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

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

Seja $$$u=1 - \frac{1}{x}$$$.

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

A integral pode ser reescrita como

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

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

$${\color{red}{\int{\sqrt{u} d u}}}={\color{red}{\int{u^{\frac{1}{2}} d u}}}={\color{red}{\frac{u^{\frac{1}{2} + 1}}{\frac{1}{2} + 1}}}={\color{red}{\left(\frac{2 u^{\frac{3}{2}}}{3}\right)}}$$

Recorde que $$$u=1 - \frac{1}{x}$$$:

$$\frac{2 {\color{red}{u}}^{\frac{3}{2}}}{3} = \frac{2 {\color{red}{\left(1 - \frac{1}{x}\right)}}^{\frac{3}{2}}}{3}$$

Portanto,

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

Simplifique:

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

Adicione a constante de integração:

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

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