Find the indefinite integral of 3 radical x / (3 radical x - 1)

Let x=u² so dx=2udu.

∫3√x / (3√x-1) dx=∫6u²du / (3u-1) = ∫ (2u+⅔+2 / (9u-3)) du=u²/2+2u/3+2ln|3u-1|+C, where constant C absorbs ln (3).

Substituting for u: x/2+2√x/3+2ln|3√x-1|+C