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: [26-Jun-2022-Shift-2]

Correct Answer is : loge(√3−1√3+1)+π3

\log _e( \sqrt{3}-1}/\sqrt{3}+1})+\frac{\pi}{3}

\log _e( \sqrt{3}+1}/\sqrt{3}-1})+\frac{\pi}{3}

\log _e( \sqrt{3}+1}/\sqrt{3}-1})-\frac{\pi}{3}

\frac{1}{2} \log _e( \sqrt{3}-1}/\sqrt{3}+1})-\frac{\pi}{6}

Indefinite Integrals

Section: Mathematics

Question : 35 of 89

Marks: +1, -0

If ∫‌

√‌

1−x

1+x

dx=g(x)+c,g(1)=0, then g(‌

) is equal to:

[26-Jun-2022-Shift-2]

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