The value of int_0^{pi /4} frac{1 + tan x}{1 | vedclass

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(D) We know that $\frac{1 + 	an x}{1 - 	an x} = 	an(\frac{\pi}{4} + x)$.Therefore,the integral becomes $I = \int_0^{\pi/4} 	an(\frac{\pi}{4} + x)

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(D) We know that $\frac{1 + 	an x}{1 - 	an x} = 	an(\frac{\pi}{4} + x)$.Therefore,the integral becomes $I = \int_0^{\pi/4} 	an(\frac{\pi}{4} + x) \, dx$.Using the formula $\int 	an(ax+b) \, dx = \frac{1}{a} \ln|\sec(ax+b)| + C$,we get:$I = [\ln|\sec(\frac{\pi}{4} + x)|]_0^{\pi/4}$.Evaluating at the limits:$I = \ln|\sec(\frac{\pi}{4} + \frac{\pi}{4})| - \ln|\sec(\frac{\pi}{4} + 0)|$.$I = \ln|\sec(\frac{\pi}{2})| - \ln|\sec(\frac{\pi}{4})|$.Since $\sec(\frac{\pi}{2})$ is undefined,let us re-evaluate using $\frac{1 + 	an x}{1 - 	an x} = 	an(\frac{\pi}{4} + x)$.Actually,the integral $\int_0^{\pi/4} 	an(\frac{\pi}{4} + x) \, dx = [\ln|\sec(\frac{\pi}{4} + x)|]_0^{\pi/4} = \ln(\sec \frac{\pi}{2}) - \ln(\sec \frac{\pi}{4})$.Since $\sec(\frac{\pi}{2}) 	o \infty$,the integral diverges. Thus,the correct option is $D$.

The value of $\int_0^{\pi /4} \frac{1 + \tan x}{1 - \tan x} \, dx$ is

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