A) \[-\frac{1}{2}\log 2\]
B) \[\frac{1}{4}\log 2\]
C) \[\frac{1}{3}\log 2\]
D) None of these
Correct Answer: A
Solution :
\[\int_{0}^{\pi /4}{\frac{1+\tan x}{1-\tan x}dx=\int_{0}^{\pi /4}{\tan \left( \frac{\pi }{4}+x \right)\,dx}}\] \[=\left[ \log \left\{ \sec \left( \frac{\pi }{4}+x \right) \right\} \right]_{0}^{-\pi /4}=-\frac{1}{2}\log 2\].
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