How to prove :$\int^{1}_{-1} \frac{\ln(2+x\sqrt 3)}{\sqrt{1-x^2}(2+x\sqrt 3)} dx=-\pi\ln\left({\frac{3}{2}}\right)$
first let :$x=\sin(\theta)$ $$I =\int^{\frac{\pi}{2}}_{-\frac{\pi}{2}} \frac{\ln(2+\sin(\theta)\sqrt 3)}{2+\sin(\theta)\sqrt 3} dx $$
let :$y=\theta +\frac{\pi}{2}$ $$I =\int^{\pi}_{0} \frac{\ln(2-\cos(y)\sqrt 3)}{2-\cos(y)\sqrt 3} dy$$
we have : $$\ln(1+r^2-2r\cos(x)) = -2\sum_{k=1}^{\infty} \frac{r^k \cos(kx)}{k}$$ Let : $r=\frac{1}{\sqrt{3}}$ therfore: $$\ln(2-\sqrt{3}\cos(y)) = \ln\left({\frac{3}{2}}\right)-2\sum_{k=1}^{\infty} \frac{ \cos(ky)}{k(\sqrt{3})^k}$$
Thus : \begin{align*} I &= \int_{0}^{\pi} \frac{\ln(2 - \sqrt{3}\cos y)}{2 - \sqrt{3}\cos y} \, dy \\ &= \ln\left(\frac{3}{2}\right) \int_{0}^{\pi} \frac{1}{2 - \sqrt{3}\cos y} \, dy - 2 \sum_{k=1}^{\infty} \frac{1}{k\sqrt{3}^k} \int_{0}^{\pi} \frac{\cos(ky)}{2 - \sqrt{3}\cos y} \, dy \\ &= \pi \ln\left(\frac{3}{2}\right) - 2 \sum_{k=1}^{\infty} \frac{1}{k\sqrt{3}^k} \cdot \frac{\pi}{\sqrt{3}^k} \\ &= \pi \ln\left(\frac{3}{2}\right) - 2\pi \sum_{k=1}^{\infty} \frac{1}{k} \left(\frac{1}{3}\right)^k \\ &= \pi \ln\left(\frac{3}{2}\right) + 2\pi \ln\left(\frac{2}{3}\right) \\ &= -\pi \ln\left(\frac{3}{2}\right) \end{align*}
we have : $$\int^{\pi}_{0} \frac{1}{2-\cos(y)\sqrt 3} dy =\pi $$
and $$\int^{\pi}_{0} \frac{cos(kx)}{2-\cos(y)\sqrt 3} dy = \frac{(-1)^k}{2}\frac{cos(kx)}{1+\cos(y)\frac{\sqrt{3}}{2}} dy$$
using identity :
$$\int^{\pi}_{0} \frac{\cos(kx)}{1+a\cos(x)} dy = \frac{\pi}{\sqrt{1-a^2}}\left({\frac{\sqrt{1-a^2}-1}{a}}\right)$$
therfore: $a=\frac{\sqrt{3}}{2}$ $$\int^{\pi}_{0} \frac{\cos(kx)}{2-\cos(y)\sqrt 3} dy = \frac{\pi}{(\sqrt{3})^k}$$
any other solution this integral ?
