1. **State the problem:** We want to evaluate the integral $$\int \frac{\sqrt{\tan x}}{\sin x \cos x} \, dx.$$\n\n2. **Rewrite the integrand:** Recall that $$\tan x = \frac{\sin x}{\cos x}.$$ So, $$\sqrt{\tan x} = \sqrt{\frac{\sin x}{\cos x}} = \frac{\sqrt{\sin x}}{\sqrt{\cos x}}.$$\n\nSubstitute this into the integral:\n$$\int \frac{\sqrt{\tan x}}{\sin x \cos x} \, dx = \int \frac{\frac{\sqrt{\sin x}}{\sqrt{\cos x}}}{\sin x \cos x} \, dx = \int \frac{\sqrt{\sin x}}{\sqrt{\cos x} \sin x \cos x} \, dx.$$\n\n3. **Simplify the integrand:**\nRewrite the denominator:\n$$\sqrt{\cos x} \sin x \cos x = \sin x \cos x \sqrt{\cos x} = \sin x \cos^{3/2} x.$$\nSo the integrand becomes:\n$$\frac{\sqrt{\sin x}}{\sin x \cos^{3/2} x} = \frac{\sqrt{\sin x}}{\sin x} \cdot \frac{1}{\cos^{3/2} x} = \frac{1}{\sqrt{\sin x}} \cdot \frac{1}{\cos^{3/2} x} = \frac{1}{\sqrt{\sin x} \cos^{3/2} x}.$$\n\n4. **Rewrite the integral:**\n$$\int \frac{1}{\sqrt{\sin x} \cos^{3/2} x} \, dx.$$\n\n5. **Use substitution:** Let $$t = \sqrt{\tan x} = \frac{\sqrt{\sin x}}{\sqrt{\cos x}}.$$ Then, $$t^2 = \tan x = \frac{\sin x}{\cos x}.$$\n\nDifferentiate both sides with respect to $$x$$:\n$$2t \frac{dt}{dx} = \sec^2 x = \frac{1}{\cos^2 x}.$$\nSo, $$\frac{dt}{dx} = \frac{1}{2t \cos^2 x}.$$\n\nRewrite $$dx$$:\n$$dx = 2t \cos^2 x \, dt.$$\n\n6. **Rewrite the integral in terms of $$t$$:**\nRecall the integrand is $$\frac{\sqrt{\tan x}}{\sin x \cos x} = \frac{t}{\sin x \cos x}.$$\nBut from step 3, we simplified to $$\frac{1}{\sqrt{\sin x} \cos^{3/2} x}$$ which is equivalent to $$\frac{t}{\sin x \cos x}$$ after substitution. To avoid confusion, let's use the original substitution approach:\n\nSince $$t = \sqrt{\tan x}$$, then $$t^2 = \tan x = \frac{\sin x}{\cos x} \Rightarrow \sin x = t^2 \cos x.$$\n\nSubstitute $$\sin x$$ in the denominator:\n$$\sin x \cos x = t^2 \cos x \cdot \cos x = t^2 \cos^2 x.$$\n\nThe integrand becomes:\n$$\frac{t}{\sin x \cos x} = \frac{t}{t^2 \cos^2 x} = \frac{1}{t \cos^2 x}.$$\n\nTherefore, the integral is:\n$$\int \frac{1}{t \cos^2 x} dx.$$\n\nUsing $$dx = 2t \cos^2 x dt$$ from step 5, substitute into the integral:\n$$\int \frac{1}{t \cos^2 x} \cdot 2t \cos^2 x dt = \int 2 dt = 2t + C.$$\n\n7. **Back-substitute $$t$$:**\nRecall $$t = \sqrt{\tan x}$$, so the final answer is:\n$$\boxed{2 \sqrt{\tan x} + C}.$$