$\int_{1}^{\sqrt{3}} \frac{d x}{1 + x^{2}}$ equals

\frac{π}{3}

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\frac{2 π}{3}

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\frac{π}{6}

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\frac{π}{12}

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Solution

The correct option is C $\frac{π}{12}$
$\int \frac{d}{1 + x^{2}} = {tan}^{- 1} x = F (x)$
By second fundamental theorem of calculus, we obtain
$\int_{1}^{\sqrt{3}} \frac{d x}{1 + x^{2}} = F (\sqrt{3}) - F (1)$
$= {tan}^{- 1} \sqrt{3} - {tan}^{- 1} 1$
$= \frac{π}{3} - \frac{π}{4} = \frac{π}{12}$

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