Recall: $dp= \rho g \ dh$ where $dp$ is the change in pressure, $ \rho$ is the constant of density and $dh$ is the change in height. This is a part of fluid dynamics (buoyancy).
I am to integrate this expression to show that if $ \rho$ is constant then the pressure at depth $h$ is given by: $p = p_o + \rho g h$.
But I'm not really getting there with the integral methods that I know !
Hint: It comes down to this: $$ \int_0^{h} dp = p(h) - p_0\\ \int_0^{h} dp=\int_0^{h}\rho g\,dh = \rho g\int_0^{h}\,dh = ? $$ Use this to solve for $p(h)$.