Show that $\int _0^1\frac{\ln \left(1+\left(\frac{1-t}{1+t}\right)^2\right)}{t\left(\ln t\right)^2}dt=\ln 2$ According to WolframAlpha the integral 
$$\int _0^1\frac{\ln \left(1+\left(\frac{1-t}{1+t}\right)^2\right)}{t\left(\ln t\right)^2}dt$$
is shown to have a decimal expansion exactly identical to $\ln 2$.
How can we prove they are equal?
This integral came about as a result of integrating by parts
$$\int_0^x\frac{\arctan t}{\ln \left(t\right)\left(1+t\right)^2}dt$$
with 
$u=\frac{1}{\ln t}$ 
$dv=\frac{\arctan t}{\left(1+t\right)^2}dt$
$du=-\frac{1}{t\ln^2 t}dt$
$v=\frac{1}{4}\ln \left(2\right)-\frac{1}{4}\ln \left(1+\left(\frac{1-t}{1+t}\right)^2\right)-\frac{1}{2}\left(\frac{1-t}{1+t}\right)\arctan \left(t\right)$
 A: Hint:
$$\int\frac{\ln(1+(\dfrac{1-t}{1+t})^2)}{t(\ln t)^2}=\int\frac{\ln2}{t(\ln t)^2}+\int\frac{\ln(1+t^2)}{t(\ln t)^2}-2\int\frac{\ln(1+t)}{t(\ln t)^2}=\int\frac{\ln2}{t(\ln t)^2}$$
A: The given integral equals
$$\begin{eqnarray*} I &=& \int_{0}^{1}\frac{\log(2)+\log(1+t^2)-2\log(1+t)}{t\log^2t}\,dt\\&\stackrel{t\mapsto e^{-x}}{=}&\int_{0}^{+\infty}\frac{\log(2)+\log(1+e^{-2x})-2\log(1+e^{-x})}{x^2}\,dx\\&\stackrel{\text{IBP}}{=}&2\int_{0}^{+\infty}\left(\frac{1}{e^x+1}-\frac{1}{e^{2x}+1}\right)\frac{dx}{x} \tag{1}\end{eqnarray*}$$
and the claim simply follows by recalling Frullani's theorem.
As an alternative, 
$$\mathcal{L}\left(\frac{1}{e^x+1}\right) = \frac{1}{2}\left(H_{\frac{s}{2}}-H_{\frac{s-1}{2}}\right)\tag{2}$$
leads to
$$ I = 2\lim_{s\to +\infty}\log\left(\frac{\Gamma\left(\frac{s+2}{4}\right)\,\Gamma\left(\frac{s+2}{2}\right)}{\Gamma\left(\frac{s+4}{4}\right)\,\Gamma\left(\frac{s+1}{2}\right)}\right) = \color{red}{\log 2}\tag{3}$$
by Stirling's inequality.
A: Using $\enspace\displaystyle\zeta(0)=-\frac{1}{2}\enspace$ and $\enspace\displaystyle\lim\limits_{h\to 0}\frac{2^h-1}{h}=\ln 2$ 
and $\enspace\displaystyle\int _0^\infty\frac{x^{h-1}}{e^{ax}+1}dx=\frac{2^{h-1}-1}{2^{h-1}a^h}\zeta(h)\Gamma(h)\enspace$ for $\enspace h,a>0\enspace$ 
and $\enspace\displaystyle t:=e^{-x}\enspace$ we get:
$\displaystyle\int _0^1\frac{\ln \left(1+\left(\frac{1-t}{1+t}\right)^2\right)}{t\left(\ln t\right)^2}dt=\int _0^\infty\frac{\ln 2+\ln(1+e^{-2x})-2\ln(1+e^{-x})}{x^2}dx$
$\displaystyle=2\int _0^\infty (\frac{1}{e^{2x}+1}-\frac{1}{e^{x}+1}) \frac{dx}{x}=2\lim\limits_{h\to 0} \int _0^\infty (\frac{x^{h-1}}{e^{2x}+1}-\frac{x^{h-1}}{e^{x}+1})dx$
$\displaystyle=2\lim\limits_{h\to 0} \zeta(h)\Gamma(h+1)\frac{2^{h-1}-1}{2^{2h-1}}\frac{2^h-1}{h}=2(-\frac{1}{2}\frac{2^{-1}-1}{2^{-1}}\ln 2)=\ln 2$
