Question

Integration by Parts Fractions

∫dy/(y^2+2y)

Answer 1

Let

$\frac{1}{{{y^2} + 2y}} = \frac{1}{{y(y + 2)}} = \frac{A}{y} + \frac{B}{{y + 2}} = \frac{{Ay + 2A + By}}{{y(y + 2)}} = \frac{{\left( {A + B} \right)y + 2A}}{{y(y + 2)}}$

Then

$\left\{ \begin{array}{l} A + B = 0\\ 2A = 1 \end{array} \right.$

$\left\{ \begin{array}{l} A = \frac{1}{2}\\ B = - \frac{1}{2} \end{array} \right.$

Then

$\int {\frac{{dy}}{{{y^2} + 2y}} = \frac{1}{2}} \int {\frac{{dy}}{y}} - \frac{1}{2}\int {\frac{{dy}}{{y + 2}} = \frac{1}{2}} \ln |y| - \frac{1}{2}\ln |y + 2| + C$