Integral de $$$\frac{2}{5 x - 1}$$$

Encontre $$$\int \frac{2}{5 x - 1}\, dx$$$.

Aplique a regra do múltiplo constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ usando $$$c=2$$$ e $$$f{\left(x \right)} = \frac{1}{5 x - 1}$$$:

$${\color{red}{\int{\frac{2}{5 x - 1} d x}}} = {\color{red}{\left(2 \int{\frac{1}{5 x - 1} d x}\right)}}$$

Seja $$$u=5 x - 1$$$.

Então $$$du=\left(5 x - 1\right)^{\prime }dx = 5 dx$$$ (veja os passos »), e obtemos $$$dx = \frac{du}{5}$$$.

Logo,

$$2 {\color{red}{\int{\frac{1}{5 x - 1} d x}}} = 2 {\color{red}{\int{\frac{1}{5 u} d u}}}$$

Aplique a regra do múltiplo constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ usando $$$c=\frac{1}{5}$$$ e $$$f{\left(u \right)} = \frac{1}{u}$$$:

$$2 {\color{red}{\int{\frac{1}{5 u} d u}}} = 2 {\color{red}{\left(\frac{\int{\frac{1}{u} d u}}{5}\right)}}$$

A integral de $$$\frac{1}{u}$$$ é $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

$$\frac{2 {\color{red}{\int{\frac{1}{u} d u}}}}{5} = \frac{2 {\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{5}$$

Recorde que $$$u=5 x - 1$$$:

$$\frac{2 \ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{5} = \frac{2 \ln{\left(\left|{{\color{red}{\left(5 x - 1\right)}}}\right| \right)}}{5}$$

Portanto,

$$\int{\frac{2}{5 x - 1} d x} = \frac{2 \ln{\left(\left|{5 x - 1}\right| \right)}}{5}$$

Adicione a constante de integração:

$$\int{\frac{2}{5 x - 1} d x} = \frac{2 \ln{\left(\left|{5 x - 1}\right| \right)}}{5}+C$$

$$$\int \frac{2}{5 x - 1}\, dx = \frac{2 \ln\left(\left|{5 x - 1}\right|\right)}{5} + C$$$A