Integral de $$$\frac{8 i}{\sqrt{x}}$$$

Encontre $$$\int \frac{8 i}{\sqrt{x}}\, dx$$$.

Aplique a regra do múltiplo constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ usando $$$c=8 i$$$ e $$$f{\left(x \right)} = \frac{1}{\sqrt{x}}$$$:

$${\color{red}{\int{\frac{8 i}{\sqrt{x}} d x}}} = {\color{red}{\left(8 i \int{\frac{1}{\sqrt{x}} d x}\right)}}$$

Aplique a regra da potência $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ com $$$n=- \frac{1}{2}$$$:

$$8 i {\color{red}{\int{\frac{1}{\sqrt{x}} d x}}}=8 i {\color{red}{\int{x^{- \frac{1}{2}} d x}}}=8 i {\color{red}{\frac{x^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1}}}=8 i {\color{red}{\left(2 x^{\frac{1}{2}}\right)}}=8 i {\color{red}{\left(2 \sqrt{x}\right)}}$$

Portanto,

$$\int{\frac{8 i}{\sqrt{x}} d x} = 16 i \sqrt{x}$$

Adicione a constante de integração:

$$\int{\frac{8 i}{\sqrt{x}} d x} = 16 i \sqrt{x}+C$$

$$$\int \frac{8 i}{\sqrt{x}}\, dx = 16 i \sqrt{x} + C$$$A