Use a suitable substitution to evaluate the following integral. 7. ∫ x^{2} \sqrt[3]{1+x^{3}} dx

Para resolver a integral \(\int x^{2} \sqrt[3]{1+x^{3}} \, dx\), podemos usar a substituição \(u = 1 + x^{3}\). 1. Derivando \(u\), temos: \[ du = 3x^{2} \, dx \quad \Rightarrow \quad dx = \frac{du}{3x^{2}} \] 2. Agora, substituímos \(x^{2}\) e \(dx\) na integral: \[ \int x^{2} \sqrt[3]{u} \cdot \frac{du}{3x^{2}} = \frac{1}{3} \int \sqrt[3]{u} \, du \] 3. A integral \(\int \sqrt[3]{u} \, du\) é: \[ \int u^{1/3} \, du = \frac{u^{4/3}}{4/3} = \frac{3}{4} u^{4/3} \] 4. Portanto, temos: \[ \frac{1}{3} \cdot \frac{3}{4} u^{4/3} = \frac{1}{4} u^{4/3} \] 5. Substituindo \(u\) de volta: \[ \frac{1}{4} (1 + x^{3})^{4/3} + C \] Assim, a integral \(\int x^{2} \sqrt[3]{1+x^{3}} \, dx\) é: \[ \frac{1}{4} (1 + x^{3})^{4/3} + C \]