MCQ
$\int\limits^\sqrt{3}_1\frac{1}{1+\text{x}^2}\text{ dx}$ is equal to:
  • $\frac{\pi}{12}$
  • B
    $\frac{\pi}{6}$
  • C
    $\frac{\pi}{4}$
  • D
    $\frac{\pi}{3}$

Answer

Correct option: A.
$\frac{\pi}{12}$
$\int\limits^\sqrt{3}_1\frac{1}{1+\text{x}^2}\text{dx}$
$=\big[\tan^{-1}\text{x}\big]^\sqrt{3}_1$
$=\frac{\pi}{3}-\frac{\pi}{4}$
$=\frac{\pi}{12}$

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