\[\begin{align*} E(X) &= \int_0^1 x \frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)} x^{\alpha-1}(1-x)^{\beta-1} dx \\ &= \frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)} \int_0^1 x^{\alpha}(1-x)^{\beta-1} dx \\ \text{And: } \int_0^1 x^{\alpha}(1-x)^{\beta-1}dx &= \frac{\Gamma(\alpha+1)\Gamma(\beta)}{\Gamma(\alpha +1, \beta)} \\ &= \frac{\alpha\Gamma(\alpha)\Gamma(\beta)}{(\alpha + \beta)\Gamma(\alpha, \beta)} \\ \therefore E(X) &= \frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)} \cdot \frac{\alpha\Gamma(\alpha)\Gamma(\beta)}{(\alpha + \beta)\Gamma(\alpha+ \beta)} \\ &= \frac{\alpha}{\alpha + \beta} \end{align*}\]
\[\begin{align*} E(X^2) &= \int_0^1 x^2 \frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)} x^{\alpha-1}(1-x)^{\beta-1} dx \\ &= \frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)} \int_0^1 x^{\alpha+1}(1-x)^{\beta-1} dx \\ \text{And: } \int_0^1 x^{\alpha+1}(1-x)^{\beta-1}dx &= \frac{\Gamma(\alpha+2)\Gamma(\beta)}{\Gamma(\alpha +2, \beta)} \\ &= \frac{\alpha(\alpha+1)\Gamma(\alpha)\Gamma(\beta)}{(\alpha+\beta+1) (\alpha+\beta)\Gamma(\alpha, \beta)} \\ \therefore E(X^2) &= \frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)} \cdot \frac{\alpha(\alpha+1)\Gamma(\alpha)\Gamma(\beta)}{(\alpha+\beta+1) (\alpha+\beta)\Gamma(\alpha + \beta)} \\ &= \frac{\alpha(\alpha+1)}{(\alpha+\beta+1) (\alpha+\beta)}\\ Var(X) &= \frac{\alpha(\alpha+1)}{(\alpha+\beta+1) (\alpha+\beta)} - \left(\frac{\alpha}{\alpha + \beta} \right)^2\\ &= \frac{(\alpha^2 + \alpha)(\alpha+\beta) - \alpha^2(\alpha+\beta+1)}{(\alpha+\beta)^2(\alpha+\beta+1)} \\ &= \frac{\alpha^3 + \alpha^2\beta +\alpha^2 +\alpha\beta - \alpha^3 -\alpha-\alpha^2\beta - \alpha^2}{(\alpha+\beta)^2(\alpha+\beta+1)} \\ &= \frac{\alpha\beta}{(\alpha+\beta)^2(\alpha+\beta+1)} \end{align*}\]
\[\begin{align*} M_X(t) &= \int_0^1 e^{tx} \frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)} x^{\alpha-1}(1-x)^{\beta-1} dx \\ &= \frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)} \int_0^1 \left(\sum_{k=0}^\infty \frac{(tx)^k}{k!} \right) x^{\alpha-1}(1-x)^{\beta-1} dx \\ &= \sum_{k=0}^\infty \frac{t^k}{k!} \cdot \frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)} \int_0^1 x^{\alpha+k-1}(1-x)^{\beta-1} dx \\ &= \sum_{k=0}^\infty \frac{t^k}{k!} \left(\frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)} \cdot \frac{\Gamma(\alpha + k)\Gamma(\beta)}{\Gamma(\alpha + \beta + k)}\right) \\ &= \sum_{k=0}^\infty \frac{t^k}{k!} \left(\frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)} \cdot \frac{\Gamma(\alpha + k)}{\Gamma(\alpha + \beta + k)}\right) \\ &= \frac{t^0}{0!} \left(\frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)} \cdot \frac{\Gamma(\alpha)}{\Gamma(\alpha + \beta)}\right) + \sum_{k=1}^\infty \frac{t^k}{k!} \left(\frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)} \cdot \frac{\Gamma(\alpha + k)}{\Gamma(\alpha + \beta + k)}\right) \\ &= 1 + \sum_{k=1}^\infty \left(\frac{\Gamma(\alpha+k)}{\Gamma(\alpha)} \cdot \frac{\Gamma(\alpha + \beta)}{\Gamma(\alpha + \beta + k)}\right) \frac{t^k}{k!} \\ &= 1 + \sum_{k=1}^\infty \left(\frac{\Gamma(\alpha)\prod_{r=0}^k(\alpha+r)}{\Gamma(\alpha)} \cdot \frac{\Gamma(\alpha + \beta)}{\Gamma(\alpha + \beta)\prod_{r=0}^k(\alpha+\beta+r)}\right) \frac{t^k}{k!} \\ &= 1 + \sum_{k=1}^\infty \left(\prod_{r=0}^{k-1} \frac{\alpha+r}{\alpha+\beta+r} \right) \frac{t^k}{k!} \end{align*}\]