# How do you integrate 1/(y(1-y))?

$\int \frac{1}{y \left(1 - y\right)} \mathrm{dy} = - \int \frac{1}{{y}^{2} - y + {\left(- \frac{1}{2}\right)}^{2} - {\left(- \frac{1}{2}\right)}^{2}} \mathrm{dy}$
$\implies - \int \frac{1}{{\left(y - \frac{1}{2}\right)}^{2} - {\left(\frac{1}{2}\right)}^{2}} \mathrm{dy} = - \frac{1}{\frac{1}{2}} ^ 2 \int \frac{1}{{\left(\frac{y - \frac{1}{2}}{\frac{1}{2}}\right)}^{2} - 1} \mathrm{dy} = - 4 \int \frac{1}{{\left(2 y - 1\right)}^{2} - 1} \mathrm{dy}$
$\implies \frac{1}{2} \cdot - 4 a r \tanh \left(2 y - 1\right) + C = - \ln | \frac{1 - 2 y + 1}{1 + 2 y - 1} | + C = \ln | \frac{y}{1 - y} | + C$