Given the p.d.f. of X, find the mean and variance of $X$. Suppose that $X$ has probability density function,
$$f(x)=\begin{cases}
\dfrac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)}x^{\alpha-1}(1-x)^{\beta-1} & \text{for }0\leq x\leq1 \\[8pt]
0 & \text{otherwise}.
\end{cases}
$$
where $\alpha,\beta>0$ and $\Gamma$ is the gamma function which is defined by
$$\Gamma(r)=\int^\infty_0 u^{r-1}e^{-u} \, du$$
This function has the property that $\Gamma(r+1)=r\Gamma(r)$ for $r>0$. Using this fact, find the mean and variance of $X$.
Hint: remember that p.d.f.'s integrate to 1; there's no need to actually do any integration in this question

Not sure. My attempt seems completely botched.
Consider
$$
E[X]=\int^\infty_{-\infty} xf(x) dx=\int^{1}_{0}x\frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)}x^{\alpha-1}(1-x)^{\beta-1} \, dx
$$
Note:
$\dfrac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)}=\dfrac{\alpha^{\beta}\Gamma(\alpha)}{\Gamma(\alpha)\Gamma(\beta)}=\dfrac{\alpha^{\beta}}{\Gamma(\beta)}$
Since $\frac{\alpha^{\beta}}{\Gamma(\beta)}$ is a constant,
$$\int^1_0 x\frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)} x^{\alpha-1}(1-x)^{\beta-1} \, dx = \frac{\alpha^\beta}{\Gamma(\beta)} \int^1_0 x x^{\alpha-1} (1-x)^{\beta-1} \, dx=\frac{\alpha^\beta}{\Gamma(\beta)} \int^1_0 x^{\alpha}(1-x)^{\beta-1} \, dx$$
How should I proceed or am approaching this completely incorrectly?
 A: Since the pdf is normalized 
$$\int^1_0x^{\alpha-1}(1-x)^{\beta-1} \, dx =  \frac{\Gamma(\alpha)\Gamma(\beta)} {\Gamma(\alpha+\beta)}
$$
So that 
$$\int^1_0x^{\alpha}(1-x)^{\beta-1} \, dx =  \frac{\Gamma(\alpha +1 )\Gamma(\beta)} {\Gamma(\alpha+\beta+1)}$$
$$E(x) = \frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)} \int^1_0x^{\alpha}(1-x)^{\beta-1} \, dx = \frac{\Gamma(\alpha+1)}{\Gamma(\alpha)} \times \frac{\Gamma(\beta)}{\Gamma(\beta)} \times \frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha + \beta+1)} = \frac{\alpha}{\alpha + \beta}
$$
A: You have
$$
\int_0^1 x^{\alpha-1} (1-x)^{\beta-1}\,dx = \frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)}. \tag 1
$$
You want
\begin{align}
\operatorname{E}(X) & = \int_0^1 x f(x)\,dx = \int_0^1 x \frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)} x^{\alpha-1} (1-x)^{\beta-1}\,dx \\[10pt]
& = \frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)} \int_0^1 x^\alpha (1-x)^{\beta-1}\,dx = \frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)} \overbrace{\int_0^1 x^{(\alpha+1)-1} (1-x)^{\beta-1}\,dx}
\end{align}
The expression under the $\overbrace{\text{overbrace}}$ is the same thing you see in $(1)$ except that it has $\alpha+1$ instead of $\alpha$.  The equality in $(1)$ holds if $\alpha$ is any positive number at all.  And $\alpha+1$ is a positive number.  Therefore
$$
\int_0^1 x^{(\alpha+1)-1} (1-x)^{\beta-1}\,dx = \frac{\Gamma(\alpha+1)\Gamma(\beta)}{\Gamma((\alpha+1)+\beta)}.
$$
So you get a product:
$$
\frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)}\cdot\frac{\Gamma(\alpha+1)\Gamma(\beta)}{\Gamma((\alpha+1)+\beta)}.
$$
Then $\Gamma(\beta)$ cancels, $\Gamma(\alpha+1)$ is replaced by $\alpha\Gamma(\alpha)$, and $\Gamma(\alpha+\beta+1)$ is replaced by $(\alpha+\beta)\Gamma(\alpha+\beta)$, and we do further cancellations, getting
$$
\frac \alpha {\alpha+\beta}.
$$
That is the expected value.
You can find $\operatorname{E}(X^2)$ similarly and then use the fact that $\operatorname{var}(X) = \operatorname{E}(X^2) - (\operatorname{E}(X))^2$.
