Integral de $$$\frac{x}{\left(x + 1\right) \left(x + 2\right)}$$$

Halla $$$\int \frac{x}{\left(x + 1\right) \left(x + 2\right)}\, dx$$$.

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

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

Integra término a término:

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

Sea $$$u=x + 1$$$.

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

Entonces,

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

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

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

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

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

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

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

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

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

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

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

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

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

Por lo tanto,

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

Añade la constante de integración:

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

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