Answer
$$\frac{{3{z^{2/3}}}}{2} - 2z + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{1 - 2\root 3 \of z }}{{\root 3 \of z }}} dz \cr
& {\text{rewrite the radicals using }}\root n \of z = {z^{1/n}} \cr
& = \int {\frac{{1 - 2{z^{1/3}}}}{{{z^{1/3}}}}} dz \cr
& {\text{distributive property}} \cr
& = \int {\left( {\frac{1}{{{z^{1/3}}}} - \frac{{2{z^{1/3}}}}{{{z^{1/3}}}}} \right)} dz \cr
& {\text{simplifying}} \cr
& = \int {{z^{ - 1/3}}} dz - 2\int {dz} \cr
& {\text{use }}\int {{z^n}dz} = \frac{{{z^{n + 1}}}}{{n + 1}} + C{\text{ and }}\int {dz} = z + C \cr
& = \frac{{{z^{ - 1/3 + 1}}}}{{ - 1/3 + 1}} - 2z + C \cr
& = \frac{{{z^{2/3}}}}{{2/3}} - 2z + C \cr
& = \frac{{3{z^{2/3}}}}{2} - 2z + C \cr} $$
