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

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

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

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

La integral puede reescribirse como

$${\color{red}{\int{\frac{x + 3}{x - 3} d x}}} = {\color{red}{\int{\frac{u + 6}{u} d u}}}$$

Expand the expression:

$${\color{red}{\int{\frac{u + 6}{u} d u}}} = {\color{red}{\int{\left(1 + \frac{6}{u}\right)d u}}}$$

Integra término a término:

$${\color{red}{\int{\left(1 + \frac{6}{u}\right)d u}}} = {\color{red}{\left(\int{1 d u} + \int{\frac{6}{u} d u}\right)}}$$

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

$$\int{\frac{6}{u} d u} + {\color{red}{\int{1 d u}}} = \int{\frac{6}{u} d u} + {\color{red}{u}}$$

Aplica la regla del factor constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ con $$$c=6$$$ y $$$f{\left(u \right)} = \frac{1}{u}$$$:

$$u + {\color{red}{\int{\frac{6}{u} d u}}} = u + {\color{red}{\left(6 \int{\frac{1}{u} d u}\right)}}$$

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

$$u + 6 {\color{red}{\int{\frac{1}{u} d u}}} = u + 6 {\color{red}{\ln{\left(\left|{u}\right| \right)}}}$$

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

$$6 \ln{\left(\left|{{\color{red}{u}}}\right| \right)} + {\color{red}{u}} = 6 \ln{\left(\left|{{\color{red}{\left(x - 3\right)}}}\right| \right)} + {\color{red}{\left(x - 3\right)}}$$

Por lo tanto,

$$\int{\frac{x + 3}{x - 3} d x} = x + 6 \ln{\left(\left|{x - 3}\right| \right)} - 3$$

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

$$\int{\frac{x + 3}{x - 3} d x} = x + 6 \ln{\left(\left|{x - 3}\right| \right)}+C$$

$$$\int \frac{x + 3}{x - 3}\, dx = \left(x + 6 \ln\left(\left|{x - 3}\right|\right)\right) + C$$$A