Integral of $$$x + \frac{1}{x^{2}}$$$

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

Integrate term by term:

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

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

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

Apply the power rule $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=-2$$$:

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

Therefore,

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

Simplify:

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

Add the constant of integration:

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

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