3. \( \int\left(2 x^{2}+3 x\right) d x= \) 4. \( \int\left(x-x^{3}\right) d x=\frac{x^{2}}{2}-\frac{x^{4}}{4}+C \) 5. \( \int \sqrt{2 x} d x= \)

To solve the integral \( \int\left(2 x^{2}+3 x\right) d x \), we apply the power rule for integration. The integral of \( x^n \) is \( \frac{x^{n+1}}{n+1} + C \). Thus, we get: \[ \int\left(2 x^{2}\right) d x = \frac{2 x^{3}}{3}, \quad \int\left(3 x\right) d x = \frac{3 x^{2}}{2} \] Putting it all together, we have: \[ \int\left(2 x^{2}+3 x\right) d x = \frac{2 x^{3}}{3} + \frac{3 x^{2}}{2} + C \] Now, for the integral \( \int \sqrt{2 x} d x \), we can use substitution to make it easier. Let \( u = 2x \), so \( du = 2 dx\) or \( dx = \frac{du}{2} \). Now, rewrite the integral: \[ \int \sqrt{2x} \, dx = \int \sqrt{u} \cdot \frac{1}{2} du = \frac{1}{2} \cdot \frac{u^{3/2}}{3/2} = \frac{1}{3} u^{3/2} + C \] Substituting back \( u = 2x \), we get: \[ \int \sqrt{2x} d x = \frac{1}{3} (2x)^{3/2} + C = \frac{2^{3/2}}{3} x^{3/2} + C = \frac{2\sqrt{2}}{3} x^{3/2} + C \]