31. \( \int_{0}^{1} x(\sqrt[3]{x}+\sqrt[4]{x}) d x \)

To evaluate the integral \( \int_{0}^{1} x(\sqrt[3]{x}+\sqrt[4]{x}) d x \), we can first simplify the integrand: 1. Rewrite the integrand: \[ x(\sqrt[3]{x} + \sqrt[4]{x}) = x \cdot x^{1/3} + x \cdot x^{1/4} = x^{1 + 1/3} + x^{1 + 1/4} = x^{4/3} + x^{5/4}. \] 2. Now, the integral becomes: \[ \int_{0}^{1} (x^{4/3} + x^{5/4}) \, dx. \] 3. Split this into two separate integrals: \[ \int_{0}^{1} x^{4/3} \, dx + \int_{0}^{1} x^{5/4} \, dx. \] 4. Evaluate each integral separately. Using the power rule for integration: \[ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C. \] For \( x^{4/3} \): \[ \int x^{4/3} \, dx = \frac{x^{7/3}}{7/3} = \frac{3}{7} x^{7/3}. \] Evaluating the limits from 0 to 1 gives: \[ \left[ \frac{3}{7} x^{7/3} \right]_{0}^{1} = \frac{3}{7} \cdot 1 - \frac{3}{7} \cdot 0 = \frac{3}{7}. \] For \( x^{5/4} \): \[ \int x^{5/4} \, dx = \frac{x^{9/4}}{9/4} = \frac{4}{9} x^{9/4}. \] Evaluating from 0 to 1 gives: \[ \left[ \frac{4}{9} x^{9/4} \right]_{0}^{1} = \frac{4}{9} \cdot 1 - \frac{4}{9} \cdot 0 = \frac{4}{9}. \] 5. Now, combine the results: \[ \int_{0}^{1} x(\sqrt[3]{x}+\sqrt[4]{x}) d x = \frac{3}{7} + \frac{4}{9}. \] 6. To add these fractions, we find a common denominator. The least common multiple of 7 and 9 is 63: \[ \frac{3}{7} = \frac{3 \times 9}{7 \times 9} = \frac{27}{63}, \] \[ \frac{4}{9} = \frac{4 \times 7}{9 \times 7} = \frac{28}{63}. \] 7. So, now we combine them: \[ \frac{27}{63} + \frac{28}{63} = \frac{55}{63}. \] Thus, the final result of the integral is: \[ \int_{0}^{1} x(\sqrt[3]{x}+\sqrt[4]{x}) d x = \frac{55}{63}. \]