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$\newcommand{\rset}{\mathbb R} \newcommand{\1}{\mathsf{1}} \newcommand{\rme}{\mathrm{e}} \newcommand{\rmd}{\mathrm{d}} \newcommand{\mcb}{\mathcal{B}} \newcommand{\PE}{\mathbb{E}} \newcommand{\eqsp}{\;} $

The Gamma distribution has the following density $$ x\mapsto \frac{b^a}{\Gamma(a)}\,x^{a-1}\,\rme^{- b x}\,\1_{\rset_+}(x)\;, \quad\text{with}\quad\Gamma(a)=\int_0^\infty x^{a-1}\,\rme^{-x}\,\rmd x \;. $$

It is easy to check that $\Gamma(k)=(k-1)!$ and, by integration by parts, $\Gamma(a+1)=a\Gamma(a)$. Now, for $\alpha,\beta>0$, the Beta distribution $\mcb(\alpha,\beta)$ has the density $$ x\mapsto\frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha) \Gamma(\beta)}\,x^{\alpha-1}\,(1-x)^{\beta-1}\1_{[0,1]}(x)\;. $$

We have to check that $B(\alpha,\beta)=\int_0^1 x^{\alpha-1}\,(1-x)^{\beta-1} \rmd x$ satisfies: $B(\alpha,\beta)=\frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)}$. Write $$ \Gamma(\alpha+\beta)B(\alpha,\beta)=\iint y^{\alpha+\beta-1}\rme^{-y} \1_{\rset^+}(y) x^{\alpha-1}(1-x)^{\beta-1} \1_{[0,1]}(x)\rmd x \rmd y $$ We first set $t=y; \ s=xy$ and then, $u=s; \ v=t-s$ and the formula is proved. If $X \sim \mcb(\alpha,\beta)$, then $$ \PE[X]=\frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)} \times \frac{ \Gamma(\alpha+1) \Gamma(\beta)}{\Gamma(\alpha+\beta+1)}=\frac{\alpha}{\alpha+\beta} $$ Similarly, $\PE[X^2]=\frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta) } \times \frac{\Gamma(\alpha+2) \Gamma(\beta)}{\Gamma(\alpha+\beta+2)}$ and we obtain $$ \mathrm{Var}[X]=\frac{\alpha \beta}{(\alpha+\beta)^2 (\alpha+\beta+1)}\eqsp. $$