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GATE Overflow Test Series | Calculus | Test 1 | Question: 30

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$I =\int_{ -\infty}^{\infty}\frac{dx}{1+x^2}$ = $[\tan^{-1} x]_{-\infty}^{\infty}$

$\quad = \tan^{-1}(\infty) - \tan ^{-1} (-\infty) = \tan^{-1} \left(\tan \frac{\pi}{2}\right) –  \tan^{-1} \left(\tan \left(\frac{-\pi}{2}\right)\right) = \frac{\pi}{2} + \frac{\pi}{2} = \pi.$
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