Integral de $$$\ln\left(1 - x^{2}\right)$$$

Halla $$$\int \ln\left(1 - x^{2}\right)\, dx$$$.

Para la integral $$$\int{\ln{\left(1 - x^{2} \right)} d x}$$$, utiliza la integración por partes $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Sean $$$\operatorname{u}=\ln{\left(1 - x^{2} \right)}$$$ y $$$\operatorname{dv}=dx$$$.

Entonces $$$\operatorname{du}=\left(\ln{\left(1 - x^{2} \right)}\right)^{\prime }dx=\frac{2 x}{x^{2} - 1} dx$$$ (los pasos pueden verse ») y $$$\operatorname{v}=\int{1 d x}=x$$$ (los pasos pueden verse »).

La integral puede reescribirse como

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

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

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

Como el grado del numerador no es menor que el grado del denominador, realiza la división larga de polinomios (los pasos pueden verse »):

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

Integra término a término:

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

Aplica la regla de la constante $$$\int c\, dx = c x$$$ con $$$c=1$$$:

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

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

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

Integra término a término:

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

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

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

Sea $$$u=x - 1$$$.

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

Por lo tanto,

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

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

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

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

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

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

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

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

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

La integral puede reescribirse como

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

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

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

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

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

Por lo tanto,

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

Añade la constante de integración:

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

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