Integral of $$$\frac{1}{x^{2} - 1}$$$

Perform partial fraction decomposition (steps can be seen here):

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

Integrate term by term:

$${\color{red}{\int{\left(- \frac{1}{2 \left(x + 1\right)} + \frac{1}{2 \left(x - 1\right)}\right)d x}}} = {\color{red}{\left(\int{\frac{1}{2 \left(x - 1\right)} d x} - \int{\frac{1}{2 \left(x + 1\right)} d x}\right)}}$$

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=\frac{1}{2}$$$ and $$$f{\left(x \right)} = \frac{1}{x - 1}$$$:

$$- \int{\frac{1}{2 \left(x + 1\right)} d x} + {\color{red}{\int{\frac{1}{2 \left(x - 1\right)} d x}}} = - \int{\frac{1}{2 \left(x + 1\right)} d x} + {\color{red}{\left(\frac{\int{\frac{1}{x - 1} d x}}{2}\right)}}$$

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

Then $$$du=\left(x - 1\right)^{\prime }dx = 1 dx$$$ (steps can be seen »), and we have that $$$dx = du$$$.

The integral can be rewritten as

$$- \int{\frac{1}{2 \left(x + 1\right)} d x} + \frac{{\color{red}{\int{\frac{1}{x - 1} d x}}}}{2} = - \int{\frac{1}{2 \left(x + 1\right)} d x} + \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{2}$$

The integral of $$$\frac{1}{u}$$$ is $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

$$- \int{\frac{1}{2 \left(x + 1\right)} d x} + \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{2} = - \int{\frac{1}{2 \left(x + 1\right)} d x} + \frac{{\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{2}$$

Recall that $$$u=x - 1$$$:

$$\frac{\ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{2} - \int{\frac{1}{2 \left(x + 1\right)} d x} = \frac{\ln{\left(\left|{{\color{red}{\left(x - 1\right)}}}\right| \right)}}{2} - \int{\frac{1}{2 \left(x + 1\right)} d x}$$

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=\frac{1}{2}$$$ and $$$f{\left(x \right)} = \frac{1}{x + 1}$$$:

$$\frac{\ln{\left(\left|{x - 1}\right| \right)}}{2} - {\color{red}{\int{\frac{1}{2 \left(x + 1\right)} d x}}} = \frac{\ln{\left(\left|{x - 1}\right| \right)}}{2} - {\color{red}{\left(\frac{\int{\frac{1}{x + 1} d x}}{2}\right)}}$$

Let $$$u=x + 1$$$.

Then $$$du=\left(x + 1\right)^{\prime }dx = 1 dx$$$ (steps can be seen »), and we have that $$$dx = du$$$.

The integral can be rewritten as

$$\frac{\ln{\left(\left|{x - 1}\right| \right)}}{2} - \frac{{\color{red}{\int{\frac{1}{x + 1} d x}}}}{2} = \frac{\ln{\left(\left|{x - 1}\right| \right)}}{2} - \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{2}$$

The integral of $$$\frac{1}{u}$$$ is $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

$$\frac{\ln{\left(\left|{x - 1}\right| \right)}}{2} - \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{2} = \frac{\ln{\left(\left|{x - 1}\right| \right)}}{2} - \frac{{\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{2}$$

Recall that $$$u=x + 1$$$:

$$\frac{\ln{\left(\left|{x - 1}\right| \right)}}{2} - \frac{\ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{2} = \frac{\ln{\left(\left|{x - 1}\right| \right)}}{2} - \frac{\ln{\left(\left|{{\color{red}{\left(x + 1\right)}}}\right| \right)}}{2}$$

Therefore,

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

Add the constant of integration:

$$\int{\frac{1}{x^{2} - 1} d x} = \frac{\ln{\left(\left|{x - 1}\right| \right)}}{2} - \frac{\ln{\left(\left|{x + 1}\right| \right)}}{2}+C$$

Answer: $$$\int{\frac{1}{x^{2} - 1} d x}=\frac{\ln{\left(\left|{x - 1}\right| \right)}}{2} - \frac{\ln{\left(\left|{x + 1}\right| \right)}}{2}+C$$$