Integral of log(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 $- du$:

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

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

         $$\int \log{\left(u \right)}\, du = - \int \log{\left(u \right)}\, du$$

         1. Use integration by parts:

            $$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$

            Let $u{\left(u \right)} = \log{\left(u \right)}$ and let $\operatorname{dv}{\left(u \right)} = 1$.

            Then $\operatorname{du}{\left(u \right)} = \frac{1}{u}$.

            To find $v{\left(u \right)}$:

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

               $$\int 1\, du = u$$

            Now evaluate the sub-integral.

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

            $$\int 1\, du = u$$

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

      Now substitute $u$ back in:

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

   Method #2

   1. Use integration by parts:

      $$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$

      Let $u{\left(x \right)} = \log{\left(1 - x \right)}$ and let $\operatorname{dv}{\left(x \right)} = 1$.

      Then $\operatorname{du}{\left(x \right)} = - \frac{1}{1 - x}$.

      To find $v{\left(x \right)}$:

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

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

      Now evaluate the sub-integral.

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

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

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

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

2. Now simplify:

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

3. Add the constant of integration:

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

The answer is:

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