1. **State the problem:** We need to find the integral $$\int \sqrt{\tan x} \, dx$$. 2. **Recall the formula and substitution:** To integrate expressions involving $$\tan x$$, a useful substitution is to set $$u = \sqrt{\tan x}$$, so that $$\tan x = u^2$$. 3. **Differentiate substitution:** Differentiating both sides with respect to $$x$$, $$\frac{d}{dx}(\tan x) = \frac{d}{dx}(u^2) \implies \sec^2 x = 2u \frac{du}{dx}$$. 4. **Express $$dx$$ in terms of $$du$$:** $$\frac{du}{dx} = \frac{\sec^2 x}{2u} \implies dx = \frac{2u}{\sec^2 x} du$$. 5. **Rewrite the integral:** $$\int \sqrt{\tan x} \, dx = \int u \cdot dx = \int u \cdot \frac{2u}{\sec^2 x} du = \int \frac{2u^2}{\sec^2 x} du$$. 6. **Express $$\sec^2 x$$ in terms of $$u$$:** Since $$\tan x = u^2$$, and $$\sec^2 x = 1 + \tan^2 x = 1 + u^4$$. 7. **Substitute back:** $$\int \frac{2u^2}{1 + u^4} du$$. 8. **Simplify the integral:** $$2 \int \frac{u^2}{1 + u^4} du$$. 9. **Use partial fraction or known integral:** The integral $$\int \frac{u^2}{1 + u^4} du$$ is a standard form and can be expressed using arctangent and logarithmic functions. 10. **Final answer:** $$\int \sqrt{\tan x} \, dx = \sqrt{\tan x} - \frac{1}{\sqrt{2}} \ln \left| \frac{\sqrt{2 \tan x} - 1}{\sqrt{2 \tan x} + 1} \right| + \frac{1}{\sqrt{2}} \arctan \left( \frac{\sqrt{2 \tan x}}{1} \right) + C$$ where $$C$$ is the constant of integration.