To evaluate the integral \( I = \int_{0}^{\pi/4} \log(1 + \tan x) \, dx \), we can proceed with the following steps: 1. Given Integral: \[ I = \int_{0}^{\pi/4} \log(1 + \tan x) \, dx \quad \text{(i)} \] 2. Use Substitution: Let \( x = \frac{\pi}{4} - t \). Then, \( dx = -dt \), and when \( x = 0 \), \( t = \frac{\pi}{4} \), and when \( x = \frac{\pi}{4} \), \( t = 0 \). The integral becomes: \[ I = \int_{\pi/4}^{0} \log\left(1 + \tan\left(\frac{\pi}{4} - t\right)\right) (-dt) \] \[ I = \int_{0}^{\pi/4} \log\left(1 + \tan\left(\frac{\pi}{4} - t\right)\right) dt \] 3. Simplify the Integrand: Using the identity \( \tan\left(\frac{\pi}{4} - t\right) = \frac{1 - \tan t}{1 + \tan t} \): \[ I = \int_{0}^{\pi/4} \log\left(1 + \frac{1 - \tan t}{1 + \tan t}\right) dt \] \[ I = \int_{0}^{\pi/4} \log\left(\frac{2}{1 + \tan t}\right) dt \] \[ I = \int_{0}^{\pi/4} \log 2 \, dt - \int_{0}^{\pi/4} \log(1 + \tan t) \, dt \] \[ I = \log 2 \cdot \left[ t \right]_{0}^{\pi/4} - I \] \[ I = \frac{\pi}{4} \log 2 - I \] 4. Solve for \( I \): \[ 2I = \frac{\pi}{4} \log 2 \] \[ I = \frac{\pi}{8} \log 2 \] Therefore, the value of the integral is: \[ \boxed{\frac{\pi}{8} \log 2} \]