Answer
$$\frac{3}{2}\ln \left| {{x^{2/3}} - 1} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{x - \root 3 \of x }}} \cr
& {\text{letting }}x = {u^3},{\text{ }}dx = 3{u^2}du \cr
& = \int {\frac{{dx}}{{x - \root 3 \of x }}} = \int {\frac{{3{u^2}du}}{{{u^3} - \root 3 \of {{u^3}} }}} \cr
& = \int {\frac{{3{u^2}du}}{{{u^3} - u}}} = \int {\frac{{3u}}{{{u^2} - 1}}} du \cr
& z = {u^2} - 1,{\text{ }}dz = 2udu,{\text{ }}udu = \frac{1}{2}dz \cr
& \int {\frac{{3u}}{{{u^2} - 1}}} du = \frac{3}{2}\int {\frac{{dz}}{z}} \cr
& = \frac{3}{2}\ln \left| z \right| + C \cr
& {\text{with }}z = {u^2} - 1, \cr
& = \frac{3}{2}\ln \left| {{u^2} - 1} \right| + C \cr
& x = {u^3} \to u = \root 3 \of x \cr
& = \frac{3}{2}\ln \left| {{{\left( {\root 3 \of x } \right)}^2} - 1} \right| + C \cr
& = \frac{3}{2}\ln \left| {{x^{2/3}} - 1} \right| + C \cr} $$
