How to show that $\int_{-\alpha}^{\alpha}\arccos\left({x\over \alpha}\right)\ln(\alpha+x)\mathrm dx=\alpha \pi \ln\left({\alpha \over 2}\right)?$ Given that:
Where $\alpha >0$

$$\int_{-\alpha}^{\alpha}\arccos\left({x\over \alpha}\right)\ln(\alpha+x)\mathrm dx=\alpha \pi \ln\left({\alpha \over 2}\right)\tag1$$

Setting $y=\arccos(x/\alpha)\implies \cos(y)={x\over \alpha}$
$dx=-\alpha \sin(y)dy$
$$\alpha\int_{0}^{\pi}y\sin(y)\ln(\alpha+\alpha\cos y)\mathrm dy\tag2$$
$$\alpha \ln\alpha \int_{0}^{\pi}y\sin y\mathrm dy+\alpha\int_{0}^{\pi}y\sin y\ln(1+\cos y)\mathrm dy\tag3$$
$$\alpha \pi \ln\alpha +\alpha\int_{0}^{\pi}y\sin y\ln(1+\cos y)\mathrm dy\tag4$$

$\displaystyle J=\int_{-\alpha}^{\alpha} \arccos\left(\dfrac{x}{\alpha}\right)\ln\left(\alpha+x\right)dx$
Perform the change of variable $y=\dfrac{x}{\alpha}$,
$\begin{align} J&=\alpha\int_{-1}^{1} \arccos\left(x\right)\ln\left(\alpha+\alpha x\right)dx\\
\end{align}$
Perform the change of variable $y=\arccos x$,
$\begin{align} J&=\alpha\int_{0}^{\pi} x\sin x\ln\left(\alpha+\alpha \cos x\right)dx\\
&=\alpha\int_{0}^{\pi} x\sin x\ln\left(1+ \cos x\right)dx+\alpha\ln \alpha\int_{0}^{\pi} x\sin x\,dx\\
&=\alpha\int_{0}^{\pi} x\sin x\ln\left(1+ \cos x\right)dx+\alpha\ln\alpha\Big[-x\cos x\Big]_{0}^{\pi}+\alpha\ln\alpha\int_{0}^{\pi}\cos x\,dx\\
&=\alpha\int_{0}^{\pi} x\sin x\ln\left(1+ \cos x\right)dx+\pi\alpha\ln\alpha+\alpha\ln\alpha\Big[\sin x\Big]_{0}^{\pi}\\
&=\alpha\int_{0}^{\pi} x\sin x\ln\left(1+ \cos x\right)dx+\pi\alpha\ln\alpha
\end{align}$
$\begin{align}
K&=\int_{0}^{\pi} x\sin x\ln\left(1+ \cos x\right)dx\\
&=\int_{0}^{\tfrac{\pi}{2}} x\sin x\ln\left(1+ \cos x\right)dx+\int_{\tfrac{\pi}{2}}^{\pi} x\sin x\ln\left(1+ \cos x\right)dx
\end{align}$
In the latter integral perform the change of variable $y=\pi-x$,
$\begin{align}K&=\int_{0}^{\tfrac{\pi}{2}} x\sin x\ln\left(1+ \cos x\right)dx+\int_{0}^{\tfrac{\pi}{2}}\left(\pi-x\right)\sin x\ln\left(1-\cos x\right)dx\\
&=\int_{0}^{\tfrac{\pi}{2}} x\sin x\ln\left(\dfrac{1+ \cos x}{1-\cos x}\right)\,dx+\pi\int_{0}^{\tfrac{\pi}{2}}\sin x\ln\left(1-\cos x\right)dx\\
&=\pi\int_{0}^{\tfrac{\pi}{2}}\sin x\ln\left(1-\cos x\right)dx-2\int_{0}^{\tfrac{\pi}{2}} x\sin x\ln\left(\tan\left(\dfrac{x}{2}\right)\right)\,dx\\
\end{align}$
In the first integral perform the change of variable $y=1-\cos x$,
$\begin{align}K&=\pi\int_{0}^1 \ln x dx-2\int_{0}^{\tfrac{\pi}{2}} x\sin x\ln\left(\tan\left(\dfrac{x}{2}\right)\right)\,dx\\
&=\pi\Big[x\ln x\Big]_0^1-\pi\int_0^1 1\,dx-2\int_{0}^{\tfrac{\pi}{2}} x\sin x\ln\left(\tan\left(\dfrac{x}{2}\right)\right)\,dx\\
&=-\pi-2\int_{0}^{\tfrac{\pi}{2}} x\sin x\ln\left(\tan\left(\dfrac{x}{2}\right)\right)\,dx
\end{align}$
In the latter integral perform the change of variable $y=\dfrac{x}{2}$,
$\begin{align}K&=-\pi-8\int_{0}^{\tfrac{\pi}{4}} x\sin(2x)\ln\left(\tan x\right)\,dx\\
&=-\pi+4\Big[x\cos(2x)\ln\left(\tan x\right)\Big]_{0}^{\tfrac{\pi}{4}}-4\int_{0}^{\tfrac{\pi}{4}} \cos(2x)\left(\ln\left(\tan x\right)+\dfrac{x(1+\tan^2 x)}{\tan x}\right)\,dx\\
&=-\pi-4\int_{0}^{\tfrac{\pi}{4}}\left(\dfrac{1-\tan^2 x}{1+\tan^2 x}\right)\left(\ln\left(\tan x\right)+\dfrac{x(1+\tan^2 x)}{\tan x}\right)\,dx\\
&=-\pi-4\int_{0}^{\tfrac{\pi}{4}}\left(\dfrac{1-\tan^2 x}{1+\tan^2 x}\right)\left(\ln\left(\tan x\right)\right)\,dx-4\int_{0}^{\tfrac{\pi}{4}}\dfrac{x}{\tan x}\,dx+4\int_{0}^{\tfrac{\pi}{4}}x\tan x\,dx\\
&=-\pi-4\int_{0}^{\tfrac{\pi}{4}}\ln\left(\tan x\right)dx+8\int_{0}^{\tfrac{\pi}{4}}\dfrac{\tan^2 x\ln\left(\tan x\right)}{1+\tan^2 x}dx-4\int_{0}^{\tfrac{\pi}{4}}\dfrac{x}{\tan x}dx+4\int_{0}^{\tfrac{\pi}{4}}x\tan x dx\\
&=-\pi-4\int_{0}^{\tfrac{\pi}{4}}\ln\left(\tan x\right)dx+\\
&4\left(\left[\left(x-\dfrac{\tan x}{1+\tan^2 x}\right)\ln(\tan x)\right]_{0}^{\tfrac{\pi}{4}}-\int_{0}^{\tfrac{\pi}{4}}\left(x-\dfrac{\tan x}{1+\tan^2 x}\right)\left(\dfrac{1+\tan^2}{\tan x}\right)dx\right)-\\
&4\int_{0}^{\tfrac{\pi}{4}}\dfrac{x}{\tan x}dx+4\int_{0}^{\tfrac{\pi}{4}}x\tan x dx\\
&=4G-8\int_{0}^{\tfrac{\pi}{4}}\dfrac{x}{\tan x}dx\\
&=4G-8\left(\Big[x\ln(\sin x)\Big]_{0}^{\tfrac{\pi}{4}}-\int_{0}^{\tfrac{\pi}{4}}\ln(\sin x)\,dx\right)\\
&=4G+\pi\ln 2+8\int_{0}^{\tfrac{\pi}{4}}\ln(\sin x)\,dx\\
&=4G+\pi\ln 2+4\left(\int_{0}^{\tfrac{\pi}{4}}\ln(\tan x)\,dx+\int_{0}^{\tfrac{\pi}{4}}\ln(\sin x\cos x)\,dx\right)\\
&=\pi\ln 2+4\int_{0}^{\tfrac{\pi}{4}}\ln(\sin x\cos x)\,dx\\
&=\pi\ln 2+4\int_{0}^{\tfrac{\pi}{4}}\ln\left(\dfrac{\sin(2x)}{2}\right)\,dx\\
&=\pi\ln 2-\pi\ln 2+4\int_{0}^{\tfrac{\pi}{4}}\ln\left(\sin(2x)\right)\,dx\\
&=4\int_{0}^{\tfrac{\pi}{4}}\ln\left(\sin(2x)\right)\,dx
\end{align}$
In the latter integral perform the change of variable $y=2x$,
$\begin{align}K&=2\int_{0}^{\tfrac{\pi}{2}}\ln\left(\sin x\right)\,dx\\
&=-\pi\ln 2
\end{align}$
Therefore,
$\boxed{\displaystyle J=\pi\alpha\ln\left(\dfrac{\alpha}{2}\right)}$
Nota Bene:
I have used the following results:
$\displaystyle \int_{0}^{\tfrac{\pi}{2}}\ln\left(\sin x\right)\,dx=-\dfrac{\pi}{2}\ln 2$
$\displaystyle G=-\int_{0}^{\tfrac{\pi}{4}}\ln\left(\tan x\right)\,dx$
