MCQ
$\int_{1}^{\sqrt{3}} \frac{d x}{1+x^{2}}$ बराबर है:
  • A
    $\frac{\pi}{12}$
  • B
    $\frac{2\pi}{3}$
  • C
    $\frac{\pi}{6}$
  • D
    $\frac{\pi}{3}$

Answer

$\int_{1}^{\sqrt{3}} \frac{1}{1+x^{2}} d x$ $=\int_{1}^{\sqrt{3}} \frac{1}{x^{2}+1^{2}} d x$
$=\left[\frac{1}{1} \tan ^{-1}\left(\frac{x}{1}\right)\right]_{1}^{\sqrt{3}}$  = $\tan ^{-1} \sqrt{3}$  - $\tan ^{-1} 1\left(\because \int \frac{d x}{1+x^{2}}=\tan ^{-1} x\right)$
$=\frac{\pi}{3}-\frac{\pi}{4} \Rightarrow \frac{4 \pi-3 \pi}{12}=\frac{\pi}{12}$

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