Integral of $$$2 \alpha i_{n} x x^{1 - n}$$$ with respect to $$$x$$$

Find $$$\int 2 \alpha i_{n} x x^{1 - n}\, dx$$$.

The input is rewritten: $$$\int{2 \alpha i_{n} x x^{1 - n} d x}=\int{2 \alpha i_{n} x^{2 - n} d x}$$$.

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=2 \alpha i_{n}$$$ and $$$f{\left(x \right)} = x^{2 - n}$$$:

$${\color{red}{\int{2 \alpha i_{n} x^{2 - n} d x}}} = {\color{red}{\left(2 \alpha i_{n} \int{x^{2 - n} d x}\right)}}$$

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

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

Therefore,

$$\int{2 \alpha i_{n} x^{2 - n} d x} = \frac{2 \alpha i_{n} x^{3 - n}}{3 - n}$$

Simplify:

$$\int{2 \alpha i_{n} x^{2 - n} d x} = - \frac{2 \alpha i_{n} x^{3 - n}}{n - 3}$$

Add the constant of integration:

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

$$$\int 2 \alpha i_{n} x x^{1 - n}\, dx = - \frac{2 \alpha i_{n} x^{3 - n}}{n - 3} + C$$$A