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 $y=\arccos(x/\alpha)\implies \cos(y)={x\over \alpha}$
$dx=-\alpha \sin(y)dy$
$$\alpha\int_{0}^{\pi}y\sin(y)\ln(\alpha+\alpha\cos y)\mathrm dy\tag2$$
$$\alpha \ln\alpha \int_{0}^{\pi}y\sin y\mathrm dy+\alpha\int_{0}^{\pi}y\sin y\ln(1+\cos y)\mathrm dy\tag3$$
$$\alpha \pi \ln\alpha +\alpha\int_{0}^{\pi}y\sin y\ln(1+\cos y)\mathrm dy\tag4$$
 A: We just need to show that
$$ \int_{0}^{\pi}y\sin(y)\log(1+\cos y)\,dy =-\pi\log 2\tag{A}$$
where by the Fourier series of $\log\cos$ we have:
$$ \forall y\in(0,\pi),\qquad \log(1+\cos y)=-\log 2+2\sum_{k\geq 1}\frac{(-1)^{k+1}\cos(ky)}{k}\tag{B} $$
and the Fourier cosine series of $y\sin y$ is straightforward to compute by IBP:
$$ \forall y\in(0,\pi),\qquad y\sin y = 1-\frac{1}{2}\cos(y)+2\sum_{k\geq 2}\frac{(-1)^{k+1}\cos(ky)}{k^2-1}.\tag{C}$$
It follows that the LHS of $(A)$ can be computed from $(B),(C)$ and the orthogonality relation
$$ \forall m,n\in\mathbb{Z},\qquad \int_{0}^{\pi}\cos(nx)\cos(mx)\,dx = \frac{\pi}{2}\,\delta(m,n).\tag{D}$$
$(A)$ boils down to proving that
$$ \sum_{k\geq 2}\frac{1}{(k-1)k(k+1)}=\frac{1}{4}\tag{E} $$
and that is trivial since the LHS of $(E)$ is telescopic.
A: $\displaystyle J=\int_{-\alpha}^{\alpha} \arccos\left(\dfrac{x}{\alpha}\right)\ln\left(\alpha+x\right)dx$
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:
$\displaystyle \int_{0}^{\tfrac{\pi}{2}}\ln\left(\sin x\right)\,dx=-\dfrac{\pi}{2}\ln 2$
$\displaystyle G=-\int_{0}^{\tfrac{\pi}{4}}\ln\left(\tan x\right)\,dx$
A: $\begin{align} J= \int_{-1}^{1} \arccos x\ln(1+x)\,dx&=\Big[(x+1)(\ln(1+x)-1)\arccos x\Big]_{-1}^1+\int_{-1}^{1}\dfrac{(x+1)(\ln(1+x)-1)}{\sqrt{1-x^2}}dx\\
=&\int_{-1}^{1}\sqrt{\dfrac{1+x}{1-x}}(\ln(1+x)-1)dx\\
\end{align}$
Perform the change of variable $y=\sqrt{\dfrac{1+x}{1-x}}$,
$\begin{align}
J&=8\int_0^{+\infty} \dfrac{x^2\ln x}{(1+x^2)^2}\,dx+4(\ln 2-1)\int_0^{+\infty} \dfrac{x^2}{(1+x^2)^2}\,dx-4\int_0^{+\infty} \dfrac{x^2\ln(1+x^2)}{(1+x^2)^2}\,dx\\
&=-4\left(\Big[\dfrac{x\ln x}{1+x^2}\Big]_0^{+\infty}-\int_0^{+\infty}\left(\dfrac{\ln x}{1+x^2}+\dfrac{1}{1+x^2}\right)\,dx\right)+4(\ln 2-1)\int_0^{+\infty} \dfrac{x^2}{(1+x^2)^2}\,dx+\\
&2\left(\Big[\dfrac{x\ln(1+x^2)}{1+x^2}\Big]_0^{+\infty}-\int_0^{+\infty}\left(\dfrac{\ln(1+x^2)}{1+x^2}+\dfrac{2x^2}{(1+x^2)^2}\right)\,dx\right)\\
&=2\pi+4(\ln 2-2)\int_0^{+\infty} \dfrac{x^2}{(1+x^2)^2}\,dx-2\int_0^{+\infty} \dfrac{\ln(1+x^2)}{1+x^2}\,dx\\
&=2\pi+2(\ln 2-2)\left(\Big[-\dfrac{x}{1+x^2}\Big]_0^{+\infty}+\int_0^{+\infty}\dfrac{1}{1+x^2}\,dx\right)-2\int_0^{+\infty} \dfrac{\ln(1+x^2)}{1+x^2}\,dx\\
&=2\pi+(\ln 2-2)\pi-2\int_0^{+\infty} \dfrac{\ln(1+x^2)}{1+x^2}\,dx\\
&=\pi\ln 2-2\int_0^{+\infty} \dfrac{\ln(1+x^2)}{1+x^2}\,dx\\
\end{align}$
In the latter integral perform the change of variable $y=\arctan x$,
$\begin{align}J&=\pi\ln 2+4\int_0^{\tfrac{\pi}{2}} \ln(\cos x)\,dx\\
&=\pi\ln 2-2\pi\ln 2\\
&=-\pi\ln 2
\end{align}$
(Thanks to Gebrane for inspiration)
