A) \[-\frac{1}{2}\]
B) 0
C) 1
D) \[2\log \frac{1}{2}\]
Correct Answer: A
Solution :
\[I=\int_{-1/2}^{1/2}{[x]dx+\int_{-1/2}^{1/2}{\log \left( \frac{1+x}{1-x} \right)dx}}\] If \[f(x)=\log \left( \frac{1+x}{1-x} \right)\], then \[f(-x)=\log \left( \frac{1-x}{1+x} \right)=-\log \left( \frac{1+x}{1-x} \right)=-f(x)\] \[\therefore I=\int_{-1/2}^{1/2}{\text{ }[x]}\text{ }dx+\text{0}\] (being integral of odd function) \[=\int_{-1/2}^{0}{-1dx+\int_{0}^{1/2}{\,\,\,\,0\,dx}}=-(x)_{-1/2}^{0}=\frac{-1}{2}.\]
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