Evaluating $\int\limits_0^{\pi/4}\log(1+\tan x)\,\mathrm dx$

$$I=\int^{\frac{\pi }{4} }_{0}\ln\left( 1+\tan x \right)\ dx=\int ^{\frac{\pi }{4} }_{0}\ln\left( 1+\tan \left( \frac{\pi }{4} -x\right) \right)\ dx=\int^{\frac{\pi }{4} }_{0}\ln\left( 2\right)\ dx-I$$ So we have $$2I=\ln\left( 2\right) \int ^{\frac{\pi }{4} }_{0}\ dx=\ln\left( 2\right) \times \frac{\pi }{4} $$ Hence we get$$ I= \boxed{\frac{\pi }{8} \ln\left( 2\right)} $$

Hint. By the change of variable $$ x=\frac{\pi}4-u, \qquad dx=-du, \qquad 1+ \tan x=? \qquad \log(1+ \tan x)=? $$