Integral of $$$\frac{x^{2} - 1}{x^{3}}$$$

Find $$$\int \frac{x^{2} - 1}{x^{3}}\, dx$$$.

Expand the expression:

$${\color{red}{\int{\frac{x^{2} - 1}{x^{3}} d x}}} = {\color{red}{\int{\left(\frac{1}{x} - \frac{1}{x^{3}}\right)d x}}}$$

Integrate term by term:

$${\color{red}{\int{\left(\frac{1}{x} - \frac{1}{x^{3}}\right)d x}}} = {\color{red}{\left(- \int{\frac{1}{x^{3}} d x} + \int{\frac{1}{x} d x}\right)}}$$

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

$$- \int{\frac{1}{x^{3}} d x} + {\color{red}{\int{\frac{1}{x} d x}}} = - \int{\frac{1}{x^{3}} d x} + {\color{red}{\ln{\left(\left|{x}\right| \right)}}}$$

Apply the power rule $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=-3$$$:

$$\ln{\left(\left|{x}\right| \right)} - {\color{red}{\int{\frac{1}{x^{3}} d x}}}=\ln{\left(\left|{x}\right| \right)} - {\color{red}{\int{x^{-3} d x}}}=\ln{\left(\left|{x}\right| \right)} - {\color{red}{\frac{x^{-3 + 1}}{-3 + 1}}}=\ln{\left(\left|{x}\right| \right)} - {\color{red}{\left(- \frac{x^{-2}}{2}\right)}}=\ln{\left(\left|{x}\right| \right)} - {\color{red}{\left(- \frac{1}{2 x^{2}}\right)}}$$

Therefore,

$$\int{\frac{x^{2} - 1}{x^{3}} d x} = \ln{\left(\left|{x}\right| \right)} + \frac{1}{2 x^{2}}$$

Add the constant of integration:

$$\int{\frac{x^{2} - 1}{x^{3}} d x} = \ln{\left(\left|{x}\right| \right)} + \frac{1}{2 x^{2}}+C$$

$$$\int \frac{x^{2} - 1}{x^{3}}\, dx = \left(\ln\left(\left|{x}\right|\right) + \frac{1}{2 x^{2}}\right) + C$$$A