The problem asks us to find the indefinite integral of the function $5x^4 - 3x^2 + \frac{3\sqrt{x}}{2}$ with respect to $x$.

1. Problem Description

The problem asks us to find the indefinite integral of the function $5x^4 - 3x^2 + \frac{3\sqrt{x}}{2}$ with respect to $x$.

2. Solution Steps

We will integrate each term separately.

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

We can rewrite $\sqrt{x}$ as $x^{\frac{1}{2}}$.

$\int 5x^4 dx = 5 \int x^4 dx$

$\int 3x^2 dx = 3 \int x^2 dx$

$\int \frac{3\sqrt{x}}{2} dx = \frac{3}{2} \int x^{\frac{1}{2}} dx$

Using the power rule for integration, we have

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

Therefore,

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

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

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

Now, we substitute these results back into the original expression:

$5 \int x^4 dx - 3 \int x^2 dx + \frac{3}{2} \int x^{\frac{1}{2}} dx = 5(\frac{x^5}{5}) - 3(\frac{x^3}{3}) + \frac{3}{2}(\frac{2}{3} x^{\frac{3}{2}}) + C$

Simplifying the expression, we have:

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

3. Final Answer

$x^5 - x^3 + x\sqrt{x} + C$