Integral de $$$\frac{1}{x^{2} - 25}$$$

Halla $$$\int \frac{1}{x^{2} - 25}\, dx$$$.

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

$${\color{red}{\int{\frac{1}{x^{2} - 25} d x}}} = {\color{red}{\int{\left(- \frac{1}{10 \left(x + 5\right)} + \frac{1}{10 \left(x - 5\right)}\right)d x}}}$$

Integra término a término:

$${\color{red}{\int{\left(- \frac{1}{10 \left(x + 5\right)} + \frac{1}{10 \left(x - 5\right)}\right)d x}}} = {\color{red}{\left(\int{\frac{1}{10 \left(x - 5\right)} d x} - \int{\frac{1}{10 \left(x + 5\right)} d x}\right)}}$$

Aplica la regla del factor constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ con $$$c=\frac{1}{10}$$$ y $$$f{\left(x \right)} = \frac{1}{x + 5}$$$:

$$\int{\frac{1}{10 \left(x - 5\right)} d x} - {\color{red}{\int{\frac{1}{10 \left(x + 5\right)} d x}}} = \int{\frac{1}{10 \left(x - 5\right)} d x} - {\color{red}{\left(\frac{\int{\frac{1}{x + 5} d x}}{10}\right)}}$$

Sea $$$u=x + 5$$$.

Entonces $$$du=\left(x + 5\right)^{\prime }dx = 1 dx$$$ (los pasos pueden verse »), y obtenemos que $$$dx = du$$$.

La integral puede reescribirse como

$$\int{\frac{1}{10 \left(x - 5\right)} d x} - \frac{{\color{red}{\int{\frac{1}{x + 5} d x}}}}{10} = \int{\frac{1}{10 \left(x - 5\right)} d x} - \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{10}$$

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

$$\int{\frac{1}{10 \left(x - 5\right)} d x} - \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{10} = \int{\frac{1}{10 \left(x - 5\right)} d x} - \frac{{\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{10}$$

Recordemos que $$$u=x + 5$$$:

$$- \frac{\ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{10} + \int{\frac{1}{10 \left(x - 5\right)} d x} = - \frac{\ln{\left(\left|{{\color{red}{\left(x + 5\right)}}}\right| \right)}}{10} + \int{\frac{1}{10 \left(x - 5\right)} d x}$$

Aplica la regla del factor constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ con $$$c=\frac{1}{10}$$$ y $$$f{\left(x \right)} = \frac{1}{x - 5}$$$:

$$- \frac{\ln{\left(\left|{x + 5}\right| \right)}}{10} + {\color{red}{\int{\frac{1}{10 \left(x - 5\right)} d x}}} = - \frac{\ln{\left(\left|{x + 5}\right| \right)}}{10} + {\color{red}{\left(\frac{\int{\frac{1}{x - 5} d x}}{10}\right)}}$$

Sea $$$u=x - 5$$$.

Entonces $$$du=\left(x - 5\right)^{\prime }dx = 1 dx$$$ (los pasos pueden verse »), y obtenemos que $$$dx = du$$$.

Por lo tanto,

$$- \frac{\ln{\left(\left|{x + 5}\right| \right)}}{10} + \frac{{\color{red}{\int{\frac{1}{x - 5} d x}}}}{10} = - \frac{\ln{\left(\left|{x + 5}\right| \right)}}{10} + \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{10}$$

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

$$- \frac{\ln{\left(\left|{x + 5}\right| \right)}}{10} + \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{10} = - \frac{\ln{\left(\left|{x + 5}\right| \right)}}{10} + \frac{{\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{10}$$

Recordemos que $$$u=x - 5$$$:

$$- \frac{\ln{\left(\left|{x + 5}\right| \right)}}{10} + \frac{\ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{10} = - \frac{\ln{\left(\left|{x + 5}\right| \right)}}{10} + \frac{\ln{\left(\left|{{\color{red}{\left(x - 5\right)}}}\right| \right)}}{10}$$

Por lo tanto,

$$\int{\frac{1}{x^{2} - 25} d x} = \frac{\ln{\left(\left|{x - 5}\right| \right)}}{10} - \frac{\ln{\left(\left|{x + 5}\right| \right)}}{10}$$

Añade la constante de integración:

$$\int{\frac{1}{x^{2} - 25} d x} = \frac{\ln{\left(\left|{x - 5}\right| \right)}}{10} - \frac{\ln{\left(\left|{x + 5}\right| \right)}}{10}+C$$

$$$\int \frac{1}{x^{2} - 25}\, dx = \left(\frac{\ln\left(\left|{x - 5}\right|\right)}{10} - \frac{\ln\left(\left|{x + 5}\right|\right)}{10}\right) + C$$$A