1. **State the problem:** Evaluate the integral $$\int_{6}^{9} \frac{\sqrt{x}}{1 + \sqrt[4]{x^3}} \, dx.$$\n\n2. **Rewrite the integral:** Let us express the integrand in terms of powers of $x$. Recall that $\sqrt{x} = x^{1/2}$ and $\sqrt[4]{x^3} = x^{3/4}$. So the integral becomes:\n$$\int_{6}^{9} \frac{x^{1/2}}{1 + x^{3/4}} \, dx.$$\n\n3. **Substitution:** Let $t = x^{1/4}$. Then $x = t^4$ and $dx = 4t^3 dt$. Also, when $x=6$, $t=6^{1/4}$; when $x=9$, $t=9^{1/4}$.\n\n4. **Rewrite the integral in terms of $t$:**\n$$\int_{t=6^{1/4}}^{9^{1/4}} \frac{(t^4)^{1/2}}{1 + (t^4)^{3/4}} \cdot 4t^3 dt = \int_{6^{1/4}}^{9^{1/4}} \frac{t^{2}}{1 + t^{3}} \cdot 4t^3 dt = \int_{6^{1/4}}^{9^{1/4}} \frac{4 t^{5}}{1 + t^{3}} dt.$$\n\n5. **Simplify the integrand:** The integral is now\n$$4 \int_{6^{1/4}}^{9^{1/4}} \frac{t^{5}}{1 + t^{3}} dt.$$\n\n6. **Divide numerator and denominator:** Note that $t^{5} = t^{3} \cdot t^{2}$, so\n$$\frac{t^{5}}{1 + t^{3}} = \frac{t^{3} t^{2}}{1 + t^{3}} = t^{2} \cdot \frac{t^{3}}{1 + t^{3}}.$$\n\n7. **Rewrite the fraction:**\n$$\frac{t^{3}}{1 + t^{3}} = 1 - \frac{1}{1 + t^{3}}.$$\n\n8. **Split the integral:**\n$$4 \int_{6^{1/4}}^{9^{1/4}} t^{2} \left(1 - \frac{1}{1 + t^{3}}\right) dt = 4 \left( \int_{6^{1/4}}^{9^{1/4}} t^{2} dt - \int_{6^{1/4}}^{9^{1/4}} \frac{t^{2}}{1 + t^{3}} dt \right).$$\n\n9. **Focus on the second integral:** Let us evaluate $$I = \int \frac{t^{2}}{1 + t^{3}} dt.$$\n\n10. **Substitution for $I$:** Let $u = 1 + t^{3}$, then $du = 3 t^{2} dt$, so $t^{2} dt = \frac{du}{3}$.\n\n11. **Rewrite $I$ in terms of $u$:**\n$$I = \int \frac{t^{2}}{u} dt = \int \frac{1}{u} \cdot t^{2} dt = \int \frac{1}{u} \cdot \frac{du}{3} = \frac{1}{3} \int \frac{1}{u} du = \frac{1}{3} \ln|u| + C = \frac{1}{3} \ln|1 + t^{3}| + C.$$\n\n12. **Evaluate the definite integrals:**\n\n- First integral:\n$$\int_{6^{1/4}}^{9^{1/4}} t^{2} dt = \left[ \frac{t^{3}}{3} \right]_{6^{1/4}}^{9^{1/4}} = \frac{(9^{1/4})^{3} - (6^{1/4})^{3}}{3} = \frac{9^{3/4} - 6^{3/4}}{3}.$$\n\n- Second integral:\n$$\int_{6^{1/4}}^{9^{1/4}} \frac{t^{2}}{1 + t^{3}} dt = \left[ \frac{1}{3} \ln|1 + t^{3}| \right]_{6^{1/4}}^{9^{1/4}} = \frac{1}{3} \left( \ln(1 + (9^{1/4})^{3}) - \ln(1 + (6^{1/4})^{3}) \right) = \frac{1}{3} \ln \frac{1 + 9^{3/4}}{1 + 6^{3/4}}.$$\n\n13. **Combine results:**\n$$4 \left( \frac{9^{3/4} - 6^{3/4}}{3} - \frac{1}{3} \ln \frac{1 + 9^{3/4}}{1 + 6^{3/4}} \right) = \frac{4}{3} \left( 9^{3/4} - 6^{3/4} - \ln \frac{1 + 9^{3/4}}{1 + 6^{3/4}} \right).$$\n\n14. **Final answer:**\n$$\boxed{\frac{4}{3} \left( 9^{3/4} - 6^{3/4} - \ln \frac{1 + 9^{3/4}}{1 + 6^{3/4}} \right)}.$$