Applying the trigonometry transformation of $$a\sin{\alpha}+b\cos{\alpha}=\sqrt{a^2+b^2}\sin{\alpha+\varphi}$$

where $\varphi$ satisfying $\tan{\varphi}=\frac{b}{a}$ gives

$$\int_0^{\frac{\pi}{4}}\frac{1}{\sin{x}+\cos{x}}d{x}=\int_0^{\frac{\pi}{4}}\frac{1}{\sqrt{2}\sin(x+\frac{\pi}{4})}d{x}$$

Letting $u=x+\frac{\pi}{4}$ and applying the result of # 4604 yield (using the result from the last solution is more convenient here)

$$\frac{1}{\sqrt{2}}\int_{-\frac{\pi}{4}}^{0}\frac{1}{\sin{u}}d{u}=\frac{1}{\sqrt{2}}\left.\left(-\ln\left|{\csc{u}+\cot{u}}\right|\right)\right|_{-\frac{\pi}{4}}^0=\boxed{\frac{\sqrt{2}}{2}\ln(\sqrt{2}+1)}$$