$$I=\int_0^{\pi/4} \frac{x}{(\sin x+\cos x)\cos x}\,dx=\int_0^{\pi/4} \frac{x}{\sqrt{2}\cos \left(\frac{\pi}{4}-x\right)\cos x}\,dx\,\,(*)$$
Also,
$$I=\int_0^{\pi/4} \frac{\frac{\pi}{4}-x}{\sqrt{2}\cos \left(\frac{\pi}{4}-x\right)\cos x}\,dx\,\,\,(**)$$
Add $(*)$ and $(**)$ to get:
$$2I=\frac{\pi}{4\sqrt{2}}\int_0^{\pi/4} \frac{dx}{\cos \left(\frac{\pi}{4}-x\right)\cos x}$$
$$\Rightarrow I=\frac{\pi}{8}\int_0^{\pi/4} \frac{\sin\left(\frac{\pi}{4}-x+x\right)}{\cos \left(\frac{\pi}{4}-x\right)\cos x}\,dx$$
Since
$$\sin\left(\frac{\pi}{4}-x+x\right)=\sin\left(\frac{\pi}{4}-x\right)\cos x+\cos \left(\frac{\pi}{4}-x\right)\sin x$$
Hence,
$$I=\frac{\pi}{8}\int_0^{\pi/4} \left(\tan\left(\frac{\pi}{4}-x\right)+\tan x\right)\,dx$$
$$\Rightarrow I=\frac{\pi}{4}\int_0^{\pi/4}\tan x\,dx$$
It can be shown that:
$$\int_0^{\pi/4} \tan x\,dx=\left(\ln(\sec x)\right|_0^{\pi/4}=\frac{1}{2}\ln 2$$
$$\Rightarrow \boxed{I=\dfrac{\pi}{8}\ln 2}$$
:)