How do you find the second moment of the beta distribution? I'm required to show $ E(Y^2) = \dfrac{\alpha(\alpha + 1)}{(\alpha + \beta + 1)(\alpha + \beta)} $ for the beta distribution using the definition of expectation.

Now so far I have $ \int\limits_0^1 {y^2  \dfrac{\Gamma\left( \alpha + \beta \right)}{\Gamma \left( \alpha \right)\Gamma \left( \beta \right)} y^{\alpha-1}(1-y)^{\beta-1} dy} $ and I simplified it so that I pulled the gamma constants out front of the integral and combined $ y^2y^{\alpha-1} $ to be $ y^{\alpha+1} $. I'm not too sure where to continue from here... can anyone help me out?
Try looking at the kernel of the integral - in other words, ignore all the constant factors, focus on the bits involving the variable you are integrating with respect to.

Do you recognise it?

It's a good idea whenever you see the PDF of a distribution to pay attention to what its kernel is too. So forget about its normalization factor. Now, if a PDF can be written as the product of a normalizing factor $N$ and a kernel $k(x)$, then because I know:

$$\int_{-\infty}^{\infty}f_X(x)dx=N\int_{-\infty}^{\infty}k(x)dx=1$$

I also know that:

$$\int_{-\infty}^{\infty}k(x)dx=\frac{1}{N}$$

So, learn to recognize your PDF kernels! If you see an integral with the same kernel, but a different constant factor, then you can easily evaluate it:

$$\int_{-\infty}^{\infty}A\cdot k(x)dx=\frac{A}{N}$$

Extra hint: the beta distribution has support [0, 1] so we only need our integrals to have those limits. If $X \sim Beta (\alpha,\, \beta)$ then $f_X(x)=\frac{\Gamma(\alpha + \beta)}{\Gamma(\alpha)\Gamma(\beta)}x^{\alpha -1}(1-x)^{\beta -1}$ so the kernel is just $x^{\alpha -1}(1-x)^{\beta -1}$.

You have $ \int\limits_0^1 {y^2  \dfrac{\Gamma\left( \alpha + \beta \right)}{\Gamma \left( \alpha \right)\Gamma \left( \beta \right)} y^{\alpha-1}(1-y)^{\beta-1} dy} $ which has kernel $y^{\alpha+1}(1-y)^{\beta-1} $. Which PDF is this the kernel of, and with what parameters?
you have to use the recurrent formula
First gross moment
µ= B(a+1,b)/B(a,b)= a/(a+b)

2nd gross moment
M20= B(a+2,b)/B(a,b) = B(a+2,b)/(B(a+1,b){ B(a+1,b)/B(a,b)= {(a+1)/(a+1+b)}{a)/(a+b)
If you need the variance

V= M20 -µ^2

This is a presnt of Agustín F. Correa from Argentina