Integral of $$$\frac{3 i n t}{x^{3}}$$$ with respect to $$$x$$$

Find $$$\int \frac{3 i n t}{x^{3}}\, dx$$$.

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=3 i n t$$$ and $$$f{\left(x \right)} = \frac{1}{x^{3}}$$$:

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

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

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

Therefore,

$$\int{\frac{3 i n t}{x^{3}} d x} = - \frac{3 i n t}{2 x^{2}}$$

Add the constant of integration:

$$\int{\frac{3 i n t}{x^{3}} d x} = - \frac{3 i n t}{2 x^{2}}+C$$

$$$\int \frac{3 i n t}{x^{3}}\, dx = - \frac{3 i n t}{2 x^{2}} + C$$$A