Values of $\int\limits_{-1 / 2}^{+1 / 2} \cos x \log \frac{1+x}{1-x} d x$ is :

0

$I=\int\limits_{-1 / 2}^{1 / 2} \cos x \log \frac{1+x}{1-x} dx$

$f(x)=\cos x \ln \frac{1+x}{1-x}$

$f(-x)=\cos(-x) \ln \frac{1-x}{1+x}$

$=-\cos (x) \ln \left(\frac{1+x}{1-x}\right)=-f(x)$

f(x) is an odd function    [Using $\int\limits_{-a}^a f(x) d x=0$  if f(x) = f(x)]

hence I = 0

Hence (3) is the correct answer.