If ∫ (1)/(x) (1 - x)/(1 + x) dx = g (x) + c, g (1) = 0 , then...

Given,

$∫ \dfrac{1}{x}\sqrt{\dfrac{1 - x}{1 + x}}dx = g\left(x\right) + c$

Put $x = \cos 2\theta$

$dx = - 2\sin 2\theta \cdot d\theta$

$= ∫ \dfrac{1}{\cos 2\theta}tan\theta\left(- 4sin\theta \cdot cos\theta\right)d\theta$

$= ∫ \dfrac{1}{\cos 2\theta}\left(- 4\sin^{2}\theta\right)d\theta$

$= - 2 ∫ \dfrac{1 - \cos 2\theta}{\cos 2\theta}d\theta$

$= - \dfrac{2}{2}\ln \left|\sec 2\theta - \tan 2\theta\right| + 2\theta + c$

$= \ln \left|\sec 2\theta - \tan 2\theta\right| + 2\theta + c$

$= \ln \left|\dfrac{1 - \sin 2\theta}{\cos 2\theta}\right| + \cos^{- 1}x + c$

$\mathop{\mathop{= \ln \left|\dfrac{1 - \sqrt{1 - {x}^{2}}}{x}\right| + \cos^{- 1}x + c}_{⏟}}_{g\left(x\right)}$

$∴ g\left(1\right) = 0 ⇒ c = 0$

So, $g\left(x\right) = \ln \left|\dfrac{1 - \sqrt{1 - {x}^{2}}}{x}\right| + \cos^{- 1}x$

$g\left(\dfrac{1}{2}\right) = \ln \left|2 - \sqrt{3}\right| + \dfrac{\pi}{3}$

$g\left(\dfrac{1}{2}\right) = \ln \left|\dfrac{\sqrt{3} - 1}{\sqrt{3} + 1}\right| + \dfrac{\pi}{3}$