1. We are asked to evaluate the definite integral $$\int_0^1 x^2 \sqrt{1 + 4x^2} \, dx.$$\n\n2. To solve this integral, we use substitution. Let $$u = 1 + 4x^2.$$ Then, $$\frac{du}{dx} = 8x,$$ so $$du = 8x \, dx.$$\n\n3. We want to express everything in terms of $u$. Notice that $$x^2 = \frac{u - 1}{4}$$ and $$x \, dx = \frac{du}{8}.$$\n\n4. Rewrite the integral in terms of $u$:$$\int_0^1 x^2 \sqrt{1 + 4x^2} \, dx = \int_{u=1}^{u=5} \frac{u - 1}{4} \sqrt{u} \cdot \frac{du}{8x}.$$ But we must be careful: since $du = 8x \, dx$, then $x \, dx = \frac{du}{8}$. So the integral becomes $$\int_{1}^{5} \frac{u - 1}{4} \sqrt{u} \cdot \frac{du}{8} = \int_1^5 \frac{u - 1}{32} u^{1/2} \, du.$$\n\n5. Simplify the integrand: $$\frac{u - 1}{32} u^{1/2} = \frac{1}{32} (u^{3/2} - u^{1/2}).$$\n\n6. The integral is now $$\frac{1}{32} \int_1^5 (u^{3/2} - u^{1/2}) \, du = \frac{1}{32} \left[ \int_1^5 u^{3/2} \, du - \int_1^5 u^{1/2} \, du \right].$$\n\n7. Integrate each term: $$\int u^{3/2} \, du = \frac{2}{5} u^{5/2}, \quad \int u^{1/2} \, du = \frac{2}{3} u^{3/2}.$$\n\n8. Substitute back: $$\frac{1}{32} \left[ \frac{2}{5} u^{5/2} - \frac{2}{3} u^{3/2} \right]_1^5 = \frac{1}{32} \left( \frac{2}{5} 5^{5/2} - \frac{2}{3} 5^{3/2} - \left( \frac{2}{5} 1^{5/2} - \frac{2}{3} 1^{3/2} \right) \right).$$\n\n9. Simplify the constants: $$= \frac{1}{32} \left( \frac{2}{5} 5^{5/2} - \frac{2}{3} 5^{3/2} - \frac{2}{5} + \frac{2}{3} \right).$$\n\n10. Calculate powers: $$5^{3/2} = 5 \sqrt{5}, \quad 5^{5/2} = 25 \sqrt{5}.$$\n\n11. Substitute: $$= \frac{1}{32} \left( \frac{2}{5} \cdot 25 \sqrt{5} - \frac{2}{3} \cdot 5 \sqrt{5} - \frac{2}{5} + \frac{2}{3} \right) = \frac{1}{32} \left( 10 \sqrt{5} - \frac{10}{3} \sqrt{5} - \frac{2}{5} + \frac{2}{3} \right).$$\n\n12. Combine like terms: $$10 \sqrt{5} - \frac{10}{3} \sqrt{5} = \frac{30}{3} \sqrt{5} - \frac{10}{3} \sqrt{5} = \frac{20}{3} \sqrt{5}.$$\n\n13. Combine constants: $$- \frac{2}{5} + \frac{2}{3} = \frac{-6 + 10}{15} = \frac{4}{15}.$$\n\n14. Final expression: $$\frac{1}{32} \left( \frac{20}{3} \sqrt{5} + \frac{4}{15} \right) = \frac{1}{32} \left( \frac{100}{15} \sqrt{5} + \frac{4}{15} \right) = \frac{1}{32} \cdot \frac{100 \sqrt{5} + 4}{15} = \frac{100 \sqrt{5} + 4}{480}.$$\n\n15. Simplify numerator and denominator by dividing numerator and denominator by 4: $$= \frac{25 \sqrt{5} + 1}{120}.$$\n\n**Final answer:** $$\int_0^1 x^2 \sqrt{1 + 4x^2} \, dx = \frac{25 \sqrt{5} + 1}{120}.$$