Uniform Convergence of $\sum_{n=0}^\infty (1-x^2)^2x^n$ on $[0,1]$; subsequent integral Let $a_n = (1-x^2)^2 x^n$. Show that $\sum_{n=0}^\infty a_n$ converges uniformly on $[0,1]$ and deduce that $\int_0^1 \frac{(1-x^2)^2}{1-x} dx = \sum_{n=1}^\infty \frac{8}{n(n+2)(n+4)}$.
Attempt: Denoting partial sums by $S_n$, we have 
\begin{eqnarray*}
S_n &=& \sum_{k=0}^n (1-x^2)^2 x^k \\
&=&(1-x^2)^2 \frac{1-x^{n+1}}{1-x} \\
&=& (1+x)(1-x^2)(1-x^{n+1}).
\end{eqnarray*}
I tried to show uniform convergence by showing convergence of $\sup_{x \in [0,1]} S_n(x)$, but I seem unable to solve the resulting first order condition to find the maximizing x. Is there a simpler way to do this?
If I could show uniform convergence, I could integrate the infinite series term by term yielding
\begin{eqnarray*}
\int \frac{(1-x^2)^2}{1-x} dx &=& \sum_{n=0}^\infty \int_0^1 (1-x^2)^2 x^n dx \\ &=&\sum_{n=0}^\infty \int_0^1 x^n - 2x^{2+n} + x^{4+n} dx
\\ &=& \dots
\\ &=&\sum_{n=0}^\infty \frac{8}{(n+1)(n+3)(n+5)} 
\\ &=&\sum_{n=1}^\infty \frac{8}{n(n+2)(n+4)} 
\end{eqnarray*}
and I would be done.

\begin{eqnarray*}
S_n &=& \sum_{k=0}^n (1-x^2)^2 x^k \\
&=&(1-x^2)^2 \frac{1-x^{n+1}}{1-x} \\
&=& (1+x)(1-x^2)(1).
\end{eqnarray*}
the uniform convergence follows from
\begin{eqnarray*}
S_m -S_n&=& \sum_{k=n+1}^m (1-x^2)^2 x^k \\
&=&(1-x^2)^2 \frac{x^{n+1}-x^{m+1}}{1-x}\xrightarrow[n \to \infty]{} 0 \\
\end{eqnarray*}
 Since $x^{m+1} \leq x^{n+1} \to 0$
Edit: To get uniform convergence one needs further to note that $x^{m+1},x^{n+1} \leq 1$ and that $1-x^2  = 1+x (1-x) $ so $$(1-x^2)^2 \frac{x^{n+1}-x^{m+1}}{1-x} = (1+x)^2(1-x)x^{n+1}-x^{m+1}.$$
So for $x>1-\delta$
$$(1+x)^2(1-x)x^{n+1}-x^{m+1}\leq 2\delta $$
and for $x<1-\delta$ $x^n \leq (1-\delta)^n \to 0$
$$(1+x)^2(1-x)x^{n+1}-x^{m+1} \leq 2 (1-\delta)^n$$
