Integral de $$$\left(2 x - 3\right)^{7}$$$

Encontre $$$\int \left(2 x - 3\right)^{7}\, dx$$$.

Seja $$$u=2 x - 3$$$.

Então $$$du=\left(2 x - 3\right)^{\prime }dx = 2 dx$$$ (veja os passos »), e obtemos $$$dx = \frac{du}{2}$$$.

A integral pode ser reescrita como

$${\color{red}{\int{\left(2 x - 3\right)^{7} d x}}} = {\color{red}{\int{\frac{u^{7}}{2} d u}}}$$

Aplique a regra do múltiplo constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ usando $$$c=\frac{1}{2}$$$ e $$$f{\left(u \right)} = u^{7}$$$:

$${\color{red}{\int{\frac{u^{7}}{2} d u}}} = {\color{red}{\left(\frac{\int{u^{7} d u}}{2}\right)}}$$

Aplique a regra da potência $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ com $$$n=7$$$:

$$\frac{{\color{red}{\int{u^{7} d u}}}}{2}=\frac{{\color{red}{\frac{u^{1 + 7}}{1 + 7}}}}{2}=\frac{{\color{red}{\left(\frac{u^{8}}{8}\right)}}}{2}$$

Recorde que $$$u=2 x - 3$$$:

$$\frac{{\color{red}{u}}^{8}}{16} = \frac{{\color{red}{\left(2 x - 3\right)}}^{8}}{16}$$

Portanto,

$$\int{\left(2 x - 3\right)^{7} d x} = \frac{\left(2 x - 3\right)^{8}}{16}$$

Adicione a constante de integração:

$$\int{\left(2 x - 3\right)^{7} d x} = \frac{\left(2 x - 3\right)^{8}}{16}+C$$

$$$\int \left(2 x - 3\right)^{7}\, dx = \frac{\left(2 x - 3\right)^{8}}{16} + C$$$A