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

Halla $$$\int \frac{x - 1}{3 - x}\, dx$$$.

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

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

La integral puede reescribirse como

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

Expand the expression:

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

Integra término a término:

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

Aplica la regla de la constante $$$\int c\, du = c u$$$ con $$$c=1$$$:

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

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

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

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

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

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

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

Por lo tanto,

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

Añadir la constante de integración (y eliminar la constante de la expresión):

$$\int{\frac{x - 1}{3 - x} d x} = - x - 2 \ln{\left(\left|{x - 3}\right| \right)}+C$$

$$$\int \frac{x - 1}{3 - x}\, dx = \left(- x - 2 \ln\left(\left|{x - 3}\right|\right)\right) + C$$$A