Integral of $$$x - 8 - \frac{4 e^{x}}{x}$$$
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Find $$$\int \left(x - 8 - \frac{4 e^{x}}{x}\right)\, dx$$$.
Solution
Integrate term by term:
$${\color{red}{\int{\left(x - 8 - \frac{4 e^{x}}{x}\right)d x}}} = {\color{red}{\left(- \int{8 d x} + \int{x d x} - \int{\frac{4 e^{x}}{x} d x}\right)}}$$
Apply the constant rule $$$\int c\, dx = c x$$$ with $$$c=8$$$:
$$\int{x d x} - \int{\frac{4 e^{x}}{x} d x} - {\color{red}{\int{8 d x}}} = \int{x d x} - \int{\frac{4 e^{x}}{x} d x} - {\color{red}{\left(8 x\right)}}$$
Apply the power rule $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=1$$$:
$$- 8 x - \int{\frac{4 e^{x}}{x} d x} + {\color{red}{\int{x d x}}}=- 8 x - \int{\frac{4 e^{x}}{x} d x} + {\color{red}{\frac{x^{1 + 1}}{1 + 1}}}=- 8 x - \int{\frac{4 e^{x}}{x} d x} + {\color{red}{\left(\frac{x^{2}}{2}\right)}}$$
Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=4$$$ and $$$f{\left(x \right)} = \frac{e^{x}}{x}$$$:
$$\frac{x^{2}}{2} - 8 x - {\color{red}{\int{\frac{4 e^{x}}{x} d x}}} = \frac{x^{2}}{2} - 8 x - {\color{red}{\left(4 \int{\frac{e^{x}}{x} d x}\right)}}$$
This integral (Exponential Integral) does not have a closed form:
$$\frac{x^{2}}{2} - 8 x - 4 {\color{red}{\int{\frac{e^{x}}{x} d x}}} = \frac{x^{2}}{2} - 8 x - 4 {\color{red}{\operatorname{Ei}{\left(x \right)}}}$$
Therefore,
$$\int{\left(x - 8 - \frac{4 e^{x}}{x}\right)d x} = \frac{x^{2}}{2} - 8 x - 4 \operatorname{Ei}{\left(x \right)}$$
Add the constant of integration:
$$\int{\left(x - 8 - \frac{4 e^{x}}{x}\right)d x} = \frac{x^{2}}{2} - 8 x - 4 \operatorname{Ei}{\left(x \right)}+C$$
Answer
$$$\int \left(x - 8 - \frac{4 e^{x}}{x}\right)\, dx = \left(\frac{x^{2}}{2} - 8 x - 4 \operatorname{Ei}{\left(x \right)}\right) + C$$$A