Proving that $\int_0^1 \frac{\log^2(x)\tanh^{-1}(x)}{1+x^2}dx=\beta(4)-\frac{\pi^2}{12}G$ I am trying to prove that
$$I=\int_0^1 \frac{\log^2(x)\tanh^{-1}(x)}{1+x^2}dx=\beta(4)-\frac{\pi^2}{12}G$$
where $\beta(s)$ is the Dirichlet Beta function and $G$ is the Catalan's constant. I managed to derive the following series involving polygamma functions but it doesn't seem to be of much help. 
$$
\begin{align*}
I &=\frac{1}{64}\sum_{n=0}^\infty \frac{\psi_2 \left(\frac{n}{2}+1 \right) -\psi_2\left(\frac{n+1}{2} \right)}{2n+1}  \\
&= \frac{1}{8}\sum_{n=1}^\infty \frac{\psi_2(n)}{2n-1}-\frac{1}{32}\sum_{n=1}^\infty\frac{\psi_2\left(\frac{n}{2}\right)}{2n-1}
\end{align*}
$$
Numerical calculations show that $I \approx 0.235593$. 
 A: The generalization of the main integral follows easily by employing the same ideas used in this post and the previous one. 

Let $n$ be a natural number. Then, we have 
  $$\int_0^1 \frac{\log^{2n}(x)\operatorname{arctanh}(x)}{1+x^2}\textrm{d}x$$
$$=\lim_{s\to0}\frac{d^{2n}}{ds^{2n}}\left(\frac{\pi}{16}\cot \left(\frac{\pi  s}{2}\right) \left(\psi \left(\frac{3}{4}-\frac{s}{4}\right)-\psi\left(\frac{1}{4}-\frac{s}{4}\right)\right)-\frac{\pi ^2 }{16} \csc \left(\frac{\pi  s}{2}\right)\right),$$
  where $\psi$ represents the Digamma function.

Another similar generalization

Let $n$ be a natural number. Then, we get 
  $$\int_0^1 \frac{\log^{2n}(x)\arctan(x)}{1-x^2}\textrm{d}x$$
$$=\frac{\pi}{4} \left(1-2^{-2 n-1}\right) \zeta (2 n+1)(2 n)!$$
$$-\lim_{s\to0}\frac{d^{2n}}{ds^{2n}}\left(\frac{\pi}{16} \csc \left(\frac{\pi  s}{2}\right) \left(\pi  \cos \left(\frac{\pi  s}{2}\right)+\psi\left(\frac{s+1}{4}\right)-\psi\left(\frac{s+3}{4}\right)\right)\right),$$
  where $\zeta$ represents the Riemann zeta function and $\psi$ denotes the Digamma function.


A solution in large steps by Cornel I. Valean  to the main integral
$$\int_0^1 \frac{\log^2(x)\operatorname{arctanh}(x)}{1+x^2}dx$$
We follow the strategy used to the auxiliary result from the previous post, and then we immediately arrive at

$$\int_0^1 \frac{\log^2(x)\operatorname{arctanh}(x)}{1+x^2}dx=\frac{1}{2}\Re\biggr\{ \int_0^{\infty } \frac{\log ^2(x) \operatorname{arctanh}(x)}{1+x^2} \textrm{d}x\biggr \}$$
$$=\frac{1}{2} \int_0^{\infty }\left(PV\int_0^1 \frac{x \log ^2(x)}{(1-y^2 x^2)(1+x^2)} \textrm{d}y\right)\textrm{d}x$$
$$=\frac{1}{2}\int_0^1\left(PV\int_0^{\infty} \frac{x \log ^2(x)}{(1-y^2 x^2)(1+x^2)} \textrm{d}x\right)\textrm{d}y$$
$$=\frac{\pi^2}{12}\int_0^1 \frac{\log(y)}{1+y^2}\textrm{d}y-\frac{1}{6}\int_0^1 \frac{\log^3(y)}{1+y^2}\textrm{d}y=\beta(4)-\frac{\pi^2}{12}G,$$
  as desired.

End of story.
A note: Using the Cauchy product $\displaystyle \frac{\operatorname{arctanh}(x)}{1+x^2}=\sum _{n=1}^{\infty } \sum _{k=1}^n \frac{(-1)^{n+k} x^{2 n-1}}{2 k-1}$, an the value of the main integral, we immediately obtain the beautiful series 

$$\sum _{n=1}^{\infty }\frac{(-1)^{n-1}}{n^3} \sum _{k=1}^n \frac{(-1)^{k-1}}{2 k-1}=4\beta(4)-\frac{\pi^2}{3}G.$$

Some kind of bonus: Using the integrals relation obained with integration by parts as shown in Shobhit Bhatnagar's post and combining it with the results obtained in this post and the previous one, we obtain the value of the other integral,

$$\int_0^1\frac{\log^2(x)\arctan(x)}{1-x^2}\textrm{d}x= -\beta(4)-\frac{\pi^2}{24}G+\frac{7\pi}{16}\zeta(3).$$

A note: It's clear the generalization $\displaystyle \int_0^1 \frac{\log^{2n}(x)\arctan(x)}{1-x^2}\textrm{d}x$ may be approached in the same way as $\displaystyle \int_0^1 \frac{\log^2(x)\operatorname{arctanh}(x)}{1+x^2}dx$.
A: I was able to solve this problem on my own.
Using integration by parts,
$$\begin{align*}
&\; \int_0^1 \frac{\log^2(x)\tanh^{-1}(x)}{1+x^2}dx \\
&= -2\int_0^1 \frac{\log(x)\tan^{-1}(x)\tanh^{-1}(x)}{x}dx-\int_0^1 \frac{\log^2(x)\tan^{-1}(x)}{1-x^2}dx \tag{1}
\end{align*}$$
I posted the solution to both these integrals on another forum. Here are the links:


*

*http://integralsandseries.prophpbb.com/topic711.html#p3975

*http://integralsandseries.prophpbb.com/topic245.html#p1680
$$\begin{align*}\int_0^1\frac{\log(x)\tan^{-1}(x)\tanh^{-1}(x)}{x}dx &= \frac{\pi^2}{16}G-\frac{7\pi\zeta(3)}{32} \tag{2}\\
\int_0^1\frac{\log^2(x)\tan^{-1}(x)}{1-x^2}dx &= -\beta(4)-\frac{\pi^2}{24}G+\frac{7\pi}{16}\zeta(3)\tag{3}
\end{align*}$$
$G$ denotes the Catalan's constant and $\beta(4)=\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^4}$. Substituting these two results in equation (1) gives:
$$\int_0^1 \frac{\log^2(x)\tanh^{-1}(x)}{1+x^2}dx=\beta(4)-\frac{\pi^2}{12}G \tag{4}$$
Proof sketch of integrals (2) and (3) : (Please see the above links for  a more detailed answer)
The idea behind evaluating (2) and (3) is breaking them down into Euler Sums. Using the taylor series expansion $\tan^{-1}(x)=\sum_{n=0}^\infty\frac{(-1)^n x^{2n+1}}{2n+1}$ and integrating term-wise, we obtain the following relations:
\begin{align*}
\int_0^1\frac{\log(x)\tan^{-1}(x)\tanh^{-1}(x)}{x}dx &= -\log(2)\frac{\pi^3}{32}-\frac{1}{2}\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^3}\left( \gamma+\psi_0(n+1)\right) \\ &\;+\frac{1}{4}\sum_{n=0}^\infty \frac{(-1)^n \psi_1(n+1)}{(2n+1)^2}  \tag{5}\\
\int_0^1 \frac{\log^2(x)\tan^{-1}(x)}{1-x^2}dx &=-\frac{1}{8}\sum_{n=0}^\infty\frac{(-1)^n\psi_2(n+1)}{2n+1}\tag{6}
\end{align*}
These Euler Sums can be evaluated using the techniques shown in the paper "Euler Sums and Contour Integral Representations" by Philippe Flajolet and Bruno Salvy. Here is it's link. 
\begin{align*}
\sum_{n=0}^\infty\frac{(-1)^n\psi_2(n+1)}{2n+1} &= 8\beta(4)+\frac{\pi^2}{3}G-\frac{7\pi}{2}\zeta(3) \\
\sum_{n=0}^\infty\frac{(-1)^n\psi_1(n+1)}{(2n+1)^2} &= 6\beta(4)+\frac{\pi^2}{4}G-\frac{7\pi}{4}\zeta(3) \\
\sum_{n=0}^\infty \frac{(-1)^n\left( \gamma+\psi_0(n+1)\right)}{(2n+1)^3} &= 3\beta(4)-\frac{7\pi}{16}\zeta(3)-\frac{\pi^3}{16}\log(2)
\end{align*}
Substituting these into equations (5) and (6) gives us the integrals (2) and (3). 
A related integral
Using similar techniques, we can show that
$$\displaystyle \int_0^1 \frac{\log^2(x)\tan^{-1}(x)}{x\left(1-x^2 \right)}dx=\beta(4)+\frac{7\pi \zeta(3)}{64}+\frac{\pi^3 \log(2)}{32}$$
