If $\int \frac{1}{x} \sqrt{\frac{1 - x}{1 + x}} d x = g (x) + c, g (1) = 0$ then g $(\frac{1}{2})$ is equal to :

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$\log_{e} (\frac{\sqrt{3} - 1}{\sqrt{3} + 1}) + \frac{π}{3}$

$\log_{e} (\frac{\sqrt{3} + 1}{\sqrt{3} - 1}) + \frac{π}{3}$

$\log_{e} (\frac{\sqrt{3} + 1}{\sqrt{3} - 1}) - \frac{π}{3}$

$\frac{1}{2} \log_{e} (\frac{\sqrt{3} - 1}{\sqrt{3} + 1}) - \frac{π}{6}$

answer is A.

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Detailed Solution

$\int \frac{1}{x} \sqrt{\frac{1 - x}{1 + x}} d x = g (x) + c P u t x = \cos 2 θ d x = - 2 \sin 2 θ \cdot d θ = \int \frac{1}{\cos 2 θ} \tan θ (- 4 \sin θ \cdot \cos θ) d θ = \int \frac{1}{\cos 2 θ} (- 4 \sin^{2} θ) d θ = - 2 \int \frac{1 - \cos 2 θ}{\cos 2 θ} d θ = - \ln | \sec 2 θ + \tan 2 θ | + 2 θ + c = \ln ∣ \sec 2 θ - \tan 2 θ | + 2 θ + c = \ln |\frac{1 - \sin 2 θ}{\cos 2 θ}| + \cos^{- 1} x + c = \ln |\frac{1 - \sqrt{1 - x^{2}}}{x}| + \cos^{- 1} x (∵ g (1) = 0) g (x) = \ln |\frac{1 - \sqrt{1 - x^{2}}}{x}| + \cos^{- 1} x g (\frac{1}{2}) = \ln | 2 - \sqrt{3} | + \frac{π}{3} g (\frac{1}{2}) = \ln |\frac{\sqrt{3} - 1}{\sqrt{3} + 1}| + \frac{π}{3}$