Answer 1

\[\int_\frac{- 1}{2}^\frac{1}{2} \cos x \log\left( \frac{1 + x}{1 - x} \right) d x\]

\[\text{Let }f(x) = \cos x \log\left( \frac{1 + x}{1 - x} \right)\]

\[\text{Consider }f(- x) = \cos\left( - x \right) \log\left( \frac{1 - x}{1 + x} \right)\]

\[ = \cos x\left\{ - \log\left( \frac{1 + x}{1 - x} \right) \right\} = - \cos x \log\left( \frac{1 + x}{1 - x} \right) = - f\left( x \right)\]

Thus f(x) is an odd function

Therefore,

\[ \int_\frac{- 1}{2}^\frac{1}{2} \cos x \log\left( \frac{1 + x}{1 - x} \right) d x = 0\]