The correct option is B π/2 =∫_ 1^1dx1+x^2 #10; ∫dx1+x^2=tan^ 1x+c #10; ∫_ 1^1dx1+x^2=(tan^ 1x+c)∫_ 1^1 #10; =tan^ 11 tan^ 1( 1) ...

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Integration Using Substitution

∫-11dx/1+x2=

Question

$\int_{- 1}^{1} \frac{d x}{1 + x^{2}} =$

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\frac{π}{2}

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\frac{π}{4}

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\frac{π}{6}

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Solution

The correct option is B $\frac{π}{2}$
$= \int_{- 1}^{1} \frac{d x}{1 + x^{2}}$
$\int \frac{d x}{1 + x^{2}} = t a n^{- 1} x + c$
$\int_{- 1}^{1} \frac{d x}{1 + x^{2}} = (t a n^{- 1} x + c) \int_{- 1}^{1}$
$= t a n^{- 1} 1 - t a n^{- 1} (- 1)$
$= \frac{π}{4} - (\frac{- π}{4})$
$= \frac{π}{2}$
So,
$\int_{- 1}^{1} \frac{d x}{1 + x^{2}} = \frac{π}{2}$

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