1. **Problem statement:** Find the length of the curve defined by $$y = \int_0^x \tan t \, dt$$ for $$0 \leq x \leq \frac{\pi}{6}$$. 2. **Formula for arc length:** The length $$L$$ of a curve $$y=f(x)$$ from $$x=a$$ to $$x=b$$ is given by $$ L = \int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx $$ 3. **Find the derivative:** Since $$y = \int_0^x \tan t \, dt$$, by the Fundamental Theorem of Calculus, $$ \frac{dy}{dx} = \tan x $$ 4. **Substitute into arc length formula:** $$ L = \int_0^{\frac{\pi}{6}} \sqrt{1 + (\tan x)^2} \, dx $$ Recall the trigonometric identity: $$ 1 + \tan^2 x = \sec^2 x $$ So, $$ L = \int_0^{\frac{\pi}{6}} \sqrt{\sec^2 x} \, dx = \int_0^{\frac{\pi}{6}} |\sec x| \, dx $$ Since $$0 \leq x \leq \frac{\pi}{6}$$, $$\sec x > 0$$, so $$ L = \int_0^{\frac{\pi}{6}} \sec x \, dx $$ 5. **Evaluate the integral:** The integral of $$\sec x$$ is $$ \int \sec x \, dx = \ln |\sec x + \tan x| + C $$ Therefore, $$ L = \left[ \ln |\sec x + \tan x| \right]_0^{\frac{\pi}{6}} = \ln(\sec \frac{\pi}{6} + \tan \frac{\pi}{6}) - \ln(\sec 0 + \tan 0) $$ 6. **Calculate values:** $$ \sec \frac{\pi}{6} = \frac{1}{\cos \frac{\pi}{6}} = \frac{1}{\frac{\sqrt{3}}{2}} = \frac{2}{\sqrt{3}} $$ $$ \tan \frac{\pi}{6} = \frac{1}{\sqrt{3}} $$ $$ \sec 0 = 1, \quad \tan 0 = 0 $$ 7. **Simplify:** $$ L = \ln \left( \frac{2}{\sqrt{3}} + \frac{1}{\sqrt{3}} \right) - \ln(1) = \ln \left( \frac{3}{\sqrt{3}} \right) = \ln(\sqrt{3}) $$ 8. **Final answer:** $$ \boxed{L = \ln(\sqrt{3}) = \frac{1}{2} \ln 3} $$