Integral of sqrt(3x+4) dx

1. Let $u = 3 x + 4$.

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

   $$\int \frac{\sqrt{u}}{3}\, du$$

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

      $$\int \sqrt{u}\, du = \frac{\int \sqrt{u}\, du}{3}$$

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

         $$\int \sqrt{u}\, du = \frac{2 u^{\frac{3}{2}}}{3}$$

      So, the result is: $\frac{2 u^{\frac{3}{2}}}{9}$

   Now substitute $u$ back in:

   $$\frac{2 \left(3 x + 4\right)^{\frac{3}{2}}}{9}$$

2. Now simplify:

   $$\frac{2 \left(3 x + 4\right)^{\frac{3}{2}}}{9}$$

3. Add the constant of integration:

   $$\frac{2 \left(3 x + 4\right)^{\frac{3}{2}}}{9}+ \mathrm{constant}$$

The answer is:

$$\frac{2 \left(3 x + 4\right)^{\frac{3}{2}}}{9}+ \mathrm{constant}$$