One can certainly find a beta distribution with the same mean and variance as $X$ but whether that is a good enough approximation depends on what you need.
If you only want the probability density function of $X$, then that is
$$\sum _{i=1}^n \left(\frac{x^{a_i-1} (1-x)^{b_i-1} \prod _{j \neq i} \frac{B_x(a_j,b_j)}{B(a_j,b_j)}}{B(a_i,b_i)}\right)$$
where $B(a_i,b_i)$ is the beta function and $B_x (a_i,b_i)$ is the incomplete beta function.
I'm not sure there's a nice compact form for the mean and variance but for specific parameters one can calculate the mean and variance which can be matched to a beta distribution. Here's some Mathematica code to do so:
n = 3;
parms = {a[1] -> 1, b[1] -> 6, a[2] -> 4, b[2] -> 7, a[3] -> 4, b[3] -> 5};
pdf[x_] :=
Sum[(x^(a[i] - 1) (1 - x)^(b[i] - 1)/Beta[a[i], b[i]]) Product[Beta[x, a[j], b[j]]/Beta[a[j], b[j]],
{j, Delete[Range[n], i]}] /. parms, {i, n}];
mean = Integrate[x pdf[x], {x, 0, 1}];
variance = Integrate[x^2 pdf[x], {x, 0, 1}] - mean^2;
sol = N[Solve[{mean == a/(a + b), variance == a b/((a + b)^2 (a + b + 1))}, {a, b}][[1]]]
(* {a -> 6.80319, b -> 6.85957} *)
Plot[{pdf[x], PDF[BetaDistribution[a, b] /. sol, x]}, {x, 0, 1},
PlotLegends -> {"Actual", "Beta approximation"}]
For the cdf of the sample maximum we have
$$\begin{align*}
F_{X_{(n)}}(x)
&=\mathsf P(\text{max}{\{X_1,...,X_n}\}\leq x)\\\\
&=\mathsf P(X_1\leq x,...,X_n\leq x)\\\\
&=\mathsf P(X\leq x)^n\\\\
&=F_X(x)^n
\end{align*}$$
where
$$ F_{X}(x)=
\begin{cases}
0 & x \lt -1 \\
\frac{x+1}{2} & -1\leq x\leq1 \\
1 & x \gt 1
\end{cases} $$
Hence taking the derivative we get
$$f_{X_{(n)}} = \frac{n(x+1)^{n-1}}{2^n} I_{[-1,1]}(x)$$
Similarly for the sample minimum we have
$$\begin{align*}
F_{X_{(1)}}(x)
&=\mathsf P(\text{min}{\{X_1,...,X_n}\}\leq x)\\\\
&=1-\mathsf P(\text{min}{\{X_1,...,X_n}\}\gt x)\\\\
&=1-\left(1-F_X(x)\right)^n\\\\
&=1-\left(1-\frac{x+1}{2}\right)^n
\end{align*}$$
Hence taking the derivative we get
$$f_{X_{(1)}} = \frac{n\left(-x+1\right)^{n-1}}{2^n} I_{[-1,1]}(x)$$
Best Answer
If you have to solve an integral like this $\int y(y-x)^{n}dy$ the more easy way is an integration by parts
$\int y(y-x)^{n}dy=\int y d\frac{(y-x)^{n+1}}{n+1}=y\frac{(y-x)^{n+1}}{n+1}-\int \frac{(y-x)^{n+1}}{n+1}dy=$
$=y\frac{(y-x)^{n+1}}{n+1}-\frac{(y-x)^{n+2}}{n+2}$