Discussion
The value of the integral \(\int_{-1/3}^{1/3} \sec(x)\,\ln\left|\frac{1+x}{1-x}\right| dx\) is:
Answer: D
The integrand \(f(x) = \sec x \cdot \ln\left|\frac{1+x}{1-x}\right|\) is an odd function. The factor sec x is even, since cos(-x) = cos x. The logarithm is odd, because replacing x by -x gives \(\ln\left|\frac{1-x}{1+x}\right| = -\ln\left|\frac{1+x}{1-x}\right|\). An even function times an odd function is odd, and the interval -1/3 to 1/3 is symmetric about zero, so the integral is 0. The other values would require a non-zero net area, which is impossible for an odd integrand.
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