1. **State the problem:** We need to find the integral $$\int \sqrt{\tan x} \, dx$$. 2. **Recall the formula and substitution:** To solve integrals involving square roots of trigonometric functions, a common approach is substitution. Let $$u = \sqrt{\tan x}$$, so $$u^2 = \tan x$$. 3. **Express everything in terms of $$u$$:** Differentiating both sides, $$\frac{d}{dx}(u^2) = \frac{d}{dx}(\tan x) \implies 2u \frac{du}{dx} = \sec^2 x$$. 4. **Rewrite $$dx$$:** Since $$\sec^2 x = 1 + \tan^2 x = 1 + u^4$$, we have $$\frac{du}{dx} = \frac{1 + u^4}{2u} \implies dx = \frac{2u}{1 + u^4} du$$. 5. **Rewrite the integral:** Substitute $$\sqrt{\tan x} = u$$ and $$dx$$ as above: $$\int \sqrt{\tan x} \, dx = \int u \cdot \frac{2u}{1 + u^4} du = \int \frac{2u^2}{1 + u^4} du$$. 6. **Simplify the integral:** $$\int \frac{2u^2}{1 + u^4} du$$. 7. **Use partial fraction decomposition:** Factor the denominator: $$1 + u^4 = (u^2 - \sqrt{2}u + 1)(u^2 + \sqrt{2}u + 1)$$. 8. **Set up partial fractions:** $$\frac{2u^2}{1 + u^4} = \frac{Au + B}{u^2 - \sqrt{2}u + 1} + \frac{Cu + D}{u^2 + \sqrt{2}u + 1}$$. 9. **Solve for coefficients:** Equate numerators and solve for $$A,B,C,D$$ (omitted here for brevity). 10. **Integrate each term:** Each integral is of the form $$\int \frac{px + q}{x^2 + bx + c} dx$$, which can be solved using substitution and logarithmic/arctangent formulas. 11. **Final answer:** After integration and back-substitution, $$\int \sqrt{\tan x} \, dx = \frac{\sqrt{2}}{2} \ln \left| \frac{\sqrt{\tan x}^2 - \sqrt{2} \sqrt{\tan x} + 1}{\sqrt{\tan x}^2 + \sqrt{2} \sqrt{\tan x} + 1} \right| + \arctan \left( \frac{\sqrt{2} \sqrt{\tan x}}{1 - \tan x} \right) + C$$, where $$C$$ is the constant of integration. This integral involves advanced techniques including substitution and partial fractions, and the final expression is in terms of logarithmic and arctangent functions of $$\sqrt{\tan x}$$.