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

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

Expand the expression:

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

Integrate term by term:

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

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

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

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

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

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

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

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

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

Therefore,

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

Simplify:

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

Add the constant of integration:

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

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