Integral of sqrt(4-3x) dx

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

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

   $$\int \left(- \frac{\sqrt{u}}{3}\right)\, du$$

   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:

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

2. Add the constant of integration:

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

The answer is:

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