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If $I_{1} = \int_{1 / e}^{\tan x} \frac{t}{1 + t^{2}} d t$ and $I_{2} = \int_{1 / e}^{\cot x} \frac{d t}{t (1 + t^{2})}$ then the value of $I_{1} + I_{2}$ is

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detailed solution

Correct option is B

Putting t=1/u in I2 we haveI2=-∫etanxu du1+u2=-∫1/etanxu du1+u2+∫1/eeu du1+u2=-I1+12∫1/ee2u du1+u2so I1+I2=12logu2+11/ee=12log⁡e2+1−log⁡e2+1e2=12×2=1

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If I1=∫1/etanxt1+t2 dt and I2=∫1/ecotxdtt1+t2 then the value of I1+I2 is

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If $I_{1} = \int_{1 / e}^{\tan x} \frac{t}{1 + t^{2}} d t$ and $I_{2} = \int_{1 / e}^{\cot x} \frac{d t}{t (1 + t^{2})}$ then the value of $I_{1} + I_{2}$ is