\beta+1} e^{-p} \, \int_{-1}^1 dr \, (1+r)^{\alpha} (1-r)^{\beta}\\ &= \Gamma(\alpha+\beta+2)  \frac1{2^{\alpha+\beta+1}} \int_0^2 dr \, r^{\alpha} (2-r)^{\beta} \\ &= \Gamma(\alpha+\beta+2) \int_0^1 dx \, x^{\alpha} (1-x)^{\beta}\end{align}$$
In the second line, I used the transformation $p=u+v$, $q=u-v$, which has Jacobian $1/2$.  Thus,

$$\int_0^1 dx \, x^{\alpha} (1-x)^{\beta} = \frac{\Gamma(\alpha+1) \Gamma(\beta+1)}{\Gamma(\alpha+\beta+2)} $$

A: Using the definition of the Beta Function:
$$B(x,y)=\int_0^1t^{x-1}(1-t)^{1-y}dt$$
By matching coefficients,we would conclude that:
$$\int_0^1x^\alpha(1-x)^\beta dx=B(\alpha+1,1-\beta)$$
