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

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

The input is rewritten: $$$\int{\left(- \frac{3 - \frac{1}{x^{2}}}{3 x}\right)d x}=\int{\frac{-1 + \frac{1}{3 x^{2}}}{x} d x}$$$.

Simplify:

$${\color{red}{\int{\frac{-1 + \frac{1}{3 x^{2}}}{x} d x}}} = {\color{red}{\int{\frac{1 - 3 x^{2}}{3 x^{3}} d x}}}$$

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=\frac{1}{3}$$$ and $$$f{\left(x \right)} = \frac{1 - 3 x^{2}}{x^{3}}$$$:

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

Expand the expression:

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

Integrate term by term:

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

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

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

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=3$$$ and $$$f{\left(x \right)} = \frac{1}{x}$$$:

$$- \frac{{\color{red}{\int{\frac{3}{x} d x}}}}{3} - \frac{1}{6 x^{2}} = - \frac{{\color{red}{\left(3 \int{\frac{1}{x} d x}\right)}}}{3} - \frac{1}{6 x^{2}}$$

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

$$- {\color{red}{\int{\frac{1}{x} d x}}} - \frac{1}{6 x^{2}} = - {\color{red}{\ln{\left(\left|{x}\right| \right)}}} - \frac{1}{6 x^{2}}$$

Therefore,

$$\int{\frac{-1 + \frac{1}{3 x^{2}}}{x} d x} = - \ln{\left(\left|{x}\right| \right)} - \frac{1}{6 x^{2}}$$

Add the constant of integration:

$$\int{\frac{-1 + \frac{1}{3 x^{2}}}{x} d x} = - \ln{\left(\left|{x}\right| \right)} - \frac{1}{6 x^{2}}+C$$

$$$\int \left(- \frac{3 - \frac{1}{x^{2}}}{3 x}\right)\, dx = \left(- \ln\left(\left|{x}\right|\right) - \frac{1}{6 x^{2}}\right) + C$$$A