Interesting integral some how related to $\int_{0}^{\frac{\pi}{2}}x\ln(\sin(x))dx$ I hope I used the correct tag (definite integral). 
I ran across an integral that is rather tough and I am wondering if anyone could give me a shove in the right direction.
$\displaystyle\int_{0}^{1}\frac{\tanh^{-1}(x)\ln(x)}{x(1-x^{2})}\text{ d}x=\frac{-7}{16}\zeta(3)-\frac{{\pi}^{2}}{8}\ln(2)$
This solution is almost exactly like the solution to $\displaystyle \int_{0}^{\frac{\pi}{2}}x\ln(\sin(x))\text{ d}x$, which is $\displaystyle \frac{7}{16}\zeta(3)-\frac{{\pi}^{2}}{8}\ln(2)$
I solved the latter integral by using the identity $\displaystyle -\ln(\sin(x))-\ln(2)=\sum_{k=1}^{\infty}\frac{\cos(2kx)}{k}$, then integrating:
$\displaystyle -\int_{0}^{\frac{\pi}{2}}x\ln(\sin(x))\text{ d}x-\frac{{\pi}^{2}}{8}\ln(2)=\int_{0}^{\frac{\pi}{2}}x\cos(2x)\text{ d}x+\frac{\int_{0}^{\frac{\pi}{2}}\cos(4x)\text{ d}x}{2}+\frac{\int_{0}^{\frac{\pi}{2}}\cos(6x)}{3}\cdot\cdot\cdot\cdot$
But, $\displaystyle \int_{0}^{\frac{\pi}{2}}x\cdot \cos(2kx)\text{ d}x=-\left(\frac{1+(-1)^{k+1}}{(2k)^{2}}\right)$
Thus: $\displaystyle \int_{0}^{\frac{\pi}{2}}x\ln(\sin(x))\text{ d}x=\frac{1}{4}\sum_{k=1}^{\infty}\frac{1+(-1)^{k+1}}{k^{3}}-\frac{{\pi}^{2}}{8}\ln(2)$
$\displaystyle =\frac{1}{2}\displaystyle\sum_{k=0}^{\infty}\frac{1}{(2k+1)^{3}}-\frac{{\pi}^{2}}{8}\ln(2)$
and so on.  Which results in the solution I mentioned in the beginning. 
Sorry for all that, but I wanted to show you what I was using in order to some how relate it to the integral I am wanting to solve. 
I have been trying and trying to relate the aforementioned $\displaystyle \tanh$ integral with this one.  The solutions are so nearly the same, I figured there has to be a way to relate them and solve the integral.  Does anyone have some ideas?. I have tried the identity $\displaystyle \tanh^{-1}(x)=\frac{1}{2}\left[\ln(1+x)-\ln(1-x)\right]$, then breaking it up:
Resulting in $\displaystyle \frac{1}{2}\int_{0}^{1}\frac{\ln(x)\ln(1+x)}{x(x^{2}-1)}\text{ d}x-\frac{1}{2}\int_{0}^{1}\frac{\ln(x)\ln(1-x)}{x(x^{2}-1)} \text{ d}x$,  then I used the various series representations for $\displaystyle \ln(1+x)$, $\displaystyle \frac{1}{1-x^{2}}$, etc. I tried double integrals, but I always get stuck.
I even broke it up per partial fraction expansion, but several of the resulting integrals were still nasty.
Does anyone have some clever ideas?.  
 A: A straightforward way could be to consider the function 
$$
I(a,b)=\int_0^1 \frac{\left(\frac{1+x}{1-x}\right)^a x^b}{(1+x)^2} \, dx=
$$
$$
\Gamma (1-a) \Gamma (b+1) \, _2\tilde{F}_1(2-a,b+1;-a+b+2;-1),\quad a>1,\  b>-1,
$$
where $_2\tilde{F}_1$ is the regularized hypergeometric function (see the first integral representation in the reference). To take the second derivative $\frac {\partial^2I(a,b)}{\partial a\partial b}$ and manually evaluate the limit $a\to1+0$, $b\to-1+0$, which gives the desired integral.
Updated
Here is another idea. The integral $\int_{0}^{\frac{\pi}{2}}x\log(\sin(x))dx$ can be evaluated in the same way as the Gauss integral $I=\int_{0}^{\frac{\pi}{2}}\log \sin x\,dx\;$. 
Namely, making change of variables $y=\pi/2-x\;$ we have  $I=\int_{0}^{\frac{\pi}{2}}\log \cos x\, dx$, so 
$$
I=\frac12 \int_{0}^{\frac{\pi}{2}}\log \frac12\sin 2x\, dx=\frac14 \int_{0}^{\pi}\log \frac12\sin x\, dx=\frac12 \int_{0}^{\frac{\pi}{2}}\log \frac12\sin x\, dx,$$
which lead to an equation on $I$ etc.
Now for the function $f(x)=\frac12\frac{\log (|x|) \log \left(\left|\frac{x+1}{1-x}\right|\right)}{2 x \left(1-x^2\right)}$ there are two changes of variables leaving in place the logarithms in $f$:
1) $y=\frac{1-x}{1+x}$,
2) $y=1/x$.
The first one can be regarded as analogue to $x\to \pi/2-y$ and the second transform the  integral segment to $[1,+\infty)$ which perhaps corresponds to integrating on $[\pi/2,\pi]$. May be combining whose observations would lead to the desired result. For example, the first one leads to 
$$
\int_0^1f(x)\,dx=\int_0^1 \frac{(y+1) \log (y) \log \left(\frac{1+y}{1-y}\right)}{4 (1-y) y}\,dx,
$$ 
and denoting 
$$
I_1=\int_0^1f(x)\,dx=\frac{1}{16} \left(-7 \zeta (3)-\pi ^2 \log (4)\right),
$$
$$
I_2=\int_1^\infty f(x)\,dx=\frac{1}{16} \left(7 \zeta (3)-\pi ^2 \log (4)\right),
$$
we have $I_1+I_2=-\frac{1}{4} \pi ^2 \log (2)\;$, $I_1-I_2=-\frac{7 \zeta (3)}{8}\;$.
