Integral of 1/(1-x) dx

1. There are multiple ways to do this integral.

   Method #1

   1. Let $u = 1 - x$.

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

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

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

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

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

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

      Now substitute $u$ back in:

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

   Method #2

   1. Rewrite the integrand:

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

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

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

      1. Let $u = x - 1$.

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

         $$\int \frac{1}{u}\, du$$

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

         Now substitute $u$ back in:

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

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

   Method #3

   1. Rewrite the integrand:

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

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

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

      1. Let $u = x - 1$.

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

         $$\int \frac{1}{u}\, du$$

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

         Now substitute $u$ back in:

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

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

2. Add the constant of integration:

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

The answer is: