Integral of ln(1-2x) dx

1. There are multiple ways to do this integral.

   Method #1

   1. Let $u = 1 - 2 x$.

      Then let $du = - 2 dx$ and substitute $- \frac{du}{2}$:

      $$\int \left(- \frac{\log{\left(u \right)}}{2}\right)\, du$$

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

         $$\int \log{\left(u \right)}\, du = - \frac{\int \log{\left(u \right)}\, du}{2}$$

         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: $- \frac{u \log{\left(u \right)}}{2} + \frac{u}{2}$

      Now substitute $u$ back in:

      $$- x - \frac{\left(1 - 2 x\right) \log{\left(1 - 2 x \right)}}{2} + \frac{1}{2}$$

   Method #2

   1. Use integration by parts:

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

      Let $u{\left(x \right)} = \log{\left(1 - 2 x \right)}$ and let $\operatorname{dv}{\left(x \right)} = 1$.

      Then $\operatorname{du}{\left(x \right)} = - \frac{2}{1 - 2 x}$.

      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:

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

      1. There are multiple ways to do this integral.

         Method #1

         1. Rewrite the integrand:

            $$\frac{x}{1 - 2 x} = - \frac{1}{2} - \frac{1}{2 \left(2 x - 1\right)}$$

         2. Integrate term-by-term:

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

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

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

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

               1. Let $u = 2 x - 1$.

                  Then let $du = 2 dx$ and substitute $\frac{du}{2}$:

                  $$\int \frac{1}{2 u}\, du$$

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

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

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

                     So, the result is: $\frac{\log{\left(u \right)}}{2}$

                  Now substitute $u$ back in:

                  $$\frac{\log{\left(2 x - 1 \right)}}{2}$$

               So, the result is: $- \frac{\log{\left(2 x - 1 \right)}}{4}$

            The result is: $- \frac{x}{2} - \frac{\log{\left(2 x - 1 \right)}}{4}$

         Method #2

         1. Rewrite the integrand:

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

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

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

            1. Rewrite the integrand:

               $$\frac{x}{2 x - 1} = \frac{1}{2} + \frac{1}{2 \left(2 x - 1\right)}$$

            2. Integrate term-by-term:

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

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

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

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

                  1. Let $u = 2 x - 1$.

                     Then let $du = 2 dx$ and substitute $\frac{du}{2}$:

                     $$\int \frac{1}{2 u}\, du$$

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

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

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

                        So, the result is: $\frac{\log{\left(u \right)}}{2}$

                     Now substitute $u$ back in:

                     $$\frac{\log{\left(2 x - 1 \right)}}{2}$$

                  So, the result is: $\frac{\log{\left(2 x - 1 \right)}}{4}$

               The result is: $\frac{x}{2} + \frac{\log{\left(2 x - 1 \right)}}{4}$

            So, the result is: $- \frac{x}{2} - \frac{\log{\left(2 x - 1 \right)}}{4}$

      So, the result is: $x + \frac{\log{\left(2 x - 1 \right)}}{2}$

2. Now simplify:

   $$- x + \frac{\left(2 x - 1\right) \log{\left(1 - 2 x \right)}}{2} + \frac{1}{2}$$

3. Add the constant of integration:

   $$- x + \frac{\left(2 x - 1\right) \log{\left(1 - 2 x \right)}}{2} + \frac{1}{2}+ \mathrm{constant}$$

The answer is:

$$- x + \frac{\left(2 x - 1\right) \log{\left(1 - 2 x \right)}}{2} + \frac{1}{2}+ \mathrm{constant}$$