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]
