Đặt $x=\dfrac{\pi}{4}-t\Rightarrow dx=-dt$
$I=-\int_{\frac{\pi}{4}}^0\ln(1+\tan(\dfrac{\pi}{4}-t)dt$
$=\int_0^\frac{\pi}{4}\ln(1+\dfrac{1-\tan t}{1+\tan t})dt$
$=\int_0^\frac{\pi}{4}\ln(\dfrac{2}{1+\tan t})dt$
$=\int_0^\frac{\pi}{4}\left[\ln2-\ln(1+\tan t)\right]dt$
$=\int_0^\frac{\pi}{4}\ln2dt-\int_0^\frac{\pi}{4}\ln(1+\tan t)dt$
$=\ln2t\bigg|_0^\dfrac{\pi}{4}-\int_0^\frac{\pi}{4}\ln(1+\tan x)dx$
$=\dfrac{\pi}{4}\ln 2-I$
$\Rightarrow 2I=\dfrac{\pi}{4}\ln2 \Rightarrow I=\dfrac{\pi}{8}\ln2$