I need to integrate next equation:
$$m \cdot dz=\left( r-z(r-1) \right)^4 \cdot \dfrac{dp}{dz} \cdot \left( p+k+\dfrac{k^2}{p}\right)dz$$
where $m$, $r$, $k$ are constants and $p=p(z)$. How to integrate this equation-its right side? Would be appropriate to cut terms $dz$ because of the term $r-z(r-1)$?
$$m= \left( r-z \left( r-1 \right) \right) ^{4} \left( {\frac {\rm d}{ {\rm d}z}}p \left( z \right) \right) \left( p \left( z \right) +k+{ \frac {{k}^{2}}{p \left( z \right) }} \right)$$
$${m \left( p \left( z \right) +k+{\frac {{k}^{2}}{p \left( z \right) }} \right) ^{-1}}= \left( r-z \left( r-1 \right) \right) ^{4}{\frac {\rm d}{{\rm d}z}}p \left( z \right)$$
$${\frac {m}{ \left( r-z \left( r-1 \right) \right) ^{4}}}= \left( p \left( z \right) +k+{\frac {{k}^{2}}{p \left( z \right) }} \right) { \frac {\rm d}{{\rm d}z}}p \left( z \right)$$
$$\int \!{\frac {m}{ \left( r-z \left( r-1 \right) \right) ^{4}}} \,{\rm d}z=\int \!p(z)+k+{\frac {{k}^{2}}{p(z)}}\,{\rm d}p(z) $$
$$-1/3\,{\frac {m}{ \left( \left( 1-r \right) z+r \right) ^{3} \left( 1 -r \right) }}=1/2\,{p(z)}^{2}+kp(z)+{k}^{2}\ln \left( p(z) \right) +C $$