Integral of (3÷(2×sqrt(3x+4))-x) dx

1. Integrate term-by-term:

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

      $$\int \left(- x\right)\, dx = - \int x\, dx$$

      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: $- \frac{x^{2}}{2}$

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

      $$\int \frac{3}{2 \sqrt{3 x + 4}}\, dx = 3 \int \frac{1}{2 \sqrt{3 x + 4}}\, dx$$

      1. Let $u = 2 \sqrt{3 x + 4}$.

         Then let $du = \frac{3 dx}{\sqrt{3 x + 4}}$ and substitute $\frac{du}{6}$:

         $$\int \frac{1}{6}\, du$$

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

            $$\text{False}$$

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

               $$\int 1\, du = u$$

            So, the result is: $\frac{u}{6}$

         Now substitute $u$ back in:

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

      So, the result is: $\sqrt{3 x + 4}$

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

2. Now simplify:

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

3. Add the constant of integration:

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

The answer is: