Integral de $$$\sqrt{x - 1}$$$

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

Seja $$$u=x - 1$$$.

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

Assim,

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

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 $$$u=x - 1$$$:

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

Portanto,

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

Adicione a constante de integração:

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

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