Evaluate: ∫₀^(π/4) log(1 + tanθ) dθ

Let $I = \int_0^{\pi/4}\log(1+\tan\theta)\,d\theta$. By the property $\int_0^a f(\theta)d\theta = \int_0^a f(a-\theta)d\theta$: $I = \int_0^{\pi/4}\log\left(1+\tan\left(\frac{\pi}{4}-\theta\right)\right)d\theta = \int_0^{\pi/4}\log\left(\frac{2}{1+\tan\theta}\right)d\theta$ $I = \int_0^{\pi/4}\log 2\,d\theta - I$ $2I = \frac{\pi}{4}\log 2$ Hence $I = \frac{\pi}{8}\log 2$.