Integral de $$$\frac{1}{x^{2} - 5 x + 6}$$$

Halla $$$\int \frac{1}{x^{2} - 5 x + 6}\, dx$$$.

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

$${\color{red}{\int{\frac{1}{x^{2} - 5 x + 6} d x}}} = {\color{red}{\int{\left(- \frac{1}{x - 2} + \frac{1}{x - 3}\right)d x}}}$$

Integra término a término:

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

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

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

Por lo tanto,

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

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

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

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

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

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 se convierte en

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

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

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

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

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

Por lo tanto,

$$\int{\frac{1}{x^{2} - 5 x + 6} d x} = \ln{\left(\left|{x - 3}\right| \right)} - \ln{\left(\left|{x - 2}\right| \right)}$$

Añade la constante de integración:

$$\int{\frac{1}{x^{2} - 5 x + 6} d x} = \ln{\left(\left|{x - 3}\right| \right)} - \ln{\left(\left|{x - 2}\right| \right)}+C$$

$$$\int \frac{1}{x^{2} - 5 x + 6}\, dx = \left(\ln\left(\left|{x - 3}\right|\right) - \ln\left(\left|{x - 2}\right|\right)\right) + C$$$A