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Evaluate : \(\int\limits_0^1\frac{dx}{1+x^2}\) |
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Answer» \(\int\limits_0^1\frac{dx}{1+x^2}\) = \([tan^{-1}x]^1_0\) = tan-11 - tan-10 = \(\frac{\pi}{4}-0\) = \(\frac{\pi}{4}\) |
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