Definite integral $\int_{0}^{1}\left(\frac{x^{2}-1}{x^{2}+1}\right)\ln\left(\operatorname{arctanh}x\right)dx$ Prove the following closed form:
$$\int_{0}^{1}\left(\frac{x^{2}-1}{x^{2}+1}\right)\ln\left(\operatorname{arctanh}x\right)dx=\ln\pi-\gamma-\left(2-\frac{\pi}{2}\right)\ln2-\pi\ln\left(\frac{\Gamma\left(\frac{3}{4}\right)}{\Gamma\left(\frac{1}{4}\right)}\sqrt{2\pi}\right).$$
I had discovered this while attempting to prove a previous problem when I attempted to integrate both sides of
$$\frac{d}{dx}(x^2-1)\arctan(x)\ln(\operatorname{arctanh}(x)) = 2x\arctan(x)\ln(\operatorname{arctanh}(x))+\left(\frac{x^2-1}{x^2+1}\right)\ln(\operatorname{arctanh}(x))-\frac{\arctan(x)}{\operatorname{arctanh}(x)}.$$
That post was nine months ago and I didn't keep a systematic record of my process. I'm sure I could figure it out if I dug through my old work, but now I'm mostly curious to see how others in the community would attack this.
 A: $$I=-\int _0^1\frac{1-x^2}{1+x^2}\:\ln \left(\operatorname{arctanh}\left(x\right)\right)\:dx$$
$$=-\int _0^1\frac{1-x^2}{1+x^2}\ln \left(\ln \left(\frac{1+x}{1-x}\right)\right)\:dx-\ln \left(\frac{1}{2}\right)\int _0^1\frac{1-x^2}{1+x^2}\:dx$$
$$=-4\int _0^1\frac{t\ln \left(-\ln \left(t\right)\right)}{\left(1+t^2\right)\left(1+t\right)^2}\:dt-\ln \left(\frac{1}{2}\right)\left(\frac{\pi }{2}-1\right)$$
$$=-2\int _0^1\frac{\ln \left(-\ln \left(t\right)\right)}{1+t^2}\:dt+2\int _0^1\frac{\ln \left(-\ln \left(t\right)\right)}{\left(1+t\right)^2}\:dt-\ln \left(\frac{1}{2}\right)\left(\frac{\pi }{2}-1\right)$$
$$=-2\int _0^{\infty }\frac{\ln \left(x\right)}{1+e^{-2x}}\:e^{-x}\:dx+2\int _0^{\infty }\frac{\ln \left(x\right)}{\left(1+e^{-x}\right)^2}\:e^{-x}\:dx+\frac{\pi }{2}\ln \left(2\right)-\ln \left(2\right)$$
$$=-2\sum _{k=0}^{\infty }\left(-1\right)^k\:\int _0^{\infty }e^{-x\left(2k+1\right)}\ln \left(x\right)\:dx+2\int _0^{\infty }\frac{\ln \left(x\right)}{\left(1+e^{-x}\right)^2}\:e^{-x}\:dx+\frac{\pi }{2}\ln \left(2\right)-\ln \left(2\right)$$
$$=2\sum _{k=0}^{\infty }\left(-1\right)^k\:\left(\frac{\ln \left(2k+1\right)+\gamma }{2k+1}\right)+2\int _0^{\infty }\frac{\ln \left(x\right)}{\left(1+e^{-x}\right)^2}\:e^{-x}\:dx+\frac{\pi }{2}\ln \left(2\right)-\ln \left(2\right)$$
$$=2\sum _{k=0}^{\infty }\frac{\left(-1\right)^k\ln \left(2k+1\right)}{2k+1}+2\gamma \sum _{k=0}^{\infty }\frac{\left(-1\right)^k}{2k+1}+2\underbrace{\int _0^{\infty }\frac{\ln \left(x\right)}{\left(1+e^{-x}\right)^2}\:e^{-x}\:dx}_{K}+\frac{\pi }{2}\ln \left(2\right)-\ln \left(2\right)$$
$$\boxed{I=-2\beta '\left(1\right)+\frac{\gamma \pi }{2}+\ln \left(\pi \right)-\gamma +\frac{\pi }{2}\ln \left(2\right)-2\ln \left(2\right)}$$
Where $\displaystyle \beta '\left(s\right)$ is the derivative of the Dirichlet beta function.
Also see here for the integral $K$.
