Series and integrals for Apery's constant Two integrals for Apery's constant $\zeta(3)$ are
$$\zeta(3)=\frac{16}{3} \int_0^1 \frac{x\log^2\left(x\right)}{1+x^2}dx$$
and
$$\zeta(3)=\frac{32}{7} \int_0^1 \frac{x\log^2\left(x\right)}{1-x^4}dx$$
How can series expressions be obtained from them?
Related questions:
A series to prove $\frac{22}{7}-\pi>0$
Series for $\zeta(3)-\frac{6}{5}$
 A: From the geometric series we have $$\frac{16}{3}\int_{0}^{1}\frac{x\log^{2}\left(1/x\right)}{1+x^{2}}dx=\frac{16}{3}\sum_{k\geq0}\left(-1\right)^{k}\int_{0}^{1}x^{2k+1}\log^{2}\left(x\right)dx$$ where the exchange of the integral with the series is justified by the dominated convergenge theorem. Then integrating by parts $$\frac{16}{3}\int_{0}^{1}\frac{x\log^{2}\left(1/x\right)}{1+x^{2}}dx=\frac{4}{3}\sum_{k\geq1}\frac{\left(-1\right)^{k-1}}{k^{3}}=\frac{4}{3}\left(1-2^{-2}\right)\sum_{k\geq1}\frac{1}{k^{3}}=\color{red}{\zeta\left(3\right)}$$ where in the penultimate equality we used the relation $$\sum_{k\geq1}\frac{\left(-1\right)^{k-1}}{k^{s}}=\left(1-2^{1-s}\right)\sum_{k\geq1}\frac{1}{k^{s}},\,\mathrm{Re}\left(s\right)>1.$$ The other integral is similar.
A: Just to complete Marco's excellent answer,
$$ \int_{0}^{1}\frac{x\log^2 x}{1-x^4}\,dx = \sum_{n\geq 0} \int_{0}^{1}x^{4n+1}\log^2(x)\,dx=\left.\sum_{n\geq 0}\frac{d^2}{ds^2}\int_{0}^{1}x^{4n+1+s}\,dx\right|_{s=0}$$
equals
$$ \sum_{n\geq 0}\frac{2}{(4n+2)^3} = \frac{1}{4}\sum_{n\geq 0}\frac{1}{(2n+1)^3}=\frac{1}{4}\cdot\frac{7}{8}\zeta(3) = \frac{7\,\zeta(3)}{32}.$$
A: A variant of Marco's answer.
$\displaystyle J=\int_0^1 \frac{x\ln^2 x}{1+x^2}\,dx$
Perform the change of variable $y=x^2$,
$\begin{align} J=\frac{1}{8}\int_0^1  \frac{\ln^2 y}{1+y}\,dy\end{align}$
Let,
$\displaystyle K=\int_0^1  \frac{\ln^2 x}{1+x}\,dx$
$\displaystyle L=\int_0^1  \frac{\ln^2 x}{1-x}\,dx$
$\begin{align}
L-K=\int_0^1 \frac{2x\ln^2 x}{1-x^2}\,dx
\end{align}$
Perform the change of variable $y=x^2$,
$\begin{align}
L-K&=\frac{1}{4}\int_0^1 \frac{\ln^2 x}{1-x}\,dx\\
&=\frac{1}{4}L
\end{align}$
Therefore, 
$\displaystyle K=\frac{3}{4}L$
Therefore,
$\displaystyle J=\frac{3}{32}L$
Let,
$\begin{align}M=\int_0^1 \frac{x\ln^2 x}{1-x^4}\,dx\\
\end{align}$
Perform the change of variable $y=x^2$,
$\begin{align}M&=\frac{1}{8}\int_0^1 \frac{\ln^2 x}{1-x^2}\,dx\\
&=\frac{1}{16}\int_0^1 \frac{\ln^2 x}{1-x}\,dx+\frac{1}{16}\int_0^1 \frac{\ln^2 x}{1+x}\,dx\\
&=\frac{1}{16}L+\frac{1}{16}K\\
&=\frac{7}{64}L
\end{align}$
$\begin{align}L&=\int_0^1 \frac{\ln^2 x}{1-x}\,dx\\
&=\int_0^1\left(\sum_{n=0}^{\infty} x^n\ln^2 x\right)\,dx\\
&=\sum_{n=0}^{\infty} \left(\int_0^1 x^n\ln^2 x\,dx\right)\\
&=2\sum_{n=0}^{\infty} \frac{1}{(n+1)^3}\\
&=2\zeta(3)
\end{align}$
(the exchange of the integral and the series is justified by Fubini-Tonelli theorem)
Therefore,
$\displaystyle J=\frac{3}{16}\zeta(3)$
$\displaystyle M=\frac{7}{32}\zeta(3)$
