Integral of $$$\left(x - \frac{1}{x}\right)^{3}$$$

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

Expand the expression:

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

Integrate term by term:

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

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

$$- \int{\frac{1}{x^{3}} d x} + \int{\frac{3}{x} d x} - \int{3 x d x} + {\color{red}{\int{x^{3} d x}}}=- \int{\frac{1}{x^{3}} d x} + \int{\frac{3}{x} d x} - \int{3 x d x} + {\color{red}{\frac{x^{1 + 3}}{1 + 3}}}=- \int{\frac{1}{x^{3}} d x} + \int{\frac{3}{x} d x} - \int{3 x d x} + {\color{red}{\left(\frac{x^{4}}{4}\right)}}$$

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

$$\frac{x^{4}}{4} + \int{\frac{3}{x} d x} - \int{3 x d x} - {\color{red}{\int{\frac{1}{x^{3}} d x}}}=\frac{x^{4}}{4} + \int{\frac{3}{x} d x} - \int{3 x d x} - {\color{red}{\int{x^{-3} d x}}}=\frac{x^{4}}{4} + \int{\frac{3}{x} d x} - \int{3 x d x} - {\color{red}{\frac{x^{-3 + 1}}{-3 + 1}}}=\frac{x^{4}}{4} + \int{\frac{3}{x} d x} - \int{3 x d x} - {\color{red}{\left(- \frac{x^{-2}}{2}\right)}}=\frac{x^{4}}{4} + \int{\frac{3}{x} d x} - \int{3 x d x} - {\color{red}{\left(- \frac{1}{2 x^{2}}\right)}}$$

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

$$\frac{x^{4}}{4} + \int{\frac{3}{x} d x} - {\color{red}{\int{3 x d x}}} + \frac{1}{2 x^{2}} = \frac{x^{4}}{4} + \int{\frac{3}{x} d x} - {\color{red}{\left(3 \int{x d x}\right)}} + \frac{1}{2 x^{2}}$$

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

$$\frac{x^{4}}{4} + \int{\frac{3}{x} d x} - 3 {\color{red}{\int{x d x}}} + \frac{1}{2 x^{2}}=\frac{x^{4}}{4} + \int{\frac{3}{x} d x} - 3 {\color{red}{\frac{x^{1 + 1}}{1 + 1}}} + \frac{1}{2 x^{2}}=\frac{x^{4}}{4} + \int{\frac{3}{x} d x} - 3 {\color{red}{\left(\frac{x^{2}}{2}\right)}} + \frac{1}{2 x^{2}}$$

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

$$\frac{x^{4}}{4} - \frac{3 x^{2}}{2} + {\color{red}{\int{\frac{3}{x} d x}}} + \frac{1}{2 x^{2}} = \frac{x^{4}}{4} - \frac{3 x^{2}}{2} + {\color{red}{\left(3 \int{\frac{1}{x} d x}\right)}} + \frac{1}{2 x^{2}}$$

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

$$\frac{x^{4}}{4} - \frac{3 x^{2}}{2} + 3 {\color{red}{\int{\frac{1}{x} d x}}} + \frac{1}{2 x^{2}} = \frac{x^{4}}{4} - \frac{3 x^{2}}{2} + 3 {\color{red}{\ln{\left(\left|{x}\right| \right)}}} + \frac{1}{2 x^{2}}$$

Therefore,

$$\int{\left(x - \frac{1}{x}\right)^{3} d x} = \frac{x^{4}}{4} - \frac{3 x^{2}}{2} + 3 \ln{\left(\left|{x}\right| \right)} + \frac{1}{2 x^{2}}$$

Simplify:

$$\int{\left(x - \frac{1}{x}\right)^{3} d x} = \frac{x^{2} \left(x^{4} - 6 x^{2} + 12 \ln{\left(\left|{x}\right| \right)}\right) + 2}{4 x^{2}}$$

Add the constant of integration:

$$\int{\left(x - \frac{1}{x}\right)^{3} d x} = \frac{x^{2} \left(x^{4} - 6 x^{2} + 12 \ln{\left(\left|{x}\right| \right)}\right) + 2}{4 x^{2}}+C$$

$$$\int \left(x - \frac{1}{x}\right)^{3}\, dx = \frac{x^{2} \left(x^{4} - 6 x^{2} + 12 \ln\left(\left|{x}\right|\right)\right) + 2}{4 x^{2}} + C$$$A