Ta có:
$\int\dfrac{\ln(\cos x)}{1+\cos2x}dx$
$=\int\dfrac{\ln(\cos x)}{2\cos^2x}dx$
$=\dfrac{1}{2}\int\ln(\cos x)d(\tan x)$
$=\dfrac{1}{2}\tan x\ln(\cos x)-\dfrac{1}{2}\int\tan xd(\ln(\cos x))$
$=\dfrac{1}{2}\tan x\ln(\cos x)+\dfrac{1}{2}\int\dfrac{\sin^2x}{\cos^2x}dx$
$=\dfrac{1}{2}\tan x\ln(\cos x)+\dfrac{1}{2}\int\left(\dfrac{1}{\cos^2x}-1\right)dx$
$=\dfrac{1}{2}(\tan x\ln(\cos x)+\tan x-x)+C$