The value of the integral $\int\limits_{-1 / 2}^{1 / 2} \cos x \log \left(\frac{1+x}{1-x}\right) d x$, is

0 $\frac{1}{2}$ $-\frac{1}{2}$ none of these

0

Let $f(x)=\cos x \log \left(\frac{1+x}{1-x}\right)$. Then, $f(-x)=\cos (-x) \log \left(\frac{1-x}{1+x}\right)=-\cos x \log \left(\frac{1+x}{1-x}\right)=-f(x)$ $\Rightarrow f(x)$ is an odd function ∴  $\int\limits_{-1/2}^{1/2} \cos x \log (\frac{1+x}{1-x})dx = 0$