Integral of $$$- \frac{19}{x^{2}} - \frac{4}{x^{5}}$$$

Find $$$\int \left(- \frac{19}{x^{2}} - \frac{4}{x^{5}}\right)\, dx$$$.

Integrate term by term:

$${\color{red}{\int{\left(- \frac{19}{x^{2}} - \frac{4}{x^{5}}\right)d x}}} = {\color{red}{\left(- \int{\frac{4}{x^{5}} d x} - \int{\frac{19}{x^{2}} d x}\right)}}$$

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=19$$$ and $$$f{\left(x \right)} = \frac{1}{x^{2}}$$$:

$$- \int{\frac{4}{x^{5}} d x} - {\color{red}{\int{\frac{19}{x^{2}} d x}}} = - \int{\frac{4}{x^{5}} d x} - {\color{red}{\left(19 \int{\frac{1}{x^{2}} d x}\right)}}$$

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

$$- \int{\frac{4}{x^{5}} d x} - 19 {\color{red}{\int{\frac{1}{x^{2}} d x}}}=- \int{\frac{4}{x^{5}} d x} - 19 {\color{red}{\int{x^{-2} d x}}}=- \int{\frac{4}{x^{5}} d x} - 19 {\color{red}{\frac{x^{-2 + 1}}{-2 + 1}}}=- \int{\frac{4}{x^{5}} d x} - 19 {\color{red}{\left(- x^{-1}\right)}}=- \int{\frac{4}{x^{5}} d x} - 19 {\color{red}{\left(- \frac{1}{x}\right)}}$$

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=4$$$ and $$$f{\left(x \right)} = \frac{1}{x^{5}}$$$:

$$- {\color{red}{\int{\frac{4}{x^{5}} d x}}} + \frac{19}{x} = - {\color{red}{\left(4 \int{\frac{1}{x^{5}} d x}\right)}} + \frac{19}{x}$$

Apply the power rule $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=-5$$$:

$$- 4 {\color{red}{\int{\frac{1}{x^{5}} d x}}} + \frac{19}{x}=- 4 {\color{red}{\int{x^{-5} d x}}} + \frac{19}{x}=- 4 {\color{red}{\frac{x^{-5 + 1}}{-5 + 1}}} + \frac{19}{x}=- 4 {\color{red}{\left(- \frac{x^{-4}}{4}\right)}} + \frac{19}{x}=- 4 {\color{red}{\left(- \frac{1}{4 x^{4}}\right)}} + \frac{19}{x}$$

Therefore,

$$\int{\left(- \frac{19}{x^{2}} - \frac{4}{x^{5}}\right)d x} = \frac{19}{x} + \frac{1}{x^{4}}$$

Add the constant of integration:

$$\int{\left(- \frac{19}{x^{2}} - \frac{4}{x^{5}}\right)d x} = \frac{19}{x} + \frac{1}{x^{4}}+C$$

$$$\int \left(- \frac{19}{x^{2}} - \frac{4}{x^{5}}\right)\, dx = \left(\frac{19}{x} + \frac{1}{x^{4}}\right) + C$$$A