Integral of $$$x \left(- \frac{x^{2}}{2} + \frac{1}{x}\right) e^{2}$$$

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

Simplify the integrand:

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

Integrate term by term:

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

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

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

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

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

Apply the constant rule $$$\int c\, dx = c x$$$ with $$$c=e^{2}$$$:

$$- \frac{x^{4} e^{2}}{8} + {\color{red}{\int{e^{2} d x}}} = - \frac{x^{4} e^{2}}{8} + {\color{red}{x e^{2}}}$$

Therefore,

$$\int{x \left(- \frac{x^{2}}{2} + \frac{1}{x}\right) e^{2} d x} = - \frac{x^{4} e^{2}}{8} + x e^{2}$$

Simplify:

$$\int{x \left(- \frac{x^{2}}{2} + \frac{1}{x}\right) e^{2} d x} = \frac{x \left(8 - x^{3}\right) e^{2}}{8}$$

Add the constant of integration:

$$\int{x \left(- \frac{x^{2}}{2} + \frac{1}{x}\right) e^{2} d x} = \frac{x \left(8 - x^{3}\right) e^{2}}{8}+C$$

$$$\int x \left(- \frac{x^{2}}{2} + \frac{1}{x}\right) e^{2}\, dx = \frac{x \left(8 - x^{3}\right) e^{2}}{8} + C$$$A