Question

how can i integrate it dy/y-2y^2

Answer 1

Partial fractions.
\[\frac{ 1 }{ y-y^2 }=\frac{1}{y(1-y)}\]
\[=\frac{ 1 }{ y }+\frac{ 1 }{ 1-y }\]
so that's what you integrate instead.

Answer 2

\[\frac{ dy }{ y(1-2y) }=(\frac{1}{y}+\frac{2}{1-2y})dy\]\[\int\limits(\frac{1}{y}+\frac{2}{1-2y})dy=\int\limits \frac{ 1 }{ y }dy+\int\limits \frac{ 2 }{ 1-2y }=\ln \left| y \right|-\ln \left[ 1-2y \right]\]\[=\ln \left| \frac{ y }{ 1-2y } \right|\]