1. **State the problem:** We want to evaluate the integral $$\int \frac{\sqrt{1+\sqrt{e^{-x}}}}{\sqrt{e^x}} \, dx.$$\n\n2. **Rewrite the integrand:** Recall that $$\sqrt{e^x} = e^{x/2}$$ and $$\sqrt{e^{-x}} = e^{-x/2}.$$ So the integrand becomes $$\frac{\sqrt{1 + e^{-x/2}}}{e^{x/2}} = e^{-x/2} \sqrt{1 + e^{-x/2}}.$$\n\n3. **Substitution:** Let $$t = e^{-x/2}.$$ Then, $$dt = -\frac{1}{2} e^{-x/2} dx = -\frac{1}{2} t \, dx,$$ which implies $$dx = -2 \frac{dt}{t}.$$\n\n4. **Rewrite the integral in terms of $t$:**\n$$\int e^{-x/2} \sqrt{1 + e^{-x/2}} \, dx = \int t \sqrt{1 + t} \cdot \left(-2 \frac{dt}{t}\right) = -2 \int \sqrt{1 + t} \, dt.$$\n\n5. **Simplify the integral:**\n$$-2 \int \sqrt{1 + t} \, dt = -2 \int (1 + t)^{1/2} \, dt.$$\n\n6. **Integrate:** Use the power rule for integration: $$\int (1 + t)^{1/2} dt = \frac{(1 + t)^{3/2}}{\frac{3}{2}} = \frac{2}{3} (1 + t)^{3/2} + C.$$\n\n7. **Substitute back:**\n$$-2 \cdot \frac{2}{3} (1 + t)^{3/2} + C = -\frac{4}{3} (1 + t)^{3/2} + C.$$\nRecall that $$t = e^{-x/2},$$ so the final answer is\n$$\boxed{-\frac{4}{3} \left(1 + e^{-x/2}\right)^{3/2} + C}.$$