Đặt $x=\dfrac{\pi}{4}-t \Rightarrow dx=-dt$
$I=\int \limits_0^{\frac{\pi}{4}} \ln \bigg [1+ \tan (\dfrac{\pi}{4}-t) \bigg ]dt=\int \ln (1 +\dfrac{1-\tan t}{1+\tan t})dt$
$\int \ln ( \dfrac{2}{1+\tan t})dt=\int \ln 2 dt-\int \ln (1+\tan t)dt=t\ln 2 \bigg |_0^{\frac{\pi}{4}} - I$
$\Rightarrow 2I= \dfrac{\pi}{4} \ln 2 \Rightarrow I=\dfrac{\pi}{8} \ln 2$