1. **State the problem:** Evaluate the definite integral $$\int_0^1 x^7 \sqrt{1 + x^4} \, dx$$. 2. **Identify a substitution:** Let $$u = 1 + x^4$$. Then, $$\frac{du}{dx} = 4x^3$$, so $$du = 4x^3 \, dx$$. 3. **Rewrite the integral in terms of $$u$$:** We have $$x^7 = x^4 \cdot x^3$$. Since $$u = 1 + x^4$$, then $$x^4 = u - 1$$. Also, $$x^3 \, dx = \frac{du}{4}$$. 4. Substitute into the integral: $$\int_0^1 x^7 \sqrt{1 + x^4} \, dx = \int_{u=1}^{u=2} (u - 1) \sqrt{u} \cdot \frac{du}{4} = \frac{1}{4} \int_1^2 (u - 1) u^{1/2} \, du$$. 5. **Simplify the integrand:** $$(u - 1) u^{1/2} = u^{3/2} - u^{1/2}$$. 6. **Rewrite the integral:** $$\frac{1}{4} \int_1^2 (u^{3/2} - u^{1/2}) \, du = \frac{1}{4} \left( \int_1^2 u^{3/2} \, du - \int_1^2 u^{1/2} \, du \right)$$. 7. **Integrate each term:** $$\int u^{3/2} \, du = \frac{u^{5/2}}{\frac{5}{2}} = \frac{2}{5} u^{5/2}$$ $$\int u^{1/2} \, du = \frac{u^{3/2}}{\frac{3}{2}} = \frac{2}{3} u^{3/2}$$ 8. **Evaluate the definite integrals:** $$\int_1^2 u^{3/2} \, du = \frac{2}{5} (2^{5/2} - 1)$$ $$\int_1^2 u^{1/2} \, du = \frac{2}{3} (2^{3/2} - 1)$$ 9. **Combine results:** $$\frac{1}{4} \left( \frac{2}{5} (2^{5/2} - 1) - \frac{2}{3} (2^{3/2} - 1) \right) = \frac{1}{4} \left( \frac{2}{5} 2^{5/2} - \frac{2}{5} - \frac{2}{3} 2^{3/2} + \frac{2}{3} \right)$$ 10. **Simplify constants:** $$= \frac{1}{4} \left( \frac{2}{5} 2^{5/2} - \frac{2}{3} 2^{3/2} + \frac{2}{3} - \frac{2}{5} \right) = \frac{1}{4} \left( \frac{2}{5} 2^{5/2} - \frac{2}{3} 2^{3/2} + \frac{6}{15} - \frac{6}{15} \right)$$ 11. **Final answer:** $$\boxed{\frac{1}{4} \left( \frac{2}{5} 2^{5/2} - \frac{2}{3} 2^{3/2} + \frac{2}{3} - \frac{2}{5} \right)}$$ This is the exact value of the integral.