Integral of (1-x)/x^2 dx

1. There are multiple ways to do this integral.

   Method #1

   1. Let $u = - x$.

      Then let $du = - dx$ and substitute $- du$:

      $$\int \left(- \frac{u + 1}{u^{2}}\right)\, du$$

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

         $$\int \frac{u + 1}{u^{2}}\, du = - \int \frac{u + 1}{u^{2}}\, du$$

         1. Rewrite the integrand:

            $$\frac{u + 1}{u^{2}} = \frac{1}{u} + \frac{1}{u^{2}}$$

         2. Integrate term-by-term:

            1. The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$.

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

               $$\int \frac{1}{u^{2}}\, du = - \frac{1}{u}$$

            The result is: $\log{\left(u \right)} - \frac{1}{u}$

         So, the result is: $- \log{\left(u \right)} + \frac{1}{u}$

      Now substitute $u$ back in:

      $$- \log{\left(- x \right)} - \frac{1}{x}$$

   Method #2

   1. Rewrite the integrand:

      $$\frac{1 - x}{x^{2}} = - \frac{1}{x} + \frac{1}{x^{2}}$$

   2. Integrate term-by-term:

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

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

         1. The integral of $\frac{1}{x}$ is $\log{\left(x \right)}$.

         So, the result is: $- \log{\left(x \right)}$

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

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

      The result is: $- \log{\left(x \right)} - \frac{1}{x}$

   Method #3

   1. Rewrite the integrand:

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

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

      $$\int \left(- \frac{x - 1}{x^{2}}\right)\, dx = - \int \frac{x - 1}{x^{2}}\, dx$$

      1. Rewrite the integrand:

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

      2. Integrate term-by-term:

         1. The integral of $\frac{1}{x}$ is $\log{\left(x \right)}$.

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

            $$\int \left(- \frac{1}{x^{2}}\right)\, dx = - \int \frac{1}{x^{2}}\, dx$$

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

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

            So, the result is: $\frac{1}{x}$

         The result is: $\log{\left(x \right)} + \frac{1}{x}$

      So, the result is: $- \log{\left(x \right)} - \frac{1}{x}$

2. Add the constant of integration:

   $$- \log{\left(- x \right)} - \frac{1}{x}+ \mathrm{constant}$$

The answer is:

$$- \log{\left(- x \right)} - \frac{1}{x}+ \mathrm{constant}$$