Integral de $$$\frac{d}{2 \sqrt{x - 3}}$$$ em relação a $$$x$$$

Encontre $$$\int \frac{d}{2 \sqrt{x - 3}}\, dx$$$.

Aplique a regra do múltiplo constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ usando $$$c=\frac{d}{2}$$$ e $$$f{\left(x \right)} = \frac{1}{\sqrt{x - 3}}$$$:

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

Seja $$$u=x - 3$$$.

Então $$$du=\left(x - 3\right)^{\prime }dx = 1 dx$$$ (veja os passos »), e obtemos $$$dx = du$$$.

Assim,

$$\frac{d {\color{red}{\int{\frac{1}{\sqrt{x - 3}} d x}}}}{2} = \frac{d {\color{red}{\int{\frac{1}{\sqrt{u}} d u}}}}{2}$$

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

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

Recorde que $$$u=x - 3$$$:

$$d \sqrt{{\color{red}{u}}} = d \sqrt{{\color{red}{\left(x - 3\right)}}}$$

Portanto,

$$\int{\frac{d}{2 \sqrt{x - 3}} d x} = d \sqrt{x - 3}$$

Adicione a constante de integração:

$$\int{\frac{d}{2 \sqrt{x - 3}} d x} = d \sqrt{x - 3}+C$$

$$$\int \frac{d}{2 \sqrt{x - 3}}\, dx = d \sqrt{x - 3} + C$$$A