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

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

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

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

Reescribe el numerador del integrando como $$$x=\frac{1}{3}\left(3 x - 2\right)+\frac{2}{3}$$$ y descompón la fracción:

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

Integra término a término:

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

Aplica la regla de la constante $$$\int c\, dx = c x$$$ con $$$c=\frac{1}{3}$$$:

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

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

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

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

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

La integral puede reescribirse como

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

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

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

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

$$x + \frac{2 {\color{red}{\int{\frac{1}{u} d u}}}}{3} = x + \frac{2 {\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{3}$$

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

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

Por lo tanto,

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

Añade la constante de integración:

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

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