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प्रश्न
Evaluate : `int _0^(pi/4) 1/(1 + "x"^2) "dx"`
उत्तर
Let I = `int _0^(pi/4) 1/(1 + "x"^2) "dx"`
`= ["tan"^(-1) "x"]_0^(pi/4)`
`= "tan"^(-1) (pi/4) - "tan"^(-1) (0)`
= 1 - 0
= 1
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