Need some hint or technique for this integral What technique we should use for this integral:
$$\int_0^1\frac{\ln x\ln (1-x^2)}{1-x^2}\text{d}x$$
Can anyone give a brief way to evaluate this?
 A: One possible, although not very short, way is to reduce the integral to derivatives of the beta integral:
$$\begin{eqnarray}
   \int_0^1 \frac{\log(x) \log(1-x^2)}{1-x^2} \mathrm{d}x &\stackrel{u=x^2}{=}& \frac{1}{4} \int_0^1 \frac{\log(u)}{\sqrt{u}} \frac{\log(1-u)}{1-u} \mathrm{d}u \\
    &=& \frac{1}{4} \lim_{\alpha\to \frac{1}{2}} \lim_{\beta \downarrow 0^+} \frac{\mathrm{d}}{\mathrm{d} \alpha} \frac{\mathrm{d}}{\mathrm{d} \beta} \int_0^1 u^{\alpha-1} (1-u)^{\beta-1}\mathrm{d}u
\\ &=& \frac{1}{4} \lim_{\alpha\to \frac{1}{2}} \lim_{\beta \downarrow 0^+} \frac{\mathrm{d}}{\mathrm{d} \alpha} \frac{\mathrm{d}}{\mathrm{d} \beta} \frac{\Gamma(\alpha) \Gamma(\beta)}{\Gamma(\alpha+\beta)}
\\
\\ &=& \frac{1}{4} \lim_{\alpha\to \frac{1}{2}} \lim_{\beta \downarrow 0^+} \frac{\mathrm{d}}{\mathrm{d} \alpha} \frac{\mathrm{d}}{\mathrm{d} \beta} \frac{\Gamma(\alpha) \Gamma(\beta+1)}{\beta \, \Gamma(\alpha+\beta)}
\end{eqnarray}
$$
Now we would do a series expansion around $\beta=0$:
$$
  \frac{\Gamma(\alpha) \Gamma(\beta+1)}{\beta \, \Gamma(\alpha+\beta)} = \frac{1}{\beta} + \left(-\gamma - \psi(a)\right) + \frac{\beta}{2} \left( \frac{\pi^2}{6} + \gamma^2+2 \gamma \psi(a) + \psi(a)^2 -\psi^{(1)}(a) \right) + \mathcal{o}(\beta)
$$
Differentation with respect to $\alpha$ kills the singular term, one we find:
$$
 \int_0^1 \frac{\log(x) \log(1-x^2)}{1-x^2} \mathrm{d}x = \frac{1}{4} \lim_{\alpha \to \frac{1}{2}} \left(  \gamma \psi^{(1)}(a) + \psi(a) \psi^{(1)}(a) - \frac{1}{2} \psi^{(2)}(a)  \right)
$$
Now, using $\psi(1/2) = -\gamma - 2 \log(2)$, $\psi^{(1)}(1/2) = \frac{\pi^2}{2}$ and $\psi^{(2)}(1/2) = -14 \zeta(3)$ we finally arrive at the answer:
$$
   \int_0^1 \frac{\log(x) \log(1-x^2)}{1-x^2} \mathrm{d}x = \frac{1}{4}  \left( 7 \zeta(3) - \pi^2 \log(2) \right)
$$
Notice also that 
$$
    \frac{\log(1-x^2)}{1-x^2} = -\sum_{n=1}^\infty H_n x^{2n}
$$
This leads, together with $\int_0^1 x^{2n} \log(x) \mathrm{d}x = -\frac{1}{(2n+1)^2}$:
$$
    \int_0^1 \frac{\log(x) \log(1-x^2)}{1-x^2} \mathrm{d}x = \sum_{n=1}^\infty \frac{H_n}{(2n+1)^2}
$$
Sum of these types have been discussed before.
Edit: In fact exactly this sum was asked about by the OP, as noted by @MhenniBenghorbal in the comments, and @joriki provided an excellent answer.
A: Not sure about an integration method by hand, but here's a Monte Carlo calculation of your integral in Matlab:
m=1000;
n=1000;
a=0;
b=1;
z=zeros(1,m);
for j=1:m
    x=a+(b-a)*rand(n,1);
    q=1/(b-a);
    f=(log(x).*log(1-x.^2))./(1-x.^2);
    i=mean(f./q);
    z(j) = i;
end
integral = mean(z)
variance = var(z)

integral = 0.3932
variance = 1.5804e-04
