Integral of //e^(3x)+3/2x dx

1. Integrate term-by-term:

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

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

      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: $\frac{3 x^{2}}{4}$

   1. Let $u = 3 x$.

      Then let $du = 3 dx$ and substitute $\frac{du}{3}$:

      $$\int \frac{e^{u}}{3}\, du$$

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

         $$\text{False}$$

         1. The integral of the exponential function is itself.

            $$\int e^{u}\, du = e^{u}$$

         So, the result is: $\frac{e^{u}}{3}$

      Now substitute $u$ back in:

      $$\frac{e^{3 x}}{3}$$

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

2. Add the constant of integration:

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

The answer is:

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