Integral de $$$\frac{1}{5 - 3 x^{2}}$$$

Halla $$$\int \frac{1}{5 - 3 x^{2}}\, dx$$$.

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

$${\color{red}{\int{\frac{1}{5 - 3 x^{2}} d x}}} = {\color{red}{\int{\left(\frac{\sqrt{15}}{10 \left(3 x + \sqrt{15}\right)} - \frac{\sqrt{15}}{10 \left(3 x - \sqrt{15}\right)}\right)d x}}}$$

Integra término a término:

$${\color{red}{\int{\left(\frac{\sqrt{15}}{10 \left(3 x + \sqrt{15}\right)} - \frac{\sqrt{15}}{10 \left(3 x - \sqrt{15}\right)}\right)d x}}} = {\color{red}{\left(- \int{\frac{\sqrt{15}}{10 \left(3 x - \sqrt{15}\right)} d x} + \int{\frac{\sqrt{15}}{10 \left(3 x + \sqrt{15}\right)} d x}\right)}}$$

Aplica la regla del factor constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ con $$$c=\frac{\sqrt{15}}{10}$$$ y $$$f{\left(x \right)} = \frac{1}{3 x - \sqrt{15}}$$$:

$$\int{\frac{\sqrt{15}}{10 \left(3 x + \sqrt{15}\right)} d x} - {\color{red}{\int{\frac{\sqrt{15}}{10 \left(3 x - \sqrt{15}\right)} d x}}} = \int{\frac{\sqrt{15}}{10 \left(3 x + \sqrt{15}\right)} d x} - {\color{red}{\left(\frac{\sqrt{15} \int{\frac{1}{3 x - \sqrt{15}} d x}}{10}\right)}}$$

Sea $$$u=3 x - \sqrt{15}$$$.

Entonces $$$du=\left(3 x - \sqrt{15}\right)^{\prime }dx = 3 dx$$$ (los pasos pueden verse »), y obtenemos que $$$dx = \frac{du}{3}$$$.

La integral se convierte en

$$\int{\frac{\sqrt{15}}{10 \left(3 x + \sqrt{15}\right)} d x} - \frac{\sqrt{15} {\color{red}{\int{\frac{1}{3 x - \sqrt{15}} d x}}}}{10} = \int{\frac{\sqrt{15}}{10 \left(3 x + \sqrt{15}\right)} d x} - \frac{\sqrt{15} {\color{red}{\int{\frac{1}{3 u} d u}}}}{10}$$

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

$$\int{\frac{\sqrt{15}}{10 \left(3 x + \sqrt{15}\right)} d x} - \frac{\sqrt{15} {\color{red}{\int{\frac{1}{3 u} d u}}}}{10} = \int{\frac{\sqrt{15}}{10 \left(3 x + \sqrt{15}\right)} d x} - \frac{\sqrt{15} {\color{red}{\left(\frac{\int{\frac{1}{u} d u}}{3}\right)}}}{10}$$

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

$$\int{\frac{\sqrt{15}}{10 \left(3 x + \sqrt{15}\right)} d x} - \frac{\sqrt{15} {\color{red}{\int{\frac{1}{u} d u}}}}{30} = \int{\frac{\sqrt{15}}{10 \left(3 x + \sqrt{15}\right)} d x} - \frac{\sqrt{15} {\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{30}$$

Recordemos que $$$u=3 x - \sqrt{15}$$$:

$$- \frac{\sqrt{15} \ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{30} + \int{\frac{\sqrt{15}}{10 \left(3 x + \sqrt{15}\right)} d x} = - \frac{\sqrt{15} \ln{\left(\left|{{\color{red}{\left(3 x - \sqrt{15}\right)}}}\right| \right)}}{30} + \int{\frac{\sqrt{15}}{10 \left(3 x + \sqrt{15}\right)} d x}$$

Aplica la regla del factor constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ con $$$c=\frac{\sqrt{15}}{10}$$$ y $$$f{\left(x \right)} = \frac{1}{3 x + \sqrt{15}}$$$:

$$- \frac{\sqrt{15} \ln{\left(\left|{3 x - \sqrt{15}}\right| \right)}}{30} + {\color{red}{\int{\frac{\sqrt{15}}{10 \left(3 x + \sqrt{15}\right)} d x}}} = - \frac{\sqrt{15} \ln{\left(\left|{3 x - \sqrt{15}}\right| \right)}}{30} + {\color{red}{\left(\frac{\sqrt{15} \int{\frac{1}{3 x + \sqrt{15}} d x}}{10}\right)}}$$

Sea $$$u=3 x + \sqrt{15}$$$.

Entonces $$$du=\left(3 x + \sqrt{15}\right)^{\prime }dx = 3 dx$$$ (los pasos pueden verse »), y obtenemos que $$$dx = \frac{du}{3}$$$.

Por lo tanto,

$$- \frac{\sqrt{15} \ln{\left(\left|{3 x - \sqrt{15}}\right| \right)}}{30} + \frac{\sqrt{15} {\color{red}{\int{\frac{1}{3 x + \sqrt{15}} d x}}}}{10} = - \frac{\sqrt{15} \ln{\left(\left|{3 x - \sqrt{15}}\right| \right)}}{30} + \frac{\sqrt{15} {\color{red}{\int{\frac{1}{3 u} d u}}}}{10}$$

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

$$- \frac{\sqrt{15} \ln{\left(\left|{3 x - \sqrt{15}}\right| \right)}}{30} + \frac{\sqrt{15} {\color{red}{\int{\frac{1}{3 u} d u}}}}{10} = - \frac{\sqrt{15} \ln{\left(\left|{3 x - \sqrt{15}}\right| \right)}}{30} + \frac{\sqrt{15} {\color{red}{\left(\frac{\int{\frac{1}{u} d u}}{3}\right)}}}{10}$$

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

$$- \frac{\sqrt{15} \ln{\left(\left|{3 x - \sqrt{15}}\right| \right)}}{30} + \frac{\sqrt{15} {\color{red}{\int{\frac{1}{u} d u}}}}{30} = - \frac{\sqrt{15} \ln{\left(\left|{3 x - \sqrt{15}}\right| \right)}}{30} + \frac{\sqrt{15} {\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{30}$$

Recordemos que $$$u=3 x + \sqrt{15}$$$:

$$- \frac{\sqrt{15} \ln{\left(\left|{3 x - \sqrt{15}}\right| \right)}}{30} + \frac{\sqrt{15} \ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{30} = - \frac{\sqrt{15} \ln{\left(\left|{3 x - \sqrt{15}}\right| \right)}}{30} + \frac{\sqrt{15} \ln{\left(\left|{{\color{red}{\left(3 x + \sqrt{15}\right)}}}\right| \right)}}{30}$$

Por lo tanto,

$$\int{\frac{1}{5 - 3 x^{2}} d x} = - \frac{\sqrt{15} \ln{\left(\left|{3 x - \sqrt{15}}\right| \right)}}{30} + \frac{\sqrt{15} \ln{\left(\left|{3 x + \sqrt{15}}\right| \right)}}{30}$$

Simplificar:

$$\int{\frac{1}{5 - 3 x^{2}} d x} = \frac{\sqrt{15} \left(- \ln{\left(\left|{3 x - \sqrt{15}}\right| \right)} + \ln{\left(\left|{3 x + \sqrt{15}}\right| \right)}\right)}{30}$$

Añade la constante de integración:

$$\int{\frac{1}{5 - 3 x^{2}} d x} = \frac{\sqrt{15} \left(- \ln{\left(\left|{3 x - \sqrt{15}}\right| \right)} + \ln{\left(\left|{3 x + \sqrt{15}}\right| \right)}\right)}{30}+C$$

$$$\int \frac{1}{5 - 3 x^{2}}\, dx = \frac{\sqrt{15} \left(- \ln\left(\left|{3 x - \sqrt{15}}\right|\right) + \ln\left(\left|{3 x + \sqrt{15}}\right|\right)\right)}{30} + C$$$A