Integral of ln(x^2-1) dx

1. Use integration by parts:

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

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

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

   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:

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

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

3. Now simplify:

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

4. Add the constant of integration:

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

The answer is: