1. **Problem:** Evaluate the integral $$\int \frac{dx}{\sqrt{x} + \sqrt[3]{x}}.$$\n\n2. **Formula and rules:** To integrate expressions involving roots, rewrite roots as fractional exponents: $$\sqrt{x} = x^{\frac{1}{2}}, \quad \sqrt[3]{x} = x^{\frac{1}{3}}.$$\n\n3. **Rewrite the integral:** $$\int \frac{dx}{x^{\frac{1}{2}} + x^{\frac{1}{3}}}.$$\n\n4. **Substitution:** Let $$t = x^{\frac{1}{6}}$$ so that $$x = t^6$$ and $$dx = 6t^5 dt.$$\n\n5. **Rewrite the denominator in terms of $t$: $$x^{\frac{1}{2}} = (t^6)^{\frac{1}{2}} = t^3,$$ and $$x^{\frac{1}{3}} = (t^6)^{\frac{1}{3}} = t^2.$$\n\n6. **Substitute into the integral:** $$\int \frac{dx}{x^{\frac{1}{2}} + x^{\frac{1}{3}}} = \int \frac{6t^5 dt}{t^3 + t^2} = \int \frac{6t^5}{t^2(t + 1)} dt = \int \frac{6t^5}{t^2(t + 1)} dt.$$\n\n7. **Simplify the integrand:** $$\frac{6t^5}{t^2(t + 1)} = \frac{6t^{\cancel{5}}}{t^{\cancel{2}}(t + 1)} = \frac{6t^{3}}{t + 1}.$$\n\n8. **Rewrite the integral:** $$\int \frac{6t^{3}}{t + 1} dt.$$\n\n9. **Perform polynomial division:** Divide $$6t^3$$ by $$t + 1$$:\n\n$$6t^3 \div (t + 1) = 6t^2 - 6t + 6 - \frac{6}{t + 1}.$$\n\n10. **Rewrite the integral as:** $$\int (6t^2 - 6t + 6 - \frac{6}{t + 1}) dt = \int 6t^2 dt - \int 6t dt + \int 6 dt - \int \frac{6}{t + 1} dt.$$\n\n11. **Integrate each term:**\n\n$$\int 6t^2 dt = 6 \cdot \frac{t^3}{3} = 2t^3,$$\n$$\int 6t dt = 6 \cdot \frac{t^2}{2} = 3t^2,$$\n$$\int 6 dt = 6t,$$\n$$\int \frac{6}{t + 1} dt = 6 \ln|t + 1|.$$\n\n12. **Combine results:** $$2t^3 - 3t^2 + 6t - 6 \ln|t + 1| + C.$$\n\n13. **Back-substitute $t = x^{1/6}$:** $$2x^{\frac{1}{2}} - 3x^{\frac{1}{3}} + 6x^{\frac{1}{6}} - 6 \ln|x^{\frac{1}{6}} + 1| + C.$$\n\n**Final answer:** $$\boxed{2x^{\frac{1}{2}} - 3x^{\frac{1}{3}} + 6x^{\frac{1}{6}} - 6 \ln\left|x^{\frac{1}{6}} + 1\right| + C}.$$