1. **State the problem:** We want to find the indefinite integral $$\int \sqrt{\tan x} \, dx$$. 2. **Recall the formula and substitution:** Integrals involving square roots of trigonometric functions often require substitution. Let us set $$u = \sqrt{\tan x}$$, so $$u^2 = \tan x$$. 3. **Differentiate to find dx in terms of du:** $$\tan x = u^2 \implies \sec^2 x \, dx = 2u \, du$$ Since $$\sec^2 x = 1 + \tan^2 x = 1 + u^4$$, we have $$dx = \frac{2u}{1 + u^4} \, du$$. 4. **Rewrite the integral in terms of u:** $$\int \sqrt{\tan x} \, dx = \int u \cdot dx = \int u \cdot \frac{2u}{1 + u^4} \, du = \int \frac{2u^2}{1 + u^4} \, du$$. 5. **Simplify the integral:** $$\int \frac{2u^2}{1 + u^4} \, du$$. 6. **Use partial fraction decomposition:** Note that $$1 + u^4 = (u^2 - \sqrt{2}u + 1)(u^2 + \sqrt{2}u + 1)$$. We express: $$\frac{2u^2}{1 + u^4} = \frac{Au + B}{u^2 - \sqrt{2}u + 1} + \frac{Cu + D}{u^2 + \sqrt{2}u + 1}$$. Solving for constants, we find: $$A = \sqrt{2}, B = 0, C = -\sqrt{2}, D = 0$$. 7. **Rewrite the integral:** $$\int \frac{2u^2}{1 + u^4} \, du = \int \frac{\sqrt{2}u}{u^2 - \sqrt{2}u + 1} \, du - \int \frac{\sqrt{2}u}{u^2 + \sqrt{2}u + 1} \, du$$. 8. **Integrate each term separately:** Use substitution for each quadratic denominator and complete the square: For the first integral: $$u^2 - \sqrt{2}u + 1 = \left(u - \frac{\sqrt{2}}{2}\right)^2 + \frac{1}{2}$$ For the second integral: $$u^2 + \sqrt{2}u + 1 = \left(u + \frac{\sqrt{2}}{2}\right)^2 + \frac{1}{2}$$ 9. **Final answer:** After integration and back-substitution, the result is $$\sqrt{2} \ln \left| \frac{u^2 - \sqrt{2}u + 1}{u^2 + \sqrt{2}u + 1} \right| + C = \sqrt{2} \ln \left| \frac{\tan x - \sqrt{2} \sqrt{\tan x} + 1}{\tan x + \sqrt{2} \sqrt{\tan x} + 1} \right| + C$$. This completes the integral.