Integral of $$$3 x - \frac{1}{x^{4}}$$$

Find $$$\int \left(3 x - \frac{1}{x^{4}}\right)\, dx$$$.

Integrate term by term:

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

Apply the power rule $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=-4$$$:

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

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

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

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

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

Therefore,

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

Simplify:

$$\int{\left(3 x - \frac{1}{x^{4}}\right)d x} = \frac{9 x^{5} + 2}{6 x^{3}}$$

Add the constant of integration:

$$\int{\left(3 x - \frac{1}{x^{4}}\right)d x} = \frac{9 x^{5} + 2}{6 x^{3}}+C$$

$$$\int \left(3 x - \frac{1}{x^{4}}\right)\, dx = \frac{9 x^{5} + 2}{6 x^{3}} + C$$$A