Integral de $$$\frac{4 x}{\sqrt{x^{2} - 1}}$$$

Encontre $$$\int \frac{4 x}{\sqrt{x^{2} - 1}}\, dx$$$.

Seja $$$u=x^{2} - 1$$$.

Então $$$du=\left(x^{2} - 1\right)^{\prime }dx = 2 x dx$$$ (veja os passos »), e obtemos $$$x dx = \frac{du}{2}$$$.

Assim,

$${\color{red}{\int{\frac{4 x}{\sqrt{x^{2} - 1}} d x}}} = {\color{red}{\int{\frac{2}{\sqrt{u}} d u}}}$$

Aplique a regra do múltiplo constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ usando $$$c=2$$$ e $$$f{\left(u \right)} = \frac{1}{\sqrt{u}}$$$:

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

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

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

Recorde que $$$u=x^{2} - 1$$$:

$$4 \sqrt{{\color{red}{u}}} = 4 \sqrt{{\color{red}{\left(x^{2} - 1\right)}}}$$

Portanto,

$$\int{\frac{4 x}{\sqrt{x^{2} - 1}} d x} = 4 \sqrt{x^{2} - 1}$$

Adicione a constante de integração:

$$\int{\frac{4 x}{\sqrt{x^{2} - 1}} d x} = 4 \sqrt{x^{2} - 1}+C$$

$$$\int \frac{4 x}{\sqrt{x^{2} - 1}}\, dx = 4 \sqrt{x^{2} - 1} + C$$$A