Answer :

Certainly! Let's go through the process step-by-step to find the integral of the given function:

[tex]\[ \int \frac{4 x^5 + 3}{2 x^2} \, dx \][/tex]

1. Simplify the integrand:

We start with the function inside the integral:

[tex]\[ \frac{4 x^5 + 3}{2 x^2} \][/tex]

We can simplify this by dividing each term in the numerator by the term in the denominator:

[tex]\[ \frac{4 x^5}{2 x^2} + \frac{3}{2 x^2} \][/tex]

Simplifying each term individually:

[tex]\[ \frac{4 x^5}{2 x^2} = 2 x^3 \][/tex]

[tex]\[ \frac{3}{2 x^2} = \frac{3}{2} x^{-2} \][/tex]

So the simplified integrand becomes:

[tex]\[ 2 x^3 + \frac{3}{2} x^{-2} \][/tex]

2. Integrate the simplified expression:

Now, we integrate [tex]\( 2 x^3 + \frac{3}{2} x^{-2} \)[/tex]:

[tex]\[ \int (2 x^3 + \frac{3}{2} x^{-2}) \, dx \][/tex]

We can split this into two separate integrals:

[tex]\[ \int 2 x^3 \, dx + \int \frac{3}{2} x^{-2} \, dx \][/tex]

For the first integral:

[tex]\[ \int 2 x^3 \, dx \][/tex]

Use the power rule ([tex]\( \int x^n \, dx = \frac{x^{n+1}}{n+1} \)[/tex]):

[tex]\[ \int 2 x^3 \, dx = 2 \cdot \frac{x^{3+1}}{3+1} = 2 \cdot \frac{x^4}{4} = \frac{x^4}{2} \][/tex]

For the second integral:

[tex]\[ \int \frac{3}{2} x^{-2} \, dx \][/tex]

Again, use the power rule:

[tex]\[ \int \frac{3}{2} x^{-2} \, dx = \frac{3}{2} \cdot \frac{x^{-2+1}}{-2+1} = \frac{3}{2} \cdot \frac{x^{-1}}{-1} = \frac{3}{2} \cdot (-x^{-1}) = -\frac{3}{2} x^{-1} \][/tex]

Which simplifies to:

[tex]\[ -\frac{3}{2} \frac{1}{x} = -\frac{3}{2x} \][/tex]

3. Combine the results:

Putting it all together, we have:

[tex]\[ \int \frac{4 x^5 + 3}{2 x^2} \, dx = \frac{x^4}{2} - \frac{3}{2x} + C \][/tex]

Where [tex]\( C \)[/tex] is the constant of integration.

So, the final answer in simplest form is:

[tex]\[ \boxed{\frac{x^4}{2} - \frac{3}{2x} + C} \][/tex]