1. **Problem statement:** Evaluate the integral $$\int \sqrt{\tan x} \, dx$$. 2. **Understanding the integral:** The integral involves the square root of the tangent function, which is not straightforward. We will use substitution to simplify it. 3. **Substitution:** Let $$t = \sqrt{\tan x}$$, so $$t^2 = \tan x$$. 4. **Differentiate both sides:** $$\frac{d}{dx}(t^2) = \frac{d}{dx}(\tan x)$$ $$2t \frac{dt}{dx} = \sec^2 x$$ 5. **Express $$dx$$ in terms of $$dt$$:** $$\frac{dt}{dx} = \frac{\sec^2 x}{2t} \implies dx = \frac{2t}{\sec^2 x} dt$$ 6. **Rewrite the integral:** $$\int \sqrt{\tan x} \, dx = \int t \cdot dx = \int t \cdot \frac{2t}{\sec^2 x} dt = \int \frac{2t^2}{\sec^2 x} dt$$ 7. **Express $$\sec^2 x$$ in terms of $$t$$:** Since $$t^2 = \tan x$$, and $$\sec^2 x = 1 + \tan^2 x = 1 + t^4$$. 8. **Substitute back:** $$\int \frac{2t^2}{1 + t^4} dt$$ 9. **Integral to solve:** $$2 \int \frac{t^2}{1 + t^4} dt$$ 10. **Use partial fraction or known integral formula:** The integral $$\int \frac{t^2}{1 + t^4} dt$$ can be solved using advanced techniques or tables, resulting in: $$\frac{1}{2 \sqrt{2}} \ln \left| \frac{t^2 - \sqrt{2} t + 1}{t^2 + \sqrt{2} t + 1} \right| + \frac{1}{\sqrt{2}} \arctan \left( \frac{\sqrt{2} t}{1 + t^2} \right) + C$$ 11. **Multiply by 2:** $$2 \times \left( \frac{1}{2 \sqrt{2}} \ln \left| \frac{t^2 - \sqrt{2} t + 1}{t^2 + \sqrt{2} t + 1} \right| + \frac{1}{\sqrt{2}} \arctan \left( \frac{\sqrt{2} t}{1 + t^2} \right) \right) + C$$ $$= \frac{1}{\sqrt{2}} \ln \left| \frac{t^2 - \sqrt{2} t + 1}{t^2 + \sqrt{2} t + 1} \right| + \frac{2}{\sqrt{2}} \arctan \left( \frac{\sqrt{2} t}{1 + t^2} \right) + C$$ 12. **Substitute back $$t = \sqrt{\tan x}$$:** $$\int \sqrt{\tan x} \, dx = \frac{1}{\sqrt{2}} \ln \left| \frac{\tan x - \sqrt{2} \sqrt{\tan x} + 1}{\tan x + \sqrt{2} \sqrt{\tan x} + 1} \right| + \sqrt{2} \arctan \left( \frac{\sqrt{2} \sqrt{\tan x}}{1 + \tan x} \right) + C$$ **Final answer:** $$\boxed{\int \sqrt{\tan x} \, dx = \frac{1}{\sqrt{2}} \ln \left| \frac{\tan x - \sqrt{2} \sqrt{\tan x} + 1}{\tan x + \sqrt{2} \sqrt{\tan x} + 1} \right| + \sqrt{2} \arctan \left( \frac{\sqrt{2} \sqrt{\tan x}}{1 + \tan x} \right) + C}$$