## Calculus with Applications (10th Edition)

$2(3u-5)^{1/2}+C$
In order to integrate this function, we have to use an «z» substitution. But first let's rewrite the function: $\int 3(3u-5)^{-1/2}$ Choose a z, derivate it: $z = 3u-5$ $dz = 3du$ Then substitute: $=\int z^{-1/2}dz$ Then integrate $=2z^{1/2} + C$ And undo the substitution: $=\boxed{2(3u-5)^{1/2}+C}$