We are asked to evaluate the indefinite integral $\int (5x^4 - 3x^2 + \frac{3\sqrt{x}}{2}) dx$.

1. Problem Description

We are asked to evaluate the indefinite integral $\int (5x^4 - 3x^2 + \frac{3\sqrt{x}}{2}) dx$.

2. Solution Steps

First, we use the linearity of the integral to split the integral into three parts:

$\int (5x^4 - 3x^2 + \frac{3\sqrt{x}}{2}) dx = \int 5x^4 dx - \int 3x^2 dx + \int \frac{3\sqrt{x}}{2} dx$.

Next, we can pull out the constants from each integral:

$\int 5x^4 dx - \int 3x^2 dx + \int \frac{3\sqrt{x}}{2} dx = 5\int x^4 dx - 3\int x^2 dx + \frac{3}{2} \int \sqrt{x} dx$.

Recall that $\sqrt{x} = x^{\frac{1}{2}}$. Thus, we have:

$5\int x^4 dx - 3\int x^2 dx + \frac{3}{2} \int x^{\frac{1}{2}} dx$.

Now we apply the power rule for integration:

$\int x^n dx = \frac{x^{n+1}}{n+1} + C$, where $n \neq -1$.

$5\int x^4 dx = 5\frac{x^{4+1}}{4+1} = 5\frac{x^5}{5} = x^5$.

$-3\int x^2 dx = -3\frac{x^{2+1}}{2+1} = -3\frac{x^3}{3} = -x^3$.

$\frac{3}{2} \int x^{\frac{1}{2}} dx = \frac{3}{2} \frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1} = \frac{3}{2} \frac{x^{\frac{3}{2}}}{\frac{3}{2}} = \frac{3}{2} \cdot \frac{2}{3} x^{\frac{3}{2}} = x^{\frac{3}{2}}$.

Combining the results, we have:

$x^5 - x^3 + x^{\frac{3}{2}} + C$.

3. Final Answer

$x^5 - x^3 + x^{\frac{3}{2}} + C$