Question

$\int \sqrt{z}\left(z^{2}-\frac{1}{4z}\right)dz =$

Answer 1

$\frac{2}{7} z^{7/2} - \frac{1}{2} z^{1/2} + C$

Answer 2

We need to evaluate the integral $\int \sqrt{z}\left(z^{2}-\frac{1}{4z}\right)dz$.

First, simplify the integrand: $\sqrt{z}\left(z^{2}-\frac{1}{4z}\right) = z^{1/2}\left(z^{2}-\frac{1}{4}z^{-1}\right)$ $= z^{1/2} \cdot z^2 - z^{1/2} \cdot \frac{1}{4}z^{-1}$

Using the exponent rule $a^m \cdot a^n = a^{m+n}$: $= z^{1/2+2} - \frac{1}{4}z^{1/2-1}$ $= z^{5/2} - \frac{1}{4}z^{-1/2}$

Now, integrate the simplified expression term by term using the power rule for integration, $\int z^n dz = \frac{z^{n+1}}{n+1} + C$ (for $n \neq -1$): $\int \left(z^{5/2} - \frac{1}{4}z^{-1/2}\right)dz = \int z^{5/2} dz - \frac{1}{4} \int z^{-1/2} dz$

For the first term, $n = 5/2$: $\int z^{5/2} dz = \frac{z^{5/2+1}}{5/2+1} + C_1 = \frac{z^{7/2}}{7/2} + C_1 = \frac{2}{7} z^{7/2} + C_1$

For the second term, $n = -1/2$: $\int z^{-1/2} dz = \frac{z^{-1/2+1}}{-1/2+1} + C_2 = \frac{z^{1/2}}{1/2} + C_2 = 2 z^{1/2} + C_2$

Combine the results: $\int \left(z^{5/2} - \frac{1}{4}z^{-1/2}\right)dz = \frac{2}{7} z^{7/2} - \frac{1}{4} (2 z^{1/2}) + C$ $= \frac{2}{7} z^{7/2} - \frac{1}{2} z^{1/2} + C$ where $C$ is the constant of integration.

In summary: Simplify the integrand using exponent rules. Apply the power rule for integration $\int z^n dz = \frac{z^{n+1}}{n+1}$ to each term.