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If $I_{1} = \int_{x}^{1} \frac{d t}{1 + t^{2}} and I_{2} = \int_{1}^{1 / x} \frac{d t}{1 + t^{2}} for x > 0$ then

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I1>I2

I1=I2

I2>I1

I2=(π/2)−tan−1⁡x

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detailed solution

Correct option is B

In I2, put t=1/u to obtainI2=∫1x −1/u2du1+1/u2=∫x1 dt1+t2=I1

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If I1=∫x1 dt1+t2 and I2=∫11/x dt1+t2 for x>0 then

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If $I_{1} = \int_{x}^{1} \frac{d t}{1 + t^{2}} and I_{2} = \int_{1}^{1 / x} \frac{d t}{1 + t^{2}} for x > 0$ then