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

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

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

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

Integrate term by term:

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

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 becomes

$$- \int{\frac{1}{x} d x} + \int{\frac{1}{x - 1} d x} + {\color{red}{\int{\frac{1}{x + 1} d x}}} = - \int{\frac{1}{x} d x} + \int{\frac{1}{x - 1} d x} + {\color{red}{\int{\frac{1}{u} d u}}}$$

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

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

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

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

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

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

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

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

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

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

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

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

Therefore,

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

Add the constant of integration:

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

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