Integral de $$$\frac{1}{y \left(1 - y\right)}$$$

Halla $$$\int \frac{1}{y \left(1 - y\right)}\, dy$$$.

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

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

Integra término a término:

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

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

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

Sea $$$u=1 - y$$$.

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

Por lo tanto,

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

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

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

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

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

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

$$\ln{\left(\left|{y}\right| \right)} - \ln{\left(\left|{{\color{red}{u}}}\right| \right)} = \ln{\left(\left|{y}\right| \right)} - \ln{\left(\left|{{\color{red}{\left(1 - y\right)}}}\right| \right)}$$

Por lo tanto,

$$\int{\frac{1}{y \left(1 - y\right)} d y} = \ln{\left(\left|{y}\right| \right)} - \ln{\left(\left|{y - 1}\right| \right)}$$

Añade la constante de integración:

$$\int{\frac{1}{y \left(1 - y\right)} d y} = \ln{\left(\left|{y}\right| \right)} - \ln{\left(\left|{y - 1}\right| \right)}+C$$

$$$\int \frac{1}{y \left(1 - y\right)}\, dy = \left(\ln\left(\left|{y}\right|\right) - \ln\left(\left|{y - 1}\right|\right)\right) + C$$$A