\( \int x ^ { 2 } \sqrt { x ^ { 3 } + 1 } d x , \quad u = x ^ { 3 } + 1 \)

To solve the integral \[ \int x^2 \sqrt{x^3 + 1} \, dx \] using the substitution \( u = x^3 + 1 \), we first need to find \( du \) in terms of \( dx \). 1. Differentiate \( u \): \[ du = 3x^2 \, dx \quad \Rightarrow \quad dx = \frac{du}{3x^2} \] 2. Now, we can express \( x^2 \) in terms of \( u \): \[ x^3 = u - 1 \quad \Rightarrow \quad x = (u - 1)^{1/3} \] Therefore, \[ x^2 = ((u - 1)^{1/3})^2 = (u - 1)^{2/3} \] 3. Substitute \( x^2 \) and \( dx \) into the integral: \[ \int x^2 \sqrt{x^3 + 1} \, dx = \int (u - 1)^{2/3} \sqrt{u} \cdot \frac{du}{3x^2} \] Since \( x^2 = (u - 1)^{2/3} \), we have: \[ dx = \frac{du}{3(u - 1)^{2/3}} \] 4. Substitute everything into the integral: \[ \int (u - 1)^{2/3} \sqrt{u} \cdot \frac{du}{3(u - 1)^{2/3}} = \frac{1}{3} \int \sqrt{u} \, du \] 5. Now, we can integrate: \[ \frac{1}{3} \int u^{1/2} \, du = \frac{1}{3} \cdot \frac{u^{3/2}}{3/2} = \frac{2}{9} u^{3/2} + C \] 6. Substitute back \( u = x^3 + 1 \): \[ = \frac{2}{9} (x^3 + 1)^{3/2} + C \] Thus, the final result is: \[ \int x^2 \sqrt{x^3 + 1} \, dx = \frac{2}{9} (x^3 + 1)^{3/2} + C \]