Integral of $$$- x \cot{\left(1 \right)} - 3$$$

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

Integrate term by term:

$${\color{red}{\int{\left(- x \cot{\left(1 \right)} - 3\right)d x}}} = {\color{red}{\left(- \int{3 d x} - \int{x \cot{\left(1 \right)} d x}\right)}}$$

Apply the constant rule $$$\int c\, dx = c x$$$ with $$$c=3$$$:

$$- \int{x \cot{\left(1 \right)} d x} - {\color{red}{\int{3 d x}}} = - \int{x \cot{\left(1 \right)} d x} - {\color{red}{\left(3 x\right)}}$$

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

$$- 3 x - {\color{red}{\int{x \cot{\left(1 \right)} d x}}} = - 3 x - {\color{red}{\cot{\left(1 \right)} \int{x d x}}}$$

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

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

Therefore,

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

Simplify:

$$\int{\left(- x \cot{\left(1 \right)} - 3\right)d x} = \frac{x \left(- x \cot{\left(1 \right)} - 6\right)}{2}$$

Add the constant of integration:

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

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