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.

To show uniform convergence of this series we can use the Weierstrass M-test. How big can $f_n(x)=(1-x^2)x^n$ be on $[0,1]?$ Set $f_n'(x) = 0.$ We get $x=\sqrt {n/(4+n)}.$ At this $x$ we have
$$ f_n(\sqrt {n/(4+n)}) = [1-n/(4+n)]^2\cdot[\sqrt {n/(4+n)}\,]^n.$$
The first factor on the right is $[4/(4+n)]^2.$ The second one is $\le1$ for all $n.$ Thus $0\le f_n(x) \le [4/(4+n)]^2, x\in [0,1].$ Since $\sum [4/(4+n)]^2<\infty,\sum f_n$ converges uniformly on $[0,1]$ by Weierstrass M.
