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Question

$\int_{0}^{1} \frac{x^{2}}{1 + x^{2}} d x$ equals

$1 - \frac{π}{4}$
$1 - \frac{π}{3}$
$\frac{π}{4}$
$\frac{π}{3}$

A

1 - \frac{π}{3}

B

\frac{π}{3}

C

\frac{π}{4}

D

1 - \frac{π}{4}

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Solution

Verified by Toppr

$\int_{0}^{1} \frac{x^{2} d x}{1 + x^{2}}$
$\int_{0}^{1} 1 d x - \int_{0}^{1} \frac{1}{1 + x^{2}} d x$
$= 1 - [{tan}^{- 1} x] |_{0}^{1}$
$= 1 - [{tan}^{- 1} 1]$
$= 1 - \frac{π}{4}$
Thus $\int_{0}^{1} \frac{x^{2}}{1 + x^{2}} d x = 1 - \frac{π}{4}$

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