Integral of -(3x-1)/4x dx

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

   $$\int \frac{x \left(1 - 3 x\right)}{4}\, dx = \frac{\int x \left(1 - 3 x\right)\, dx}{4}$$

   1. Rewrite the integrand:

      $$x \left(1 - 3 x\right) = - 3 x^{2} + x$$

   2. Integrate term-by-term:

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

         $$\int \left(- 3 x^{2}\right)\, dx = - 3 \int x^{2}\, dx$$

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

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

         So, the result is: $- x^{3}$

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

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

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

   So, the result is: $- \frac{x^{3}}{4} + \frac{x^{2}}{8}$

2. Now simplify:

   $$\frac{x^{2} \cdot \left(1 - 2 x\right)}{8}$$

3. Add the constant of integration:

   $$\frac{x^{2} \cdot \left(1 - 2 x\right)}{8}+ \mathrm{constant}$$

The answer is:

$$\frac{x^{2} \cdot \left(1 - 2 x\right)}{8}+ \mathrm{constant}$$