Integral of x/(1-x) dx

1. There are multiple ways to do this integral.

   Method #1

   1. Rewrite the integrand:

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

   2. Integrate term-by-term:

      1. The integral of a constant is the constant times the variable of integration:

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

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

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

   Method #2

   1. Rewrite the integrand:

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

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

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

      1. Rewrite the integrand:

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

      2. Integrate term-by-term:

         1. The integral of a constant is the constant times the variable of integration:

            $$\int 1\, dx = x$$

         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:

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

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

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

2. Add the constant of integration:

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

The answer is:

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