For each $n\in\mathbb{Z}_{\ge0}$, define the function $\mathcal{J}_{n}:(0,1)\rightarrow\mathbb{R}$ via the doubly improper integral
$$\mathcal{J}_{n}{\left(a\right)}:=\int_{a}^{1}\mathrm{d}x\,\frac{x^{n}}{\sqrt{1-x^{2}}}\ln{\left(\frac{x+a}{x-a}\right)}.$$
I seek closed form solutions to the $n=0,1$ cases in terms of elementary functions and dilogarithms (or closely related functions), as well as a solution (in principle) for higher $n$ by means of a recurrence relation.
On
The case $n=1$ is pleasant to work $$I_1=\int \frac{x }{\sqrt{1-x^2}} \log \left(\frac{x+a}{x-a}\right)\,dx$$ $$u=\log \left(\frac{x+a}{x-a}\right) \qquad \implies \qquad du=\frac{2 a}{a^2-x^2}\,dx$$ $$dv=\frac{x }{\sqrt{1-x^2}} \,dx \qquad \implies \qquad v=-\sqrt{1-x^2}$$ $$I_1=-\sqrt{1-x^2}\log \left(\frac{x+a}{x-a}\right)+2a \int \frac{\sqrt{1-x^2}}{a^2-x^2}\,dx$$ $$\int \frac{\sqrt{1-x^2}}{a^2-x^2}\,dx=\frac 1{2a}\left(\int \frac{\sqrt{1-x^2}}{x+a}\,dx-\int\frac{\sqrt{1-x^2}}{x-a}\,dx\right)$$
Using twice $$\int\frac{\sqrt{1-x^2}}{x+b}\,dx=\sqrt{1-x^2}-2b\tan ^{-1}\left(\sqrt{\frac{1-x}{1+x}}\right)+$$ $$2\sqrt{b^2-1}\tan ^{-1}\left(\sqrt{\frac{b-1}{b+1}}\sqrt{\frac{1-x}{1+x}}\right)$$ then
$$\int_0^1 \frac{x }{\sqrt{1-x^2}} \log \left(\frac{x+a}{x-a}\right)\,dx=-\sqrt{1-a^{2}}\log{\left(1-a^{2}\right)}+4a\tan^{-1}{\left(\sqrt{\frac{1-a}{1+a}}\right)}$$
For general $n$ $$dv=\frac{x^n}{\sqrt{1-x^2}}\,dx\qquad \implies \qquad v=\frac{x^{n+1}}{n+1}\, \,\,_2F_1\left(\frac{1}{2},\frac{n+1}{2};\frac{n+3}{2};x^2\right)$$ would give $$I_n=\frac{\sqrt{\pi }\,\, \Gamma \left(\frac{n+3}{2}\right)}{(n+1) \Gamma \left(\frac{n+2}{2}\right)}\log \left(\frac{1+a}{1-a}\right)+\frac{2a}{n+1}\int_0^1 \frac {x^{n+1}}{x^2-a^2}\,\,_2F_1\left(\frac{1}{2},\frac{n+1}{2};\frac{n+3}{2};x^2\right)\,dx$$ Here, I am stuck except for $n=0$ which is $$I_0=\frac{\pi}{2} \log \left(\frac{1+a}{1-a}\right)-2a\int_0^1 \frac{ \sin ^{-1}(x)}{x^2-a^2}\,dx$$ $$J=\int_0^1 \frac{ \sin ^{-1}(x)}{x^2-a^2}\,dx=$$ is a nasty expression $$\frac{4 J}{\sqrt{-\frac{1}{a^2}}}=4 \text{Li}_2\left(-\frac{1}{\sqrt{1-\frac{1}{a^2}}+\sqrt{-\frac{1}{a^2}}} \right)-4 \text{Li}_2\left(\frac{1}{\sqrt{1-\frac{1}{a^2}}+\sqrt{-\frac{1}{a^2}}}\right) +$$ $$4 \log \left(\sqrt{1-\frac{1}{a^2}}+\sqrt{-\frac{1}{a^2}}\right) \log \left(\frac{\sqrt{a^2} \left(\sqrt{-\frac{1}{a^2}}-1\right)+\sqrt{a^2-1}}{\sqrt{a^2} \left(\sqrt{-\frac{1}{a^2}}+1\right)+\sqrt{a^2-1}}\right)-2 \pi \tan ^{-1}\left(\sqrt{-\frac{1}{a^2}}\right)+\pi ^2$$
Let's tackle the problem of obtaining a recurrence relation first.
Note that
$$\left(x^{n+1}-a^{n+1}\right)\sqrt{1-x^{2}}\ln{\left(\frac{x+a}{x-a}\right)}\bigg{|}_{x=1}=0;~~~\small{n\in\mathbb{Z}_{\ge0}\land a\in(0,1)}.$$
It can also be shown that
$$\lim_{x\to a^{+}}\left(x^{n+1}-a^{n+1}\right)\sqrt{1-x^{2}}\ln{\left(\frac{x+a}{x-a}\right)}=0;~~~\small{n\in\mathbb{Z}_{\ge0}\land a\in(0,1)},$$
where we've used the summation formula
$$\left(x-a\right)\sum_{k=0}^{n}x^{n-k}a^{k}=x^{n+1}-a^{n+1};~~~\small{n\in\mathbb{Z}_{\ge0}\land(x,a)\in\mathbb{R}^{2}},$$
as well as the simpler limit
$$\lim_{x\to a^{+}}\left(x-a\right)\ln{\left(x-a\right)};~~~\small{a\in\mathbb{R}}.$$
By the fundamental theorem of calculus, we then have
$$0=\int_{a}^{1}\mathrm{d}x\,\frac{d}{dx}\left[\left(x^{n+1}-a^{n+1}\right)\sqrt{1-x^{2}}\ln{\left(\frac{x+a}{x-a}\right)}\right];~~~\small{n\in\mathbb{Z}_{\ge0}\land a\in(0,1)}.$$
For $n\in\mathbb{Z}_{\ge0}\land a\in(0,1)$, we find
$$\begin{align} 0 &=\int_{a}^{1}\mathrm{d}x\,\frac{d}{dx}\left[\left(x^{n+1}-a^{n+1}\right)\sqrt{1-x^{2}}\ln{\left(\frac{x+a}{x-a}\right)}\right]\\ &=\int_{a}^{1}\mathrm{d}x\,\bigg{[}\frac{d}{dx}\left(x^{n+1}-a^{n+1}\right)\sqrt{1-x^{2}}\ln{\left(\frac{x+a}{x-a}\right)}\\ &~~~~~+\left(x^{n+1}-a^{n+1}\right)\frac{d}{dx}\sqrt{1-x^{2}}\ln{\left(\frac{x+a}{x-a}\right)}\\ &~~~~~+\left(x^{n+1}-a^{n+1}\right)\sqrt{1-x^{2}}\frac{d}{dx}\ln{\left(\frac{x+a}{x-a}\right)}\bigg{]}\\ &=\int_{a}^{1}\mathrm{d}x\,\bigg{[}\left(n+1\right)x^{n}\sqrt{1-x^{2}}\ln{\left(\frac{x+a}{x-a}\right)}\\ &~~~~~+\left(x^{n+1}-a^{n+1}\right)\left(-\frac{x}{\sqrt{1-x^{2}}}\right)\ln{\left(\frac{x+a}{x-a}\right)}\\ &~~~~~+\left(x^{n+1}-a^{n+1}\right)\sqrt{1-x^{2}}\left(-\frac{2a}{x^{2}-a^{2}}\right)\bigg{]}\\ &=\int_{a}^{1}\mathrm{d}x\,\bigg{[}\frac{\left(n+1\right)x^{n}\left(1-x^{2}\right)}{\sqrt{1-x^{2}}}\ln{\left(\frac{x+a}{x-a}\right)}+\frac{\left(a^{n+1}-x^{n+1}\right)x}{\sqrt{1-x^{2}}}\ln{\left(\frac{x+a}{x-a}\right)}\\ &~~~~~-\frac{2a\left(x^{n+1}-a^{n+1}\right)}{x^{2}-a^{2}}\sqrt{1-x^{2}}\bigg{]}\\ &=\int_{a}^{1}\mathrm{d}x\,\bigg{[}\frac{\left(n+1\right)x^{n}-\left(n+1\right)x^{n+2}}{\sqrt{1-x^{2}}}\ln{\left(\frac{x+a}{x-a}\right)}+\frac{a^{n+1}x-x^{n+2}}{\sqrt{1-x^{2}}}\ln{\left(\frac{x+a}{x-a}\right)}\\ &~~~~~-\frac{2a\left(x-a\right)\sum_{k=0}^{n}x^{n-k}a^{k}}{x^{2}-a^{2}}\sqrt{1-x^{2}}\bigg{]}\\ &=\int_{a}^{1}\mathrm{d}x\,\bigg{[}\frac{\left(n+1\right)x^{n}-\left(n+2\right)x^{n+2}+a^{n+1}x}{\sqrt{1-x^{2}}}\ln{\left(\frac{x+a}{x-a}\right)}\\ &~~~~~-\frac{2a\sum_{k=0}^{n}x^{n-k}a^{k}}{x+a}\sqrt{1-x^{2}}\bigg{]}\\ &=\int_{a}^{1}\mathrm{d}x\,\frac{\left(n+1\right)x^{n}-\left(n+2\right)x^{n+2}+a^{n+1}x}{\sqrt{1-x^{2}}}\ln{\left(\frac{x+a}{x-a}\right)}\\ &~~~~~-\int_{a}^{1}\mathrm{d}x\,\frac{2a\sum_{k=0}^{n}x^{n-k}a^{k}}{x+a}\sqrt{1-x^{2}}\\ &=\left(n+1\right)\int_{a}^{1}\mathrm{d}x\,\frac{x^{n}}{\sqrt{1-x^{2}}}\ln{\left(\frac{x+a}{x-a}\right)}-\left(n+2\right)\int_{a}^{1}\mathrm{d}x\,\frac{x^{n+2}}{\sqrt{1-x^{2}}}\ln{\left(\frac{x+a}{x-a}\right)}\\ &~~~~~+a^{n+1}\int_{a}^{1}\mathrm{d}x\,\frac{x}{\sqrt{1-x^{2}}}\ln{\left(\frac{x+a}{x-a}\right)}-2a\int_{a}^{1}\mathrm{d}x\,\frac{\left(\sum_{k=0}^{n}x^{n-k}a^{k}\right)\sqrt{1-x^{2}}}{x+a}\\ &=\left(n+1\right)\mathcal{J}_{n}{\left(a\right)}-\left(n+2\right)\mathcal{J}_{n+2}{\left(a\right)}+a^{n+1}\mathcal{J}_{1}{\left(a\right)}\\ &~~~~~-2a\int_{a}^{1}\mathrm{d}x\,\frac{\left(\sum_{k=0}^{n}x^{n-k}a^{k}\right)\sqrt{1-x^{2}}}{x+a},\\ \end{align}$$
$$\implies\mathcal{J}_{n+2}{\left(a\right)}=\frac{n+1}{n+2}\mathcal{J}_{n}{\left(a\right)}+\frac{a^{n+1}}{n+2}\mathcal{J}_{1}{\left(a\right)}-\frac{2a}{n+2}\int_{a}^{1}\mathrm{d}x\,\frac{\left(\sum_{k=0}^{n}x^{n-k}a^{k}\right)\sqrt{1-x^{2}}}{x+a},$$
where the remaining integral is clearly elementary.
We consider the $n=1$ case next, which is in fact elementary.
Suppose $a\in(0,1)$. It can be shown that
$$\lim_{x\to a^{+}}\left(\sqrt{1-a^{2}}-\sqrt{1-x^{2}}\right)\ln{\left(\frac{x+a}{x-a}\right)}=0.$$
Then,
$$\begin{align} \mathcal{J}_{1}{\left(a\right)} &=\int_{a}^{1}\mathrm{d}x\,\frac{x}{\sqrt{1-x^{2}}}\ln{\left(\frac{x+a}{x-a}\right)}\\ &=\int_{a}^{1}\mathrm{d}x\,\frac{d}{dx}\left(\sqrt{1-a^{2}}-\sqrt{1-x^{2}}\right)\ln{\left(\frac{x+a}{x-a}\right)}\\ &=\left(\sqrt{1-a^{2}}-\sqrt{1-x^{2}}\right)\ln{\left(\frac{x+a}{x-a}\right)}\bigg{|}_{x=1}\\ &~~~~~-\lim_{x\to a^{+}}\left(\sqrt{1-a^{2}}-\sqrt{1-x^{2}}\right)\ln{\left(\frac{x+a}{x-a}\right)}\\ &~~~~~-\int_{a}^{1}\mathrm{d}x\,\left(\sqrt{1-a^{2}}-\sqrt{1-x^{2}}\right)\frac{d}{dx}\ln{\left(\frac{x+a}{x-a}\right)};~~~\small{I.B.P.}\\ &=\sqrt{1-a^{2}}\ln{\left(\frac{1+a}{1-a}\right)}-0-\int_{a}^{1}\mathrm{d}x\,\left(\sqrt{1-a^{2}}-\sqrt{1-x^{2}}\right)\left(-\frac{2a}{x^{2}-a^{2}}\right)\\ &=\sqrt{1-a^{2}}\ln{\left(\frac{1+a}{1-a}\right)}+\int_{a}^{1}\mathrm{d}x\,\frac{2a}{x^{2}-a^{2}}\cdot\frac{(1-a^{2})-(1-x^{2})}{\sqrt{1-a^{2}}+\sqrt{1-x^{2}}}\\ &=\sqrt{1-a^{2}}\ln{\left(\frac{1+a}{1-a}\right)}+\int_{a}^{1}\mathrm{d}x\,\frac{2a}{\sqrt{1-a^{2}}+\sqrt{1-x^{2}}}\\ &=\sqrt{1-a^{2}}\ln{\left(\frac{1+a}{1-a}\right)}+\int_{a}^{1}\mathrm{d}x\,\frac{2a}{\sqrt{1-a^{2}}+\left(1+x\right)\sqrt{\frac{1-x}{1+x}}}\\ &=\sqrt{1-a^{2}}\ln{\left(\frac{1+a}{1-a}\right)}+\int_{\frac{1-a}{1+a}}^{0}\mathrm{d}y\,\frac{(-2)}{\left(1+y\right)^{2}}\cdot\frac{2a}{\sqrt{1-a^{2}}+\left(\frac{2}{1+y}\right)\sqrt{y}};~~~\small{\left[\frac{1-x}{1+x}=y\right]}\\ &=\sqrt{1-a^{2}}\ln{\left(\frac{1+a}{1-a}\right)}+4a\int_{0}^{\frac{1-a}{1+a}}\mathrm{d}y\,\frac{1}{\left(1+y\right)\left[\left(1+y\right)\sqrt{1-a^{2}}+2\sqrt{y}\right]}\\ &=\sqrt{1-a^{2}}\ln{\left(\frac{1+a}{1-a}\right)}+4a\int_{0}^{\sqrt{\frac{1-a}{1+a}}}\mathrm{d}t\,\frac{2t}{\left(1+t^{2}\right)\left[\left(1+t^{2}\right)\sqrt{1-a^{2}}+2t\right]};~~~\small{\left[\sqrt{y}=t\right]}\\ &=\sqrt{1-a^{2}}\ln{\left(\frac{1+a}{1-a}\right)}+4a\int_{0}^{\sqrt{\frac{1-a}{1+a}}}\mathrm{d}t\,\frac{2t+\left(1+t^{2}\right)\sqrt{1-a^{2}}-\left(1+t^{2}\right)\sqrt{1-a^{2}}}{\left(1+t^{2}\right)\left[2t+\left(1+t^{2}\right)\sqrt{1-a^{2}}\right]}\\ &=\sqrt{1-a^{2}}\ln{\left(\frac{1+a}{1-a}\right)}+4a\int_{0}^{\sqrt{\frac{1-a}{1+a}}}\mathrm{d}t\,\frac{1}{1+t^{2}}\\ &~~~~~-4a\int_{0}^{\sqrt{\frac{1-a}{1+a}}}\mathrm{d}t\,\frac{\sqrt{1-a^{2}}}{2t+\left(1+t^{2}\right)\sqrt{1-a^{2}}}\\ &=\sqrt{1-a^{2}}\ln{\left(\frac{1+a}{1-a}\right)}+4a\arctan{\left(\sqrt{\frac{1-a}{1+a}}\right)}\\ &~~~~~-4a\int_{0}^{\sqrt{\frac{1-a}{1+a}}}\mathrm{d}t\,\frac{1}{t^{2}+2\left(\frac{1}{\sqrt{1-a^{2}}}\right)t+1}\\ &=\sqrt{1-a^{2}}\ln{\left(\frac{1+a}{1-a}\right)}+4a\arctan{\left(\sqrt{\frac{1-a}{1+a}}\right)}\\ &~~~~~-4a\int_{0}^{\sqrt{\frac{1-a}{1+a}}}\mathrm{d}t\,\frac{1}{\left(t+\frac{1-a}{\sqrt{1-a^{2}}}\right)\left(t+\frac{1+a}{\sqrt{1-a^{2}}}\right)}\\ &=\sqrt{1-a^{2}}\ln{\left(\frac{1+a}{1-a}\right)}+4a\arctan{\left(\sqrt{\frac{1-a}{1+a}}\right)}\\ &~~~~~-4a\int_{0}^{\sqrt{\frac{1-a}{1+a}}}\mathrm{d}t\,\frac{1}{\left(t+\sqrt{\frac{1-a}{1+a}}\right)\left(t+\frac{1}{\sqrt{\frac{1-a}{1+a}}}\right)}\\ &=\sqrt{1-a^{2}}\ln{\left(\frac{1+a}{1-a}\right)}+4a\arctan{\left(\sqrt{\frac{1-a}{1+a}}\right)}\\ &~~~~~-4a\int_{0}^{b}\mathrm{d}t\,\frac{1}{\left(t+b\right)\left(t+b^{-1}\right)};~~~\small{\left[\sqrt{\frac{1-a}{1+a}}=:b\in(0,1)\right]}\\ &=\sqrt{1-a^{2}}\ln{\left(\frac{1+a}{1-a}\right)}+4a\arctan{\left(\sqrt{\frac{1-a}{1+a}}\right)}\\ &~~~~~-4a\int_{0}^{b}\mathrm{d}t\,\frac{1}{\left(b^{-1}-b\right)}\left[\frac{1}{\left(t+b\right)}-\frac{1}{\left(t+b^{-1}\right)}\right]\\ &=\sqrt{1-a^{2}}\ln{\left(\frac{1+a}{1-a}\right)}+4a\arctan{\left(\sqrt{\frac{1-a}{1+a}}\right)}\\ &~~~~~-\frac{4ab}{1-b^{2}}\int_{0}^{b}\mathrm{d}t\,\left[\frac{1}{\left(t+b\right)}-\frac{b}{\left(bt+1\right)}\right]\\ &=\sqrt{1-a^{2}}\ln{\left(\frac{1+a}{1-a}\right)}+4a\arctan{\left(\sqrt{\frac{1-a}{1+a}}\right)}\\ &~~~~~-\frac{4ab}{1-b^{2}}\int_{0}^{b}\mathrm{d}t\,\frac{d}{dt}\left[\ln{\left(t+b\right)}-\ln{\left(bt+1\right)}\right]\\ &=\sqrt{1-a^{2}}\ln{\left(\frac{1+a}{1-a}\right)}+4a\arctan{\left(\sqrt{\frac{1-a}{1+a}}\right)}\\ &~~~~~-\frac{4ab}{1-b^{2}}\int_{0}^{b}\mathrm{d}t\,\frac{d}{dt}\ln{\left(\frac{t+b}{bt+1}\right)}\\ &=\sqrt{1-a^{2}}\ln{\left(\frac{1+a}{1-a}\right)}+4a\arctan{\left(\sqrt{\frac{1-a}{1+a}}\right)}\\ &~~~~~-\frac{4ab}{1-b^{2}}\left[\ln{\left(\frac{2b}{b^{2}+1}\right)}-\ln{\left(b\right)}\right]\\ &=\sqrt{1-a^{2}}\ln{\left(\frac{1+a}{1-a}\right)}+4a\arctan{\left(\sqrt{\frac{1-a}{1+a}}\right)}-\frac{4ab}{1-b^{2}}\ln{\left(\frac{2}{1+b^{2}}\right)}\\ &=\sqrt{1-a^{2}}\ln{\left(\frac{1+a}{1-a}\right)}+4a\arctan{\left(\sqrt{\frac{1-a}{1+a}}\right)}-2\left(1+a\right)\sqrt{\frac{1-a}{1+a}}\ln{\left(1+a\right)}\\ &=-\sqrt{1-a^{2}}\ln{\left(1-a^{2}\right)}+4a\arctan{\left(\sqrt{\frac{1-a}{1+a}}\right)}.\\ \end{align}$$
Finally, we consider the $n=0$ case.
Suppose $a\in(0,1)$, and set $\sqrt{\frac{1-a}{1+a}}=:b\in(0,1)$. We then find
$$\begin{align} \mathcal{J}_{0}{\left(a\right)} &=\int_{a}^{1}\mathrm{d}x\,\frac{1}{\sqrt{1-x^{2}}}\ln{\left(\frac{x+a}{x-a}\right)}\\ &=\int_{a}^{1}\mathrm{d}x\,\frac{1}{\left(1+x\right)\sqrt{\frac{1-x}{1+x}}}\ln{\left(\frac{x+a}{x-a}\right)}\\ &=\int_{\frac{1-a}{1+a}}^{0}\mathrm{d}y\,\frac{(-2)}{\left(1+y\right)^{2}}\cdot\frac{1}{\left(\frac{2}{1+y}\right)\sqrt{y}}\ln{\left(\frac{\left(\frac{1-y}{1+y}\right)+a}{\left(\frac{1-y}{1+y}\right)-a}\right)};~~~\small{\left[\frac{1-x}{1+x}=y\right]}\\ &=\int_{0}^{\frac{1-a}{1+a}}\mathrm{d}y\,\frac{1}{\left(1+y\right)\sqrt{y}}\ln{\left(\frac{(1-y)+a(1+y)}{(1-y)-a(1+y)}\right)}\\ &=\int_{0}^{\frac{1-a}{1+a}}\mathrm{d}y\,\frac{1}{\left(1+y\right)\sqrt{y}}\ln{\left(\frac{(1+a)-(1-a)y}{(1-a)-(1+a)y}\right)}\\ &=\int_{0}^{\sqrt{\frac{1-a}{1+a}}}\mathrm{d}t\,\frac{2t}{\left(1+t^{2}\right)t}\ln{\left(\frac{(1+a)-(1-a)t^{2}}{(1-a)-(1+a)t^{2}}\right)};~~~\small{\left[\sqrt{y}=t\right]}\\ &=\int_{0}^{\sqrt{\frac{1-a}{1+a}}}\mathrm{d}t\,\frac{2}{1+t^{2}}\ln{\left(\frac{1-\left(\frac{1-a}{1+a}\right)t^{2}}{\left(\frac{1-a}{1+a}\right)-t^{2}}\right)}\\ &=\int_{0}^{b}\mathrm{d}t\,\frac{2}{1+t^{2}}\ln{\left(\frac{1-b^{2}t^{2}}{b^{2}-t^{2}}\right)}.\\ \end{align}$$
Setting $2\arctan{\left(b\right)}=:\beta\in\left(0,\frac{\pi}{2}\right)$, we have $b=\tan{\left(\frac{\beta}{2}\right)}$, and so
$$\begin{align} \mathcal{J}_{0}{\left(a\right)} &=\int_{0}^{b}\mathrm{d}t\,\frac{2}{1+t^{2}}\ln{\left(\frac{1-b^{2}t^{2}}{b^{2}-t^{2}}\right)}\\ &=\int_{0}^{2\arctan{\left(b\right)}}\mathrm{d}\theta\,\ln{\left(\frac{1-b^{2}\tan^{2}{\left(\frac{\theta}{2}\right)}}{b^{2}-\tan^{2}{\left(\frac{\theta}{2}\right)}}\right)};~~~\small{\left[2\arctan{\left(t\right)}=\theta\right]}\\ &=\int_{0}^{\beta}\mathrm{d}\theta\,\ln{\left(\frac{1-\tan^{2}{\left(\frac{\beta}{2}\right)}\tan^{2}{\left(\frac{\theta}{2}\right)}}{\tan^{2}{\left(\frac{\beta}{2}\right)}-\tan^{2}{\left(\frac{\theta}{2}\right)}}\right)}\\ &=\int_{0}^{\beta}\mathrm{d}\theta\,\ln{\left(\cot{\left(\frac{\beta-\theta}{2}\right)}\cot{\left(\frac{\beta+\theta}{2}\right)}\right)}\\ &=\int_{0}^{\beta}\mathrm{d}\theta\,\ln{\left(\cot{\left(\frac{\beta-\theta}{2}\right)}\right)}+\int_{0}^{\beta}\mathrm{d}\theta\,\ln{\left(\cot{\left(\frac{\beta+\theta}{2}\right)}\right)}\\ &=\int_{0}^{\beta}\mathrm{d}\varphi\,\ln{\left(\cot{\left(\frac{\varphi}{2}\right)}\right)};~~~\small{\left[\beta-\theta=\varphi\right]}\\ &~~~~~+\int_{\beta}^{2\beta}\mathrm{d}\varphi\,\ln{\left(\cot{\left(\frac{\varphi}{2}\right)}\right)};~~~\small{\left[\beta+\theta=\varphi\right]}\\ &=\int_{0}^{2\beta}\mathrm{d}\varphi\,\ln{\left(\cot{\left(\frac{\varphi}{2}\right)}\right)}\\ &=-\int_{0}^{2\beta}\mathrm{d}\varphi\,\ln{\left(2\sin{\left(\frac{\varphi}{2}\right)}\right)}+\int_{0}^{2\beta}\mathrm{d}\varphi\,\ln{\left(2\cos{\left(\frac{\varphi}{2}\right)}\right)}\\ &=-\int_{0}^{2\beta}\mathrm{d}\varphi\,\ln{\left(2\sin{\left(\frac{\varphi}{2}\right)}\right)}+\int_{\pi-2\beta}^{\pi}\mathrm{d}\omega\,\ln{\left(2\cos{\left(\frac{\pi-\omega}{2}\right)}\right)};~~~\small{\left[\varphi=\pi-\omega\right]}\\ &=-\int_{0}^{2\beta}\mathrm{d}\varphi\,\ln{\left(2\sin{\left(\frac{\varphi}{2}\right)}\right)}+\int_{\pi-2\beta}^{\pi}\mathrm{d}\omega\,\ln{\left(2\sin{\left(\frac{\omega}{2}\right)}\right)}\\ &=-\int_{0}^{2\beta}\mathrm{d}\varphi\,\ln{\left(2\sin{\left(\frac{\varphi}{2}\right)}\right)}-\int_{0}^{\pi-2\beta}\mathrm{d}\omega\,\ln{\left(2\sin{\left(\frac{\omega}{2}\right)}\right)}\\ &~~~~~+\int_{0}^{\pi}\mathrm{d}\omega\,\ln{\left(2\sin{\left(\frac{\omega}{2}\right)}\right)}\\ &=\operatorname{Cl}_{2}{\left(2\beta\right)}+\operatorname{Cl}_{2}{\left(\pi-2\beta\right)},\\ \end{align}$$
where the Clausen function (of order $2$) is defined here by the integral representation
$$\operatorname{Cl}_{2}{\left(\theta\right)}:=-\int_{0}^{\theta}\mathrm{d}\varphi\,\ln{\left(\left|2\sin{\left(\frac{\varphi}{2}\right)}\right|\right)};~~~\small{\theta\in\mathbb{R}}.$$
And we're done!