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प्रश्न
Evaluate: `int_0^1 1/(1 + x^2) "d"x`
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उत्तर
`int_0^1 1/(1 + x^2) "d"x = [tan^-1 x]_0^1`
= tan–11 – tan–10
= `pi/4 - 0`
= `pi/4`
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