Integral de $$$\frac{x - 5}{x \left(x - 2\right)}$$$

Halla $$$\int \frac{x - 5}{x \left(x - 2\right)}\, dx$$$.

Realizar la descomposición en fracciones parciales (los pasos pueden verse »):

$${\color{red}{\int{\frac{x - 5}{x \left(x - 2\right)} d x}}} = {\color{red}{\int{\left(- \frac{3}{2 \left(x - 2\right)} + \frac{5}{2 x}\right)d x}}}$$

Integra término a término:

$${\color{red}{\int{\left(- \frac{3}{2 \left(x - 2\right)} + \frac{5}{2 x}\right)d x}}} = {\color{red}{\left(\int{\frac{5}{2 x} d x} - \int{\frac{3}{2 \left(x - 2\right)} d x}\right)}}$$

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

$$\int{\frac{5}{2 x} d x} - {\color{red}{\int{\frac{3}{2 \left(x - 2\right)} d x}}} = \int{\frac{5}{2 x} d x} - {\color{red}{\left(\frac{3 \int{\frac{1}{x - 2} d x}}{2}\right)}}$$

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

Entonces $$$du=\left(x - 2\right)^{\prime }dx = 1 dx$$$ (los pasos pueden verse »), y obtenemos que $$$dx = du$$$.

La integral puede reescribirse como

$$\int{\frac{5}{2 x} d x} - \frac{3 {\color{red}{\int{\frac{1}{x - 2} d x}}}}{2} = \int{\frac{5}{2 x} d x} - \frac{3 {\color{red}{\int{\frac{1}{u} d u}}}}{2}$$

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

$$\int{\frac{5}{2 x} d x} - \frac{3 {\color{red}{\int{\frac{1}{u} d u}}}}{2} = \int{\frac{5}{2 x} d x} - \frac{3 {\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{2}$$

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

$$- \frac{3 \ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{2} + \int{\frac{5}{2 x} d x} = - \frac{3 \ln{\left(\left|{{\color{red}{\left(x - 2\right)}}}\right| \right)}}{2} + \int{\frac{5}{2 x} d x}$$

Aplica la regla del factor constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ con $$$c=\frac{5}{2}$$$ y $$$f{\left(x \right)} = \frac{1}{x}$$$:

$$- \frac{3 \ln{\left(\left|{x - 2}\right| \right)}}{2} + {\color{red}{\int{\frac{5}{2 x} d x}}} = - \frac{3 \ln{\left(\left|{x - 2}\right| \right)}}{2} + {\color{red}{\left(\frac{5 \int{\frac{1}{x} d x}}{2}\right)}}$$

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

$$- \frac{3 \ln{\left(\left|{x - 2}\right| \right)}}{2} + \frac{5 {\color{red}{\int{\frac{1}{x} d x}}}}{2} = - \frac{3 \ln{\left(\left|{x - 2}\right| \right)}}{2} + \frac{5 {\color{red}{\ln{\left(\left|{x}\right| \right)}}}}{2}$$

Por lo tanto,

$$\int{\frac{x - 5}{x \left(x - 2\right)} d x} = \frac{5 \ln{\left(\left|{x}\right| \right)}}{2} - \frac{3 \ln{\left(\left|{x - 2}\right| \right)}}{2}$$

Añade la constante de integración:

$$\int{\frac{x - 5}{x \left(x - 2\right)} d x} = \frac{5 \ln{\left(\left|{x}\right| \right)}}{2} - \frac{3 \ln{\left(\left|{x - 2}\right| \right)}}{2}+C$$

$$$\int \frac{x - 5}{x \left(x - 2\right)}\, dx = \left(\frac{5 \ln\left(\left|{x}\right|\right)}{2} - \frac{3 \ln\left(\left|{x - 2}\right|\right)}{2}\right) + C$$$A