Integral de $$$\frac{1}{x^{2} - 3}$$$

Halla $$$\int \frac{1}{x^{2} - 3}\, dx$$$.

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

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

Integra término a término:

$${\color{red}{\int{\left(- \frac{\sqrt{3}}{6 \left(x + \sqrt{3}\right)} + \frac{\sqrt{3}}{6 \left(x - \sqrt{3}\right)}\right)d x}}} = {\color{red}{\left(\int{\frac{\sqrt{3}}{6 \left(x - \sqrt{3}\right)} d x} - \int{\frac{\sqrt{3}}{6 \left(x + \sqrt{3}\right)} d x}\right)}}$$

Aplica la regla del factor constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ con $$$c=\frac{\sqrt{3}}{6}$$$ y $$$f{\left(x \right)} = \frac{1}{x + \sqrt{3}}$$$:

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

Sea $$$u=x + \sqrt{3}$$$.

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

La integral se convierte en

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

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

$$\int{\frac{\sqrt{3}}{6 \left(x - \sqrt{3}\right)} d x} - \frac{\sqrt{3} {\color{red}{\int{\frac{1}{u} d u}}}}{6} = \int{\frac{\sqrt{3}}{6 \left(x - \sqrt{3}\right)} d x} - \frac{\sqrt{3} {\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{6}$$

Recordemos que $$$u=x + \sqrt{3}$$$:

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

Aplica la regla del factor constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ con $$$c=\frac{\sqrt{3}}{6}$$$ y $$$f{\left(x \right)} = \frac{1}{x - \sqrt{3}}$$$:

$$- \frac{\sqrt{3} \ln{\left(\left|{x + \sqrt{3}}\right| \right)}}{6} + {\color{red}{\int{\frac{\sqrt{3}}{6 \left(x - \sqrt{3}\right)} d x}}} = - \frac{\sqrt{3} \ln{\left(\left|{x + \sqrt{3}}\right| \right)}}{6} + {\color{red}{\left(\frac{\sqrt{3} \int{\frac{1}{x - \sqrt{3}} d x}}{6}\right)}}$$

Sea $$$u=x - \sqrt{3}$$$.

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

Entonces,

$$- \frac{\sqrt{3} \ln{\left(\left|{x + \sqrt{3}}\right| \right)}}{6} + \frac{\sqrt{3} {\color{red}{\int{\frac{1}{x - \sqrt{3}} d x}}}}{6} = - \frac{\sqrt{3} \ln{\left(\left|{x + \sqrt{3}}\right| \right)}}{6} + \frac{\sqrt{3} {\color{red}{\int{\frac{1}{u} d u}}}}{6}$$

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

$$- \frac{\sqrt{3} \ln{\left(\left|{x + \sqrt{3}}\right| \right)}}{6} + \frac{\sqrt{3} {\color{red}{\int{\frac{1}{u} d u}}}}{6} = - \frac{\sqrt{3} \ln{\left(\left|{x + \sqrt{3}}\right| \right)}}{6} + \frac{\sqrt{3} {\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{6}$$

Recordemos que $$$u=x - \sqrt{3}$$$:

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

Por lo tanto,

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

Simplificar:

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

Añade la constante de integración:

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

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