The value of \(\int_{-\frac{1}{2}}^{\frac{1}{2}}\displaystyle \log_e\left(\frac{1-x}{1+x}\right)dx\) is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 0
Let
\( \displaystyle f(x) = \log_e\left(\frac{1-x}{1+x}\right) \)
Replacing x by −x:
\( \displaystyle f(-x) = \log_e\left(\frac{1+x}{1-x}\right) \)
\( \displaystyle = -\log_e\left(\frac{1-x}{1+x}\right) = -f(x) \)
So f is an odd function. For an odd function on an interval symmetric about 0,
\( \displaystyle \int_{-a}^{a} f(x)\,dx = 0 \)
Hence the value of the integral is 0.