$I=\int\limits_{0}^{\frac{\pi}{4}}\dfrac{x+\sin2x}{1+\cos2x}dx=\int\limits_{0}^{\frac{\pi}{4}}\dfrac{x+\sin2x}{2\cos^2x}dx=\int\limits_{0}^{\frac{\pi}{4}}\left ( \dfrac{x}{2\cos^2x}+\dfrac{\sin2x}{2\cos^2x} \right )dx$
$=\int\limits_{0}^{\frac{\pi}{4}}\ \dfrac{x}{2\cos^2x}dx+\int\limits_{0}^{\frac{\pi}{4}}\dfrac{\sin2x}{2\cos^2x} dx$
$=\frac12\int\limits_{0}^{\frac{\pi}{4}}xd(\tan x) - \frac12\int\limits_{0}^{\frac{\pi}{4}}\frac{d(\cos^2x)}{\cos^2x}$
$=\frac12\left[ {x\tan x} \right]_{0}^{\frac{\pi}{4}}-\frac12\int\limits_{0}^{\frac{\pi}{4}}\tan xdx - \frac12\left[ {\ln\left| {\cos^2x} \right|} \right]_{0}^{\frac{\pi}{4}}$.
$=\frac12\left[ {x\tan x} \right]_{0}^{\frac{\pi}{4}}-\frac12\int\limits_{0}^{\frac{\pi}{4}}\frac{\sin x}{\cos x}dx - \frac12\left[ {\ln\left| {\cos^2x} \right|} \right]_{0}^{\frac{\pi}{4}}$.
$=\frac12\left[ {x\tan x} \right]_{0}^{\frac{\pi}{4}}+\frac12\int\limits_{0}^{\frac{\pi}{4}}\frac{d(\cos x)}{\cos x}dx - \frac12\left[ {\ln\left| {\cos^2x} \right|} \right]_{0}^{\frac{\pi}{4}}$.
$=\frac12\left[ {x\tan x} \right]_{0}^{\frac{\pi}{4}}+\frac12\left[ {\ln\left| {\cos x} \right|} \right]_{0}^{\frac{\pi}{4}} - \frac12\left[ {\ln\left| {\cos^2x} \right|} \right]_{0}^{\frac{\pi}{4}}$.