Integral de $$$- \frac{3}{1 - 3 x}$$$

Halla $$$\int \left(- \frac{3}{1 - 3 x}\right)\, dx$$$.

Aplica la regla del factor constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ con $$$c=-3$$$ y $$$f{\left(x \right)} = \frac{1}{1 - 3 x}$$$:

$${\color{red}{\int{\left(- \frac{3}{1 - 3 x}\right)d x}}} = {\color{red}{\left(- 3 \int{\frac{1}{1 - 3 x} d x}\right)}}$$

Sea $$$u=1 - 3 x$$$.

Entonces $$$du=\left(1 - 3 x\right)^{\prime }dx = - 3 dx$$$ (los pasos pueden verse »), y obtenemos que $$$dx = - \frac{du}{3}$$$.

Por lo tanto,

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

Aplica la regla del factor constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ con $$$c=- \frac{1}{3}$$$ y $$$f{\left(u \right)} = \frac{1}{u}$$$:

$$- 3 {\color{red}{\int{\left(- \frac{1}{3 u}\right)d u}}} = - 3 {\color{red}{\left(- \frac{\int{\frac{1}{u} d u}}{3}\right)}}$$

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

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

Recordemos que $$$u=1 - 3 x$$$:

$$\ln{\left(\left|{{\color{red}{u}}}\right| \right)} = \ln{\left(\left|{{\color{red}{\left(1 - 3 x\right)}}}\right| \right)}$$

Por lo tanto,

$$\int{\left(- \frac{3}{1 - 3 x}\right)d x} = \ln{\left(\left|{3 x - 1}\right| \right)}$$

Añade la constante de integración:

$$\int{\left(- \frac{3}{1 - 3 x}\right)d x} = \ln{\left(\left|{3 x - 1}\right| \right)}+C$$

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