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

1. Rewrite the integrand:

   $$\frac{x - 4}{x^{3}} = \frac{1}{x^{2}} - \frac{4}{x^{3}}$$

2. Integrate term-by-term:

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

      $$\int \frac{1}{x^{2}}\, dx = - \frac{1}{x}$$

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

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

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

         $$\int \frac{1}{x^{3}}\, dx = - \frac{1}{2 x^{2}}$$

      So, the result is: $\frac{2}{x^{2}}$

   The result is: $- \frac{1}{x} + \frac{2}{x^{2}}$

3. Now simplify:

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

4. Add the constant of integration:

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

The answer is:

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