If $X$ ~ $\text{Beta}(\alpha, \beta)$, find $\Bbb{E}\left[ X(1-X)\right]$ I am currently reading through John I.Marden's Mathematical Statistics: Old School on my own and I am trying to solve the following problem about expectation.

If $X$ ~ $\text{Beta}(\alpha, \beta)$, find $\Bbb{E}\left[ X(1-X)\right]$ in terms of a rational polynomial in $\alpha ,\beta$


My Attempt
I started with defining $\displaystyle \Bbb{E}\left[ X(1-X)\right] = \int_0^1 x(1-x) \frac{\Gamma(\alpha + \beta)}{\Gamma(\alpha)\Gamma(\beta)} \,x^{\alpha - 1}\, (1-x)^{\beta - 1} \, dx$
From here I moved out the constant terms and bring together common factors:
$$\frac{\Gamma(\alpha + \beta)}{\Gamma(\alpha)\Gamma(\beta)}\int_0^1  x^{\alpha}\, (1-x)^{\beta} \, dx$$
Now using the binomial theorem, we have $$\frac{\Gamma(\alpha + \beta)}{\Gamma(\alpha)\Gamma(\beta)}\int_0^1  x^{\alpha}\, \sum_{i=0}^\beta \left( \binom{\beta}{i}(-x)^i \right) \, dx$$
$$ = \frac{\Gamma(\alpha + \beta)}{\Gamma(\alpha)\Gamma(\beta)}\int_0^1  x^\alpha - \beta x^{\alpha + 1} + \binom{\beta}{2}x^{\alpha + 2} - \cdots + (-x)^{\alpha + \beta} \, dx $$
$$ = \frac{\Gamma(\alpha + \beta)}{\Gamma(\alpha)\Gamma(\beta)} \left( \frac{1}{\alpha+1} - \frac{\beta}{\alpha+2}+\frac{\binom{\beta}{2}}{\alpha+3}- \cdots +\frac{(-1)^{\alpha+\beta+1}}{\alpha + \beta + 1} \right) $$
I am not sure if this is the answer that is wanted or not.
Any help would be greatly appreciated!
 A: $\require{cancel}$
For anyone interested in following along the comments
We have $\displaystyle \Bbb{E}\left[ X(1-X)\right] = \int_0^1 x(1-x) \frac{\Gamma(\alpha + \beta)}{\Gamma(\alpha)\Gamma(\beta)} \,x^{\alpha - 1}\, (1-x)^{\beta - 1} \, dx = \frac{\Gamma(\alpha + \beta)}{\Gamma(\alpha)\Gamma(\beta)}\int_0^1  x^{\alpha}\, (1-x)^{\beta} \, dx$
If we look at a Beta Distribution with parameters $\alpha+1$ and $\beta + 1$, we know that because probability density functions must have an area of $1$ under the curve, which leads to: $$1 = \frac{\Gamma(\alpha + \beta +2)}{\Gamma(\alpha+1)\Gamma(\beta+1)}\int_0^1  x^{\alpha}\, (1-x)^{\beta} \, dx$$
$$\implies  \int_0^1  x^{\alpha}\, (1-x)^{\beta} \, dx = \frac{\Gamma(\alpha+1)\Gamma(\beta+1)}{\Gamma(\alpha + \beta +2)}$$
Substituting in, we have that $$\Bbb{E}\left[ X(1-X)\right] = \frac{\Gamma(\alpha + \beta)}{\Gamma(\alpha)\Gamma(\beta)}\frac{\Gamma(\alpha+1)\Gamma(\beta+1)}{\Gamma(\alpha + \beta +2)} $$
Now we just need to use Gamma function properties to reduce:
$$\frac{\Gamma(\alpha + \beta)}{\Gamma(\alpha)\Gamma(\beta)}\frac{\Gamma(\alpha+1)\Gamma(\beta+1)}{\Gamma(\alpha + \beta +2)} =\frac{\Gamma(\alpha + \beta)}{\cancel{\Gamma(\alpha)\Gamma(\beta)}}\frac{\alpha \, \cancel{\Gamma(\alpha)} \, \beta \,\cancel{\Gamma(\beta)}}{\Gamma(\alpha + \beta +2)} =  \frac{\Gamma(\alpha + \beta)}{1}\frac{\alpha \,  \, \beta \,}{\Gamma(\alpha + \beta +2)} $$
$$=\frac{\cancel{\Gamma(\alpha + \beta)}}{1}\frac{\alpha \,  \, \beta \,}{(\alpha + \beta + 1)(\alpha + \beta) \cancel{\Gamma(\alpha+\beta)}} $$
$$\bbox[yellow]{= \frac{\alpha \beta}{(\alpha + \beta)^2+ \alpha + \beta} = \Bbb{E}\left[ X(1-X)\right]}$$
