Integraal van $$$\frac{1}{y \left(1 - y\right)}$$$

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

Voer een ontbinding in partiële breuken uit (stappen zijn te zien »):

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

Integreer termgewijs:

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

De integraal van $$$\frac{1}{y}$$$ is $$$\int{\frac{1}{y} d y} = \ln{\left(\left|{y}\right| \right)}$$$:

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

Zij $$$u=1 - y$$$.

Dan $$$du=\left(1 - y\right)^{\prime }dy = - dy$$$ (de stappen zijn te zien »), en dan geldt dat $$$dy = - du$$$.

Dus,

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

Pas de constante-veelvoudregel $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ toe met $$$c=-1$$$ en $$$f{\left(u \right)} = \frac{1}{u}$$$:

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

De integraal van $$$\frac{1}{u}$$$ is $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

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

We herinneren eraan dat $$$u=1 - y$$$:

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

Dus,

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

Voeg de integratieconstante toe:

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

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