$$
\begin{array}{l}
\text { Given that I }=\int_{0}^{\frac{\pi}{4}} \log \left(\frac{\sin x+\cos x}{\cos x}\right) d x \\
=\int_{0}^{\frac{\pi}{4}} \log (1+\tan x) d x
\end{array}
$$
As we know that
$$
\begin{array}{l}
\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x \\
\text { Now, } I=\int_{0}^{\frac{\pi}{4}} \log \left(1+\tan \left(\frac{\Pi}{4}-x\right)\right) d x \\
=\int_{0}^{\frac{\pi}{4}} \log \left(1+\frac{1-\tan x}{1+\tan x}\right) d x \\
=\int_{0}^{\frac{\pi}{4}} \log \left(\frac{2}{1+\tan x}\right) d x \\
=\int_{0}^{\frac{\pi}{4}} \log 2 d x-\int_{0}^{\frac{\pi}{4}} \log (1+\tan x) d x \\
\Rightarrow I=(\log 2) \int_{0}^{\frac{\pi}{4}} d x-I \\
\Rightarrow 2 I=(\log 2) \frac{\Pi}{4} \Rightarrow I=\frac{\Pi}{8}(\log 2)
\end{array}
$$