#### Answer

$$\ln(\sqrt{2}+1)$$

#### Work Step by Step

Since
\begin{aligned}
s &=\int_{0}^{\frac{\pi}{4}} \sqrt{1+f^{\prime}(x)^{2}}dx \\
=& \int_{0}^{\frac{\pi}{4}} \sqrt{1+(-\tan x)^{2}}dx \\
&=\int_{0} \frac{\pi}{4} \sqrt{\sec ^{2} x}dx \\
&=\int_{0}^{\frac{\pi}{4}} \sec x dx\\
&=\ln |\sec x+\tan x|\bigg|_{0}^{\frac{\pi}{4}} \\
&=\ln(\sqrt{2}+1)
\end{aligned}