Answer
$$\frac{3}{2}\ln \left| {{x^{2/3}} - 1} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{x - \root 3 \of x }}} \cr
& \cr
& {\text{Taking the hint on the book making a substitution of the form }}u = {x^{1/n}} \cr
& {\text{Then}}{\text{,}} \cr
& {\text{Take }}\,\,\,u = {x^{1/3}} \to \,\,\,\,\,\,\,\,{u^3} = x,\,\,\,\,\,\,\,\,3{u^2}du = dx\,\,\,\, \cr
& \cr
& {\text{write the integrand in terms of }}u \cr
& \int {\frac{{dx}}{{x - \root 3 \of x }}} = \int {\frac{{3{u^2}du}}{{{u^3} - u}}} \cr
& = \int {\frac{{3{u^2}du}}{{u\left( {{u^2} - 1} \right)}}} \cr
& = \int {\frac{{3u}}{{{u^2} - 1}}du} \cr
& = \frac{3}{2}\int {\frac{{2u}}{{{u^2} - 1}}du} \cr
& \cr
& {\text{integrating}} \cr
& = \frac{3}{2}\ln \left| {{u^2} - 1} \right| + C \cr
& \cr
& {\text{write the integrand in terms of }}x;{\text{ replace }}u = {x^{1/3}} \cr
& = \frac{3}{2}\ln \left| {{{\left( {{x^{1/3}}} \right)}^2} - 1} \right| + C \cr
& = \frac{3}{2}\ln \left| {{x^{2/3}} - 1} \right| + C \cr} $$
