Integral of (x*ln(1-3*x)) dx

1. Use integration by parts:

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

   Let $u{\left(x \right)} = \log{\left(1 - 3 x \right)}$ and let $\operatorname{dv}{\left(x \right)} = x$.

   Then $\operatorname{du}{\left(x \right)} = - \frac{3}{1 - 3 x}$.

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

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

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

   Now evaluate the sub-integral.

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

   $$\int \left(- \frac{3 x^{2}}{2 \left(1 - 3 x\right)}\right)\, dx = - \frac{3 \int \frac{x^{2}}{1 - 3 x}\, dx}{2}$$

   1. There are multiple ways to do this integral.

      Method #1

      1. Rewrite the integrand:

         $$\frac{x^{2}}{1 - 3 x} = - \frac{x}{3} - \frac{1}{9} - \frac{1}{9 \left(3 x - 1\right)}$$

      2. Integrate term-by-term:

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

            $$\int \left(- \frac{x}{3}\right)\, dx = - \frac{\int x\, dx}{3}$$

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

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

            So, the result is: $- \frac{x^{2}}{6}$

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

            $$\int \left(- \frac{1}{9}\right)\, dx = - \frac{x}{9}$$

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

            $$\int \left(- \frac{1}{9 \left(3 x - 1\right)}\right)\, dx = - \frac{\int \frac{1}{3 x - 1}\, dx}{9}$$

            1. Let $u = 3 x - 1$.

               Then let $du = 3 dx$ and substitute $\frac{du}{3}$:

               $$\int \frac{1}{3 u}\, du$$

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

                  $$\int \frac{1}{u}\, du = \frac{\int \frac{1}{u}\, du}{3}$$

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

                  So, the result is: $\frac{\log{\left(u \right)}}{3}$

               Now substitute $u$ back in:

               $$\frac{\log{\left(3 x - 1 \right)}}{3}$$

            So, the result is: $- \frac{\log{\left(3 x - 1 \right)}}{27}$

         The result is: $- \frac{x^{2}}{6} - \frac{x}{9} - \frac{\log{\left(3 x - 1 \right)}}{27}$

      Method #2

      1. Rewrite the integrand:

         $$\frac{x^{2}}{1 - 3 x} = - \frac{x^{2}}{3 x - 1}$$

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

         $$\int \left(- \frac{x^{2}}{3 x - 1}\right)\, dx = - \int \frac{x^{2}}{3 x - 1}\, dx$$

         1. Rewrite the integrand:

            $$\frac{x^{2}}{3 x - 1} = \frac{x}{3} + \frac{1}{9} + \frac{1}{9 \left(3 x - 1\right)}$$

         2. Integrate term-by-term:

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

               $$\int \frac{x}{3}\, dx = \frac{\int x\, dx}{3}$$

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

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

               So, the result is: $\frac{x^{2}}{6}$

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

               $$\int \frac{1}{9}\, dx = \frac{x}{9}$$

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

               $$\int \frac{1}{9 \left(3 x - 1\right)}\, dx = \frac{\int \frac{1}{3 x - 1}\, dx}{9}$$

               1. Let $u = 3 x - 1$.

                  Then let $du = 3 dx$ and substitute $\frac{du}{3}$:

                  $$\int \frac{1}{3 u}\, du$$

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

                     $$\int \frac{1}{u}\, du = \frac{\int \frac{1}{u}\, du}{3}$$

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

                     So, the result is: $\frac{\log{\left(u \right)}}{3}$

                  Now substitute $u$ back in:

                  $$\frac{\log{\left(3 x - 1 \right)}}{3}$$

               So, the result is: $\frac{\log{\left(3 x - 1 \right)}}{27}$

            The result is: $\frac{x^{2}}{6} + \frac{x}{9} + \frac{\log{\left(3 x - 1 \right)}}{27}$

         So, the result is: $- \frac{x^{2}}{6} - \frac{x}{9} - \frac{\log{\left(3 x - 1 \right)}}{27}$

   So, the result is: $\frac{x^{2}}{4} + \frac{x}{6} + \frac{\log{\left(3 x - 1 \right)}}{18}$

3. Add the constant of integration:

   $$\frac{x^{2} \log{\left(1 - 3 x \right)}}{2} - \frac{x^{2}}{4} - \frac{x}{6} - \frac{\log{\left(3 x - 1 \right)}}{18}+ \mathrm{constant}$$

The answer is:

$$\frac{x^{2} \log{\left(1 - 3 x \right)}}{2} - \frac{x^{2}}{4} - \frac{x}{6} - \frac{\log{\left(3 x - 1 \right)}}{18}+ \mathrm{constant}$$