1. We are asked to solve the integral \(\int x^5 \sqrt[3]{2 - x^3} \, dx\). 2. To solve this integral, we use substitution. Let \(u = 2 - x^3\). 3. Then, \(\frac{du}{dx} = -3x^2\), so \(du = -3x^2 dx\) or \(dx = \frac{du}{-3x^2}\). 4. Rewrite the integral in terms of \(u\): \[ \int x^5 u^{1/3} dx = \int x^5 u^{1/3} \frac{du}{-3x^2} = -\frac{1}{3} \int x^{3} u^{1/3} du \] 5. Since \(u = 2 - x^3\), we have \(x^3 = 2 - u\). Substitute this: \[ -\frac{1}{3} \int (2 - u) u^{1/3} du = -\frac{1}{3} \int (2u^{1/3} - u^{4/3}) du \] 6. Now integrate term by term: \[ -\frac{1}{3} \left( 2 \int u^{1/3} du - \int u^{4/3} du \right) \] 7. Use the power rule for integration \(\int u^n du = \frac{u^{n+1}}{n+1} + C\): \[ \int u^{1/3} du = \frac{u^{4/3}}{4/3} = \frac{3}{4} u^{4/3} \] \[ \int u^{4/3} du = \frac{u^{7/3}}{7/3} = \frac{3}{7} u^{7/3} \] 8. Substitute back: \[ -\frac{1}{3} \left( 2 \cdot \frac{3}{4} u^{4/3} - \frac{3}{7} u^{7/3} \right) + C = -\frac{1}{3} \left( \frac{3}{2} u^{4/3} - \frac{3}{7} u^{7/3} \right) + C \] 9. Simplify: \[ -\frac{1}{3} \cdot \frac{3}{2} u^{4/3} + \frac{1}{3} \cdot \frac{3}{7} u^{7/3} + C = -\frac{1}{2} u^{4/3} + \frac{1}{7} u^{7/3} + C \] 10. Substitute back \(u = 2 - x^3\): \[ \boxed{ -\frac{1}{2} (2 - x^3)^{4/3} + \frac{1}{7} (2 - x^3)^{7/3} + C } This is the solution to the first integral. q_count is 2 because there are two integrals given, but only the first is solved here as per instructions.