1. **State the problem:** We need to find the integral of $\sqrt{\tan x}$ with respect to $x$, i.e., $\int \sqrt{\tan x} \, dx$. 2. **Recall the formula and substitution:** This integral is non-trivial and typically requires substitution. Let $t = \sqrt{\tan x}$, so $t^2 = \tan x$. 3. **Express $dx$ in terms of $dt$:** Differentiating both sides, $2t dt = \sec^2 x \, dx$. Since $\sec^2 x = 1 + \tan^2 x = 1 + t^4$, we have: $$dx = \frac{2t}{1 + t^4} dt$$ 4. **Rewrite the integral:** $$\int \sqrt{\tan x} \, dx = \int t \cdot dx = \int t \cdot \frac{2t}{1 + t^4} dt = \int \frac{2 t^2}{1 + t^4} dt$$ 5. **Simplify the integral:** $$\int \frac{2 t^2}{1 + t^4} dt$$ 6. **Use partial fraction decomposition:** The denominator factors as $1 + t^4 = (t^2 - \sqrt{2} t + 1)(t^2 + \sqrt{2} t + 1)$. We can write: $$\frac{2 t^2}{1 + t^4} = \frac{A t + B}{t^2 - \sqrt{2} t + 1} + \frac{C t + D}{t^2 + \sqrt{2} t + 1}$$ 7. **Solve for coefficients:** Solving for $A, B, C, D$ yields: $$A = \sqrt{2}, B = 0, C = -\sqrt{2}, D = 0$$ 8. **Rewrite integral:** $$\int \frac{2 t^2}{1 + t^4} dt = \int \frac{\sqrt{2} t}{t^2 - \sqrt{2} t + 1} dt - \int \frac{\sqrt{2} t}{t^2 + \sqrt{2} t + 1} dt$$ 9. **Integrate each term:** Use substitution for each integral: - For $I_1 = \int \frac{\sqrt{2} t}{t^2 - \sqrt{2} t + 1} dt$, let $u = t^2 - \sqrt{2} t + 1$, then $du = 2 t - \sqrt{2} dt$. - For $I_2 = \int \frac{\sqrt{2} t}{t^2 + \sqrt{2} t + 1} dt$, let $v = t^2 + \sqrt{2} t + 1$, then $dv = 2 t + \sqrt{2} dt$. 10. **Final answer:** After integration and back-substitution, the integral is: $$\int \sqrt{\tan x} \, dx = \frac{\sqrt{2}}{2} \ln \left| \frac{t^2 - \sqrt{2} t + 1}{t^2 + \sqrt{2} t + 1} \right| + C = \frac{\sqrt{2}}{2} \ln \left| \frac{\tan x - \sqrt{2} \sqrt{\tan x} + 1}{\tan x + \sqrt{2} \sqrt{\tan x} + 1} \right| + C$$ where $C$ is the constant of integration.