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

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.
