Integral of ln(1-x^2) dx

1. Use integration by parts:

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

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

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

   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:

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

   1. There are multiple ways to do this integral.

      Method #1

      1. Rewrite the integrand:

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

      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:

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

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

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

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

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

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

      Method #2

      1. Rewrite the integrand:

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

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

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

         1. Rewrite the integrand:

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

         2. Integrate term-by-term:

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

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

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

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

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

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

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

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

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

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

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

3. Add the constant of integration:

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

The answer is:

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