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Answer
\displaystyle [br]\begin{align*}\int_0^1 \frac{\log \frac{3+x}{3-x}}{\sqrt{x(1-x)}} \, \mathrm{d}x &=\int_0^{\pi/2} \log\left(\frac{3 + \sin^2 x }{3-\sin^2 x}\right)\frac{2\sin x\cos x \, \mathrm{d}x}{\sqrt{\sin^2 x(1-\sin^2 x)}} \\ &= 2 \int_0^{\pi/2}\log\left(\frac{3 + \sin^2 x}{3-\sin^2 x}\right) \, \mathrm{d}x \end{align*}
\displaystyle [br]\begin{align*}\int_0^1 \frac{\log \frac{3+x}{3-x}}{\sqrt{x(1-x)}} \, \mathrm{d}x &=\int_0^{\pi/2} \log\left(\frac{3 + \sin^2 x }{3-\sin^2 x}\right)\frac{2\sin x\cos x \, \mathrm{d}x}{\sqrt{\sin^2 x(1-\sin^2 x)}} \\ &= 2 \int_0^{\pi/2}\log\left(\frac{3 + \sin^2 x}{3-\sin^2 x}\right) \, \mathrm{d}x \end{align*}
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