Integral of (x^2+3*sqrt(x))/(2*x) dx

1. There are multiple ways to do this integral.

   Method #1

   1. Let $u = \sqrt{x}$.

      Then let $du = \frac{dx}{2 \sqrt{x}}$ and substitute $du$:

      $$\int \left(u^{3} + 3\right)\, du$$

      1. Integrate term-by-term:

         1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$:

            $$\int u^{3}\, du = \frac{u^{4}}{4}$$

         1. The integral of a constant is the constant times the variable of integration:

            $$\int 3\, du = 3 u$$

         The result is: $\frac{u^{4}}{4} + 3 u$

      Now substitute $u$ back in:

      $$3 \sqrt{x} + \frac{x^{2}}{4}$$

   Method #2

   1. Rewrite the integrand:

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

   2. Integrate term-by-term:

      1. The integral of a constant times a function is the constant times the integral of the function:

         $$\int \frac{x}{2}\, dx = \frac{\int x\, dx}{2}$$

         1. The integral of $x^{n}$ is $\frac{x^{n + 1}}{n + 1}$ when $n \neq -1$:

            $$\int x\, dx = \frac{x^{2}}{2}$$

         So, the result is: $\frac{x^{2}}{4}$

      1. The integral of a constant times a function is the constant times the integral of the function:

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

         1. The integral of $x^{n}$ is $\frac{x^{n + 1}}{n + 1}$ when $n \neq -1$:

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

         So, the result is: $3 \sqrt{x}$

      The result is: $3 \sqrt{x} + \frac{x^{2}}{4}$

2. Add the constant of integration:

   $$3 \sqrt{x} + \frac{x^{2}}{4}+ \mathrm{constant}$$

The answer is:

$$3 \sqrt{x} + \frac{x^{2}}{4}+ \mathrm{constant}$$