Question

\(\displaystyle \int_{-1}^{1}\log\!\left(\frac{3+x}{\,3-x\,}\right)dx=\ \ ?\)

• A. \(0\)
• B. \(1\)
• C. \(2\log 3\)
• D. \(3\log 2\)

Hint:

If $f(-x)=-f(x)$, then $\int_{-a}^{a} f(x)\,dx=0$.

Answer 1

The Correct Option is A

Let \(f(x)=\ln\!\left(\dfrac{3+x}{3-x}\right)\). Check symmetry: \[ f(-x)=\ln\!\left(\frac{3-x}{3+x}\right)=-\ln\!\left(\frac{3+x}{3-x}\right)=-f(x). \] So \(f\) is an odd function. The integral of an odd function over a symmetric interval \([-a,a]\) is \(0\). Hence the value is \(0\).