## Thomas' Calculus 13th Edition

$-\frac{(5p^2+2p)^{5/2}}{15p+3}$
Given $q=(5p^2+2p)^{-3/2}$, we have $dq=-\frac{3}{2}(5p^2+2p)^{-5/2}(10pdp+2dp)=-3(5p^2+2p)^{-5/2}(5p+1)dp$ Thus $\frac{dp}{dq}=-\frac{(5p^2+2p)^{5/2}}{15p+3}$