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

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

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

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

Reescribe y separa la fracción:

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

Integra término a término:

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

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

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

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

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

La integral se convierte en

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

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

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

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

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

Por lo tanto,

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

Simplificar:

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

Añade la constante de integración:

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

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