1. **State the problem:** We want to find the indefinite integral $$\int \sqrt{\tan x} \, dx$$. 2. **Recall the formula and substitution:** This integral is not straightforward. We use the substitution $$t = \sqrt{\tan x}$$, so $$t^2 = \tan x$$. 3. **Express dx in terms of dt:** Differentiating both sides, $$\frac{d}{dx}(t^2) = \frac{d}{dx}(\tan x) \Rightarrow 2t \frac{dt}{dx} = \sec^2 x$$. Since $$\sec^2 x = 1 + \tan^2 x = 1 + t^4$$, we have $$\frac{dt}{dx} = \frac{1 + t^4}{2t} \Rightarrow dx = \frac{2t}{1 + t^4} dt$$. 4. **Rewrite the integral:** Substitute $$\sqrt{\tan x} = t$$ and $$dx$$ as above: $$\int \sqrt{\tan x} \, dx = \int t \cdot \frac{2t}{1 + t^4} dt = \int \frac{2t^2}{1 + t^4} dt$$. 5. **Simplify the integral:** We now need to evaluate $$\int \frac{2t^2}{1 + t^4} dt$$. 6. **Partial fraction decomposition:** Factor the denominator: $$1 + t^4 = (t^2 - \sqrt{2} t + 1)(t^2 + \sqrt{2} t + 1)$$. Set $$\frac{2t^2}{1 + t^4} = \frac{A t + B}{t^2 - \sqrt{2} t + 1} + \frac{C t + D}{t^2 + \sqrt{2} t + 1}$$. 7. **Solve for coefficients:** Equate numerators and solve for $$A,B,C,D$$ (omitted detailed algebra here for brevity). 8. **Integrate each term:** Each term is of the form $$\int \frac{E t + F}{t^2 + p t + 1} dt$$, which can be integrated using substitution and arctangent formulas. 9. **Final answer:** After integration and back-substitution $$t = \sqrt{\tan x}$$, the integral is $$\int \sqrt{\tan x} \, dx = \frac{1}{\sqrt{2}} \ln \left| \frac{\sqrt{\tan x}^2 - \sqrt{2} \sqrt{\tan x} + 1}{\sqrt{\tan x}^2 + \sqrt{2} \sqrt{\tan x} + 1} \right| + \frac{1}{\sqrt{2}} \arctan \left( \frac{\sqrt{2} \sqrt{\tan x}}{1 - \tan x} \right) + C$$, where $$C$$ is the constant of integration. This completes the solution.