Q87 / 150 · MHT CET 2022 (10 Aug Shift 2) · Mathematics

$\int_{-1 / 2}^{1 / 2} \log \left(\frac{1+x}{1-x}\right) \mathrm{d} x=$

  • Correct answer
    A
    $0$
  • B
    $\frac{1}{2}$
  • C
    $-1$
  • D
    $-\frac{1}{2}$

Solution

$\int_{\frac{-1}{2}}^{1 / 2} \log \left(\frac{1+x}{1-x}\right) \mathrm{d} x=0\left[\because \int_{-a}^a f(x) \mathrm{d} x=0\right.$ if $f(x)$ is odd $]$

Hence, the correct answer is option: 'A'.

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