Integral of x*ln(1-x) dx

1. Use integration by parts:

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

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

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

   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:

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

   1. There are multiple ways to do this integral.

      Method #1

      1. Rewrite the integrand:

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

      2. Integrate term-by-term:

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

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

            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: $- \frac{x^{2}}{2}$

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

      Method #2

      1. Rewrite the integrand:

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

      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:

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

         2. Integrate term-by-term:

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

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

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

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

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

3. Add the constant of integration:

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

The answer is:

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