antidervative of $\ln\left|\tan{\left(\dfrac{\cos^{-1}{\left(\left|x\right|\right)}}{2} + \dfrac{\pi}{4} \right)}\right| $ Question
How can the  antiderivative of
$$  
  \ln\left|\tan{\left(\dfrac{\cos^{-1}{\left(\left|x\right|\right)}}{2} + \dfrac{\pi}{4}  \right)}\right|  $$ be obtained?

A hint from trigonometry
Consider the Figure below, where $\triangle ABC$ is right-angled, $\overline{AB} = |x|$ and $\overline{AC}=1$, so that, by definition
$$\alpha = \angle CAB = \arccos |x|.$$
Draw the bisector of $A$, that intersects $BC$ in $D$. From $D$ draw the perpendicular to $AD$. Let, on it, $E$ be a point such that $ED \cong AD$. Finally draw from $E$ the perpendicular to $AB$, that intersects $AB$ in $H$. 
Clearly we have
\begin{eqnarray}
\angle EAB &=& \frac{\alpha}2+\frac{\pi}4=\\
&=&\frac{\arccos |x|}2 + \frac{\pi}4,
\end{eqnarray} 
and
\begin{eqnarray}
\tan \angle EAB &=& \tan \left(\frac{\arccos |x|}2 + \frac{\pi}4\right)=\\
&=& \frac{\overline{EH}}{\overline{AH}}.
\end{eqnarray}
So our first aim is to write this ratio in terms of $x$ "directly".

First we calculate
$$\overline{AD} = \frac{\overline{AB}}{\cos\frac{\alpha}2}=\frac{\sqrt 2 |x|}{\sqrt{1+|x|}},$$
by the bisection formula, and
$$\overline{BD} = |x|\sqrt{\frac{1-|x|}{1+|x|}},$$
by Pythagorean Theorem on $\triangle ABD$.
Let then $F$ be the intersection between $EH$ and $AD$. We have $\triangle EFD\sim\triangle ABD$, and, therefore
$$\overline{EF} = \frac{2|x|}{1+|x|}.$$
Pythagorean Theorem on $\triangle AFD$ gives
$$\overline{FD} = \frac{|x|\sqrt{2(1-|x|)}}{1+|x|}.$$
As a consequence
$$\overline{AF} = \frac{\sqrt 2 |x| (\sqrt{1+|x|}-\sqrt{1-|x|})}{1+|x|}.$$
Similarity $\triangle AFH\sim\triangle ABD$ yields
$$\overline{FH} = \frac{|x|\sqrt{1-|x|}(\sqrt{1+|x|}-\sqrt{1-|x|})}{1+|x|}$$
and
$$\overline{AH} = \frac{|x|(\sqrt{1+|x|}-\sqrt{1-|x|})}{\sqrt{1+|x|}}.$$
The tangent we are looking for is thus
\begin{eqnarray}
\tan \angle EAB  &=& \frac{\frac{2|x|}{1+|x|}+\frac{|x|\sqrt{1-|x|}(\sqrt{1+|x|}-\sqrt{1-|x|})}{1+|x|}}{\frac{|x|(\sqrt{1+|x|}-\sqrt{1-|x|})}{\sqrt{1+|x|}}} =\\
&=&\frac{1+|x|+\sqrt{1-x^2}}{1+|x|-\sqrt{1-x^2}}=\\
&=&\frac{\left(1+|x|+\sqrt{1-x^2}\right)^2}{(1+|x|)^2-1+x^2}=\\
&=&\frac{1+\sqrt{1-x^2}}{|x|}.
\end{eqnarray}
Therefore, integrating by parts leads to the result
\begin{eqnarray}
\mathcal I &=& \int\log\left|\tan\left(\frac{\arccos x}2+\frac{\pi}4\right)\right| dx=\\
&=& \int \log\left|\frac{1+\sqrt{1-x^2}}{|x|}\right| dx=\\
&=& \int \log\left(\frac{1+\sqrt{1-x^2}}{|x|}\right) dx=\\
&=& \boxed{x\log\left(\frac{1+\sqrt{1-x^2}}{|x|}\right) + \arcsin x + C}.
\end{eqnarray}
