The value of int_{1/e}^{tan x} frac{t , dt}{1 | vedclass

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(B) Let $I = \int_{1/e}^{	an x} \frac{t \, dt}{1 + t^2} + \int_{1/e}^{\cot x} \frac{dt}{t(1 + t^2)}$.For the first integral,let $u = 1 + t^2$,then $d

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$1$

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(B) Let $I = \int_{1/e}^{	an x} \frac{t \, dt}{1 + t^2} + \int_{1/e}^{\cot x} \frac{dt}{t(1 + t^2)}$.For the first integral,let $u = 1 + t^2$,then $du = 2t \, dt$,so $\int \frac{t \, dt}{1 + t^2} = \frac{1}{2} \ln(1 + t^2)$.Evaluating from $1/e$ to $	an x$: $\frac{1}{2} [\ln(1 + 	an^2 x) - \ln(1 + 1/e^2)] = \frac{1}{2} [\ln(\sec^2 x) - \ln(1 + 1/e^2)]$.For the second integral,$\int \frac{dt}{t(1 + t^2)} = \int (\frac{1}{t} - \frac{t}{1 + t^2}) \, dt = \ln|t| - \frac{1}{2} \ln(1 + t^2)$.Evaluating from $1/e$ to $\cot x$: $[\ln(\cot x) - \frac{1}{2} \ln(1 + \cot^2 x)] - [\ln(1/e) - \frac{1}{2} \ln(1 + 1/e^2)]$.Since $1 + \cot^2 x = \csc^2 x$,this becomes $[\ln(\cot x) - \frac{1}{2} \ln(\csc^2 x)] - [\ln(1/e) - \frac{1}{2} \ln(1 + 1/e^2)]$.Combining both parts:$I = \frac{1}{2} \ln(\sec^2 x) - \frac{1}{2} \ln(1 + 1/e^2) + \ln(\cot x) - \frac{1}{2} \ln(\csc^2 x) - \ln(1/e) + \frac{1}{2} \ln(1 + 1/e^2)$.$I = \ln(\sec x) + \ln(\cot x) - \ln(\csc x) - \ln(1/e)$.$I = \ln(\frac{\sec x \cdot \cot x}{\csc x}) - \ln(1/e) = \ln(1) - \ln(1/e) = 0 - (-1) = 1$.

The value of $\int_{1/e}^{\tan x} \frac{t \, dt}{1 + t^2} + \int_{1/e}^{\cot x} \frac{dt}{t(1 + t^2)} = $

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