Integral de $$$\frac{1}{\left(4 x + 1\right)^{10}}$$$

Encontre $$$\int \frac{1}{\left(4 x + 1\right)^{10}}\, dx$$$.

Seja $$$u=4 x + 1$$$.

Então $$$du=\left(4 x + 1\right)^{\prime }dx = 4 dx$$$ (veja os passos »), e obtemos $$$dx = \frac{du}{4}$$$.

A integral torna-se

$${\color{red}{\int{\frac{1}{\left(4 x + 1\right)^{10}} d x}}} = {\color{red}{\int{\frac{1}{4 u^{10}} d u}}}$$

Aplique a regra do múltiplo constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ usando $$$c=\frac{1}{4}$$$ e $$$f{\left(u \right)} = \frac{1}{u^{10}}$$$:

$${\color{red}{\int{\frac{1}{4 u^{10}} d u}}} = {\color{red}{\left(\frac{\int{\frac{1}{u^{10}} d u}}{4}\right)}}$$

Aplique a regra da potência $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ com $$$n=-10$$$:

$$\frac{{\color{red}{\int{\frac{1}{u^{10}} d u}}}}{4}=\frac{{\color{red}{\int{u^{-10} d u}}}}{4}=\frac{{\color{red}{\frac{u^{-10 + 1}}{-10 + 1}}}}{4}=\frac{{\color{red}{\left(- \frac{u^{-9}}{9}\right)}}}{4}=\frac{{\color{red}{\left(- \frac{1}{9 u^{9}}\right)}}}{4}$$

Recorde que $$$u=4 x + 1$$$:

$$- \frac{{\color{red}{u}}^{-9}}{36} = - \frac{{\color{red}{\left(4 x + 1\right)}}^{-9}}{36}$$

Portanto,

$$\int{\frac{1}{\left(4 x + 1\right)^{10}} d x} = - \frac{1}{36 \left(4 x + 1\right)^{9}}$$

Adicione a constante de integração:

$$\int{\frac{1}{\left(4 x + 1\right)^{10}} d x} = - \frac{1}{36 \left(4 x + 1\right)^{9}}+C$$

$$$\int \frac{1}{\left(4 x + 1\right)^{10}}\, dx = - \frac{1}{36 \left(4 x + 1\right)^{9}} + C$$$A