Integral de $$$\frac{12}{3 x - 2}$$$

Halla $$$\int \frac{12}{3 x - 2}\, dx$$$.

Aplica la regla del factor constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ con $$$c=12$$$ y $$$f{\left(x \right)} = \frac{1}{3 x - 2}$$$:

$${\color{red}{\int{\frac{12}{3 x - 2} d x}}} = {\color{red}{\left(12 \int{\frac{1}{3 x - 2} d x}\right)}}$$

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

Entonces $$$du=\left(3 x - 2\right)^{\prime }dx = 3 dx$$$ (los pasos pueden verse »), y obtenemos que $$$dx = \frac{du}{3}$$$.

Por lo tanto,

$$12 {\color{red}{\int{\frac{1}{3 x - 2} d x}}} = 12 {\color{red}{\int{\frac{1}{3 u} d u}}}$$

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 $$$f{\left(u \right)} = \frac{1}{u}$$$:

$$12 {\color{red}{\int{\frac{1}{3 u} d u}}} = 12 {\color{red}{\left(\frac{\int{\frac{1}{u} d u}}{3}\right)}}$$

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

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

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

$$4 \ln{\left(\left|{{\color{red}{u}}}\right| \right)} = 4 \ln{\left(\left|{{\color{red}{\left(3 x - 2\right)}}}\right| \right)}$$

Por lo tanto,

$$\int{\frac{12}{3 x - 2} d x} = 4 \ln{\left(\left|{3 x - 2}\right| \right)}$$

Añade la constante de integración:

$$\int{\frac{12}{3 x - 2} d x} = 4 \ln{\left(\left|{3 x - 2}\right| \right)}+C$$

$$$\int \frac{12}{3 x - 2}\, dx = 4 \ln\left(\left|{3 x - 2}\right|\right) + C$$$A